{"id":599958,"date":"2022-05-10T09:44:19","date_gmt":"2022-05-10T09:44:19","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=599958"},"modified":"2022-05-12T09:38:46","modified_gmt":"2022-05-12T09:38:46","slug":"ml-aggarwal-solutions-for-class-10-maths-chapter-6-factorization","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ml-aggarwal-solutions-for-class-10-maths-chapter-6-factorization\/","title":{"rendered":"ML Aggarwal Solutions for Class 10 Maths Chapter 6- Factorization"},"content":{"rendered":"\n

Class 10: Maths Chapter 6 solutions. Complete Class 10 Maths Chapter 6 Notes.<\/p>\n\n\n\n

ML Aggarwal Solutions for Class 10 Maths Chapter 6- Factorization<\/h2>\n\n\n\n

ML Aggarwal 10th Maths Chapter 6, Class 10 Maths Chapter 6 solutions<\/p>\n\n\n\n

Exercise 6.1<\/h4>\n\n\n\n

1. Find the remainder (without division) on dividing f(x) by (x \u2013 2) where<\/strong><\/p>\n\n\n\n

(i) f(x) = 5x2<\/sup> \u2013 7x + 4<\/strong><\/p>\n\n\n\n

Solutions:-<\/strong><\/p>\n\n\n\n

Let us assume x \u2013 2 = 0<\/p>\n\n\n\n

Then, x = 2<\/p>\n\n\n\n

Given, f(x) = 5x2<\/sup> \u2013 7x + 4<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(2)= (5 \u00d7 22<\/sup>) \u2013 (7 \u00d7 2) + 4<\/p>\n\n\n\n

= (5 \u00d7 4) \u2013 14 + 4<\/p>\n\n\n\n

= 20 \u2013 14 + 4<\/p>\n\n\n\n

= 24 \u2013 14<\/p>\n\n\n\n

= 10<\/p>\n\n\n\n

Therefore, the remainder is 10.<\/p>\n\n\n\n

(ii) f(x) = 2x3<\/sup> \u2013 7x2<\/sup> + 3<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume x \u2013 2 = 0<\/p>\n\n\n\n

Then, x = 2<\/p>\n\n\n\n

Given, f(x) = 2x3<\/sup> \u2013 7x2<\/sup> + 3<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(2)= (2 \u00d7 23<\/sup>) \u2013 (7 \u00d7 22<\/sup>) + 3<\/p>\n\n\n\n

= (2 \u00d7 8) \u2013 (7 \u00d7 4) + 3<\/p>\n\n\n\n

= 16 \u2013 28 + 3<\/p>\n\n\n\n

= 19 \u2013 28<\/p>\n\n\n\n

= -9<\/p>\n\n\n\n

Therefore, the remainder is -9.<\/p>\n\n\n\n

2. Using the remainder theorem, find the remainder on dividing f(x) by (x + 3) where<\/strong><\/p>\n\n\n\n

(i) f(x) = 2x2<\/sup> \u2013 5x + 1<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume x + 3 = 0<\/p>\n\n\n\n

Then, x = -3<\/p>\n\n\n\n

Given, f(x) = 2x2<\/sup> \u2013 5x + 1<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(-3)= (2 \u00d7 -32<\/sup>) \u2013 (5 \u00d7 (-3)) + 1<\/p>\n\n\n\n

= (2 \u00d7 9) \u2013 (-15) + 1<\/p>\n\n\n\n

= 18 + 15 + 1<\/p>\n\n\n\n

= 34<\/p>\n\n\n\n

Therefore, the remainder is 34.<\/p>\n\n\n\n

(ii) f(x) = 3x3<\/sup> + 7x2<\/sup> \u2013 5x + 1
Solution:-<\/strong><\/p>\n\n\n\n

Let us assume x + 3 = 0<\/p>\n\n\n\n

Then, x = -3<\/p>\n\n\n\n

Given, f(x) = 3x3<\/sup> + 7x2<\/sup> \u2013 5x + 1<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(-3)= (3 \u00d7 -33<\/sup>) + (7 \u00d7 -32<\/sup>) \u2013 (5 \u00d7 -3) + 1<\/p>\n\n\n\n

= (3 \u00d7 -27) + (7 \u00d7 9) \u2013 (-15) + 1<\/p>\n\n\n\n

= \u2013 81 + 63 + 15 + 1<\/p>\n\n\n\n

= -81 + 79<\/p>\n\n\n\n

= -2<\/p>\n\n\n\n

Therefore, the remainder is -2.<\/p>\n\n\n\n

3. Find the remainder (without division) on dividing f(x) by (2x + 1) where,<\/strong><\/p>\n\n\n\n

(i) f(x) = 4x2<\/sup> + 5x + 3<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume 2x + 1 = 0<\/p>\n\n\n\n

Then, 2x = -1<\/p>\n\n\n\n

X = -\u00bd<\/p>\n\n\n\n

Given, f(x) = 4x2<\/sup> + 5x + 3<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f (-\u00bd) = 4 (-\u00bd)2<\/sup> + 5 (-\u00bd) + 3<\/p>\n\n\n\n

= (4 \u00d7 \u00bc) + (-5\/2) + 3<\/p>\n\n\n\n

= 1 \u2013 5\/2 + 3<\/p>\n\n\n\n

= 4 \u2013 5\/2<\/p>\n\n\n\n

= (8 \u2013 5)\/2<\/p>\n\n\n\n

= 3\/2 = 1\u00bd<\/p>\n\n\n\n

Therefore, the remainder is 1\u00bd.<\/p>\n\n\n\n

(ii) f(x) = 3x3<\/sup> \u2013 7x2<\/sup> + 4x + 11<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume 2x + 1 = 0<\/p>\n\n\n\n

Then, 2x = -1<\/p>\n\n\n\n

X = -\u00bd<\/p>\n\n\n\n

Given, f(x) = 3x3<\/sup> \u2013 7x2<\/sup> + 4x + 11<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(-\u00bd) = (3 \u00d7 (-\u00bd)3<\/sup>) \u2013 (7 \u00d7 (-\u00bd)2<\/sup> + (4 \u00d7 -\u00bd) + 11<\/p>\n\n\n\n

= 3 \u00d7 (-1\/8) \u2013 (7 \u00d7 \u00bc) + (- 2) + 11<\/p>\n\n\n\n

= -3\/8 \u2013 7\/4 \u2013 2 + 11<\/p>\n\n\n\n

= \u2013 3\/8 \u2013 7\/4 + 9<\/p>\n\n\n\n

= (-3 \u2013 14 + 72)\/8<\/p>\n\n\n\n

= 55\/8<\/p>\n\n\n\n

=
\"ML<\/p>\n\n\n\n

4. Using remainder theorem, find the value of k if on dividing 2x3<\/sup> + 3x2<\/sup> \u2013 kx + 5 by x \u2013 2 leaves a remainder 7.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume, x \u2013 2 = 0<\/p>\n\n\n\n

Then, x = 2<\/p>\n\n\n\n

Given, 2x3<\/sup> + 3x2<\/sup> \u2013 kx + 5<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(2) = (2 \u00d7 23<\/sup>) + (3 \u00d7 22<\/sup>) \u2013 (k \u00d7 2) + 5<\/p>\n\n\n\n

= (2 \u00d7 8) + (3 \u00d7 4) \u2013 2k + 5<\/p>\n\n\n\n

= 16 + 12 \u2013 2k + 5<\/p>\n\n\n\n

= 33 \u2013 2k<\/p>\n\n\n\n

Form the question it is given that, remainder is 7.<\/p>\n\n\n\n

So, 7 = 33 \u2013 2k<\/p>\n\n\n\n

2k = 33 \u2013 7<\/p>\n\n\n\n

2k = 26<\/p>\n\n\n\n

K = 26\/2<\/p>\n\n\n\n

K = 13<\/p>\n\n\n\n

Therefore, the value of k is 13.<\/p>\n\n\n\n

5. Using remainder theorem, find the value of \u2018a\u2019 if the division of x3<\/sup> + 5x2<\/sup> \u2013 ax + 6 by (x \u2013 1) leaves the remainder 2a.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume x -1 = 0<\/p>\n\n\n\n

Then, x = 1<\/p>\n\n\n\n

Given, f(x) = x3<\/sup> + 5x2<\/sup> \u2013 ax + 6<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(1) = 13<\/sup> <\/sub>+ (5 \u00d7 12<\/sup>) \u2013 (a \u00d7 1) + 6<\/p>\n\n\n\n

= 1 + 5 \u2013 a + 6<\/p>\n\n\n\n

= 12 \u2013 a<\/p>\n\n\n\n

From the question it is given that, remainder is 2a<\/p>\n\n\n\n

So, 2a = 12 \u2013 a<\/p>\n\n\n\n

2a + a = 12<\/p>\n\n\n\n

3a = 12<\/p>\n\n\n\n

a = 12\/3<\/p>\n\n\n\n

a = 4<\/p>\n\n\n\n

Therefore, the value of a is 4.<\/p>\n\n\n\n

6. (i) What number must be divided be subtracted from 2x2<\/sup> \u2013 5x so that the resulting polynomial leaves the remainder 2, when divided by 2x + 1?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

let us assume \u2018p\u2019 be subtracted from 2x2<\/sup> \u2013 5x<\/p>\n\n\n\n

So, dividing 2x2<\/sup> \u2013 5x by 2x + 1,<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Hence, remainder is 3 \u2013 p<\/p>\n\n\n\n

From the question it is given that, remainder is 2.<\/p>\n\n\n\n

3 \u2013 p = 2<\/p>\n\n\n\n

p = 3 \u2013 2<\/p>\n\n\n\n

p = 1<\/p>\n\n\n\n

Therefore, 1 is to be subtracted.<\/p>\n\n\n\n

(ii) What number must be added to 2x3<\/sup> \u2013 7x2<\/sup> + 2x so that the resulting polynomial leaves the remainder \u2013 2 when divided by 2x \u2013 3?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

let us assume \u2018p\u2019 be subtracted from 2x3<\/sup> \u2013 7x2<\/sup> + 2x,<\/p>\n\n\n\n

So, dividing it by 2x \u2013 3,<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Hence, remainder is p \u2013 6<\/p>\n\n\n\n

From the question it is given that, remainder is \u2013 2.<\/p>\n\n\n\n

P \u2013 6 = \u2013 2<\/p>\n\n\n\n

P = -2 + 6<\/p>\n\n\n\n

P = 4<\/p>\n\n\n\n

Therefore, 4 is to be added.<\/p>\n\n\n\n

7. (i) When divided by x \u2013 3 the polynomials x3<\/sup> \u2013 px2<\/sup> + x + 6 and 2x3<\/sup> \u2013 x2<\/sup> \u2013 (p + 3) x \u2013 6 leave the same remainder. Find the value of \u2018p\u2019.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question it is given that, by dividing x3<\/sup> \u2013 px2<\/sup> + x + 6 and 2x3<\/sup> \u2013 x2<\/sup> \u2013 (p + 3)x \u2013 6 by x \u2013 3 = 0, then x = 3.<\/p>\n\n\n\n

Let us assume p(x) = x3<\/sup> \u2013 px2<\/sup> + x + 6<\/p>\n\n\n\n

Now, substitute the value of x in p(x),<\/p>\n\n\n\n

p(3) = 33<\/sup> \u2013 (p \u00d7 32<\/sup>) + 3 + 6<\/p>\n\n\n\n

= 27 \u2013 9p + 9<\/p>\n\n\n\n

= 36 \u2013 9p<\/p>\n\n\n\n

Then, q(x) = 2x3<\/sup> \u2013 x2<\/sup> \u2013 (p + 3)x \u2013 6<\/p>\n\n\n\n

Now, substitute the value of x in q(x),<\/p>\n\n\n\n

q(3) = (2 \u00d7 33<\/sup>) \u2013 (3)2<\/sup> \u2013 (p + 3) \u00d7 3 \u2013 6<\/p>\n\n\n\n

= (2 \u00d7 27) \u2013 9 \u2013 3p \u2013 9 \u2013 6<\/p>\n\n\n\n

= 54 \u2013 24 \u2013 3p<\/p>\n\n\n\n

= 30 \u2013 3p<\/p>\n\n\n\n

Given, the remainder in each case is same,<\/p>\n\n\n\n

So, 36 \u2013 9p = 30 \u2013 3p<\/p>\n\n\n\n

36 \u2013 30 = 9p \u2013 3p<\/p>\n\n\n\n

6 = 6p<\/p>\n\n\n\n

p = 6\/6<\/p>\n\n\n\n

p = 1<\/p>\n\n\n\n

Therefore, value of p is 1.<\/p>\n\n\n\n

(ii) Find \u2018a\u2019 if the two polynomials ax3<\/sup> + 3x2<\/sup> \u2013 9 and 2x3<\/sup> + 4x + a, leaves the same remainder when divided by x + 3.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume p(x) = ax3<\/sup> + 3x2<\/sup> \u2013 9 and q(x) = 2x3<\/sup> + 4x + a<\/p>\n\n\n\n

From the question it is given that, both p(x) and q(x) leaves the same remainder when divided by x + 3.<\/p>\n\n\n\n

Let us assume that, x + 3 = 0<\/p>\n\n\n\n

Then, x = -3<\/p>\n\n\n\n

Now, substitute the value of x in p(x) and in q(x),<\/p>\n\n\n\n

So, p(-3) = q(-3)<\/p>\n\n\n\n

a(-3)3<\/sup> + 3(-3)2<\/sup> \u2013 9 = 2(-3)3<\/sup> + 4(-3) + a<\/p>\n\n\n\n

-27a + 27 \u2013 9 = \u2013 54 \u2013 12 + a<\/p>\n\n\n\n

-27a + 18 = \u2013 66 + a<\/p>\n\n\n\n

-27a \u2013 a = -66 \u2013 18<\/p>\n\n\n\n

-28 a = -84<\/p>\n\n\n\n

a = 84\/28<\/p>\n\n\n\n

Therefore, a = 3<\/p>\n\n\n\n

(iii) The polynomials ax3<\/sup> + 3x2<\/sup> \u2013 3 and 2x3<\/sup> \u2013 5x + a when divided by x \u2013 4 leave the remainder r1<\/sub> and r2<\/sub> respectively. If 2r1<\/sub> = r2<\/sub>, then find the value of a.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us assume p(x) = ax3<\/sup> + 3x2<\/sup> \u2013 3 and q(x) = 2x3<\/sup> \u2013 5x + a<\/p>\n\n\n\n

From the question it is given that, both p(x) and q(x) leaves the remainder r1<\/sub> and r2<\/sub> respectively when divided by x \u2013 4.<\/p>\n\n\n\n

Also, given relation 2r1<\/sub> = r2<\/sub><\/p>\n\n\n\n

Let us assume that, x \u2013 4 = 0<\/p>\n\n\n\n

Then, x = 4<\/p>\n\n\n\n

Now, substitute the value of x in p(x) and in q(x),<\/p>\n\n\n\n

By factor theorem, r1<\/sub> = p(x) and r2<\/sub> = q(x)<\/p>\n\n\n\n

So, 2 \u00d7 p(4) = q(4)<\/p>\n\n\n\n

2[a(4)3<\/sup> + 3(4)2<\/sup> \u2013 3] = 2(4)3<\/sup> \u2013 5(4) + a<\/p>\n\n\n\n

2[64a + 48 \u2013 3] = 128 \u2013 20 + a<\/p>\n\n\n\n

128a + 96 \u2013 6 = 128 \u2013 20 + a<\/p>\n\n\n\n

128a + 90 = 108 + a<\/p>\n\n\n\n

128a \u2013 a = 108 \u2013 90<\/p>\n\n\n\n

127a = 18<\/p>\n\n\n\n

a = 18\/127<\/p>\n\n\n\n

Therefore, the value of a = 18\/127.<\/p>\n\n\n\n

8. Using remainder theorem, find the remainders obtained when x3<\/sup> + (kx + 8)x + k Is divided by x + 1 and x \u2013 2. Hence, find k if the sum of two remainders is 1.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us assume p(x) = x3<\/sup> + (kx + 8)x + k<\/p>\n\n\n\n

From the question it is given that, the sum of the remainders when p(x) is divided by (x + 1) and (x \u2013 2) is 1.<\/p>\n\n\n\n

Let us assume that, x + 1 = 0<\/p>\n\n\n\n

Then, x = -1<\/p>\n\n\n\n

Also, when x \u2013 2 = 0<\/p>\n\n\n\n

Then, x = 2<\/p>\n\n\n\n

Now, by remainder theorem we have<\/p>\n\n\n\n

p(-1) + p(2) = 1<\/p>\n\n\n\n

(-1)3<\/sup> + [k(-1) + 8](-1) + k + (2)3<\/sup> + [k(2) + 8](2) + k = 1<\/p>\n\n\n\n

-1 + k \u2013 8 + k + 8 + 4k + 16 + k = 1<\/p>\n\n\n\n

7k + 15 = 1<\/p>\n\n\n\n

7k = 1 \u2013 15<\/p>\n\n\n\n

k = -14\/7<\/p>\n\n\n\n

k = -2<\/p>\n\n\n\n

Therefore, k = -2.<\/p>\n\n\n\n

9. By factor theorem, show that (x + 3) and (2x \u2013 1) are factors of 2x2<\/sup> + 5x \u2013 3.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume, x + 3 = 0<\/p>\n\n\n\n

Then, x = \u2013 3<\/p>\n\n\n\n

Given, f(x) = 2x2<\/sup> + 5x \u2013 3<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(-3) = (2 \u00d7 (-3)2<\/sup>) + (5 \u00d7 -3) \u2013 3<\/p>\n\n\n\n

= (2 \u00d7 9) + (-15) \u2013 3<\/p>\n\n\n\n

= 18 \u2013 15 \u2013 3<\/p>\n\n\n\n

= 18 \u2013 18<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Now, 2x \u2013 1 = 0<\/p>\n\n\n\n

Then, 2x = 1<\/p>\n\n\n\n

x = \u00bd<\/p>\n\n\n\n

Given, f(x) = 2x2<\/sup> + 5x \u2013 3<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(\u00bd) = (2 \u00d7 (\u00bd)2<\/sup>) + (5 \u00d7 \u00bd) \u2013 3<\/p>\n\n\n\n

= (2 \u00d7 (\u00bc)) + 5\/2 \u2013 3<\/p>\n\n\n\n

= \u00bd + 5\/2 \u2013 3<\/p>\n\n\n\n

= (1 + 5)\/2 \u2013 3<\/p>\n\n\n\n

= 6\/2 \u2013 3<\/p>\n\n\n\n

= 3 -3<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Hence, it is proved that, (x + 3) and (2x \u2013 1) are factors of 2x2<\/sup> + 5x \u2013 3.<\/p>\n\n\n\n

10. Without actual division, prove that x4<\/sup> + 2x3<\/sup> \u2013 2x2<\/sup> + 2x + 3 is exactly divisible by x2<\/sup> + 2x \u2013 3.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Consider x2<\/sup> + 2x \u2013 3<\/p>\n\n\n\n

By factor method, x2<\/sup> + 3x \u2013 x \u2013 3<\/p>\n\n\n\n

= x (x + 3) \u2013 1(x + 3)<\/p>\n\n\n\n

= (x \u2013 1) (x + 3)<\/p>\n\n\n\n

So, f(x) = x4<\/sup> + 2x3<\/sup> \u2013 2x2<\/sup> + 2x + 3<\/p>\n\n\n\n

Now take, x + 3 = 0<\/p>\n\n\n\n

X = -3<\/p>\n\n\n\n

Then, f(-3) = (-3)4<\/sup> + 2 \u00d7 -(33<\/sup>) \u2013 (2 \u00d7 (-3)2<\/sup>) + (2 \u00d7 -3) + 3<\/p>\n\n\n\n

= 81 \u2013 54 \u2013 18 \u2013 6 \u2013 3<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Therefore, (x + 3) is a factor of f(x)<\/p>\n\n\n\n

And also, take x \u2013 1 = 0<\/p>\n\n\n\n

X = 1<\/p>\n\n\n\n

Then, f(1) = 14<\/sup> + 2(1)3<\/sup> \u2013 2(1)2<\/sup> + 2(1) \u2013 3<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Therefore, (x \u2013 1) is a factor of f(x)<\/p>\n\n\n\n

By comparing both results, p(x) is exactly divisible by x2<\/sup> + 2x \u2013 3.<\/p>\n\n\n\n

11. Show that (x \u2013 2) is a factor of 3x2<\/sup> \u2013 x \u2013 10. Hence factories 3x2<\/sup> \u2013 x \u2013 10.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume x \u2013 2 = 0<\/p>\n\n\n\n

Then, x = 2<\/p>\n\n\n\n

Given, f(x) = 3x2<\/sup> \u2013 x \u2013 10<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(2) = (3 \u00d7 22<\/sup>) \u2013 2 \u2013 10<\/p>\n\n\n\n

= (3 \u00d7 4) \u2013 2 \u2013 10<\/p>\n\n\n\n

= 12 \u2013 2 \u2013 10<\/p>\n\n\n\n

= 12 \u2013 12<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Therefore, (x \u2013 2) is a factor of f(x)<\/p>\n\n\n\n

Then, dividing (3x2<\/sup> \u2013 x \u2013 10) by (x \u2013 2), we get<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, 3x2<\/sup> \u2013 x \u2013 10 = (x \u2013 2) (3x + 5)<\/p>\n\n\n\n

12. Using the factor theorem, show that (x \u2013 2) is a factor of x3<\/sup> + x2<\/sup> \u2013 4x \u2013 4. Hence factorize the polynomial completely.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume, x \u2013 2 = 0<\/p>\n\n\n\n

Then, x = 2<\/p>\n\n\n\n

Given, f(x) = x3<\/sup> + x2<\/sup> \u2013 4x \u2013 4<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(2) = (2)3<\/sup> + (2)2<\/sup> \u2013 4(2) \u2013 4<\/p>\n\n\n\n

= 8 \u2013 4 \u2013 8 \u2013 4<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Therefore, by factor theorem (x \u2013 2) is a factor of x3<\/sup> + x2<\/sup> \u2013 4x \u2013 4<\/p>\n\n\n\n

Then, dividing f(x) by (x \u2013 2), we get<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Therefore, x3<\/sup> + x2<\/sup> \u2013 4x \u2013 4 = (x \u2013 2) (x2<\/sup> + 3x + 2)<\/p>\n\n\n\n

= (x \u2013 2) (x2<\/sup> + 2x + x + 2)<\/p>\n\n\n\n

= (x \u2013 2) (x(x + 2) + 1(x + 2))<\/p>\n\n\n\n

= (x \u2013 2) (x + 2) (x + 1)<\/p>\n\n\n\n

13. Show that 2x + 7 is a factor of 2x3<\/sup> + 5x2<\/sup> \u2013 11x \u2013 14. Hence factorize the given expression completely, using the factor theorem.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume 2x + 7 = 0<\/p>\n\n\n\n

Then, 2x = -7<\/p>\n\n\n\n

X = -7\/2<\/p>\n\n\n\n

Given, f(x) = 2x3<\/sup> + 5x2<\/sup> \u2013 11x \u2013 14<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(-7\/2) = 2(-7\/2)3<\/sup> + 5(-7\/2)2<\/sup> + 11(-7\/2) \u2013 14<\/p>\n\n\n\n

= 2(-343\/8) + 5(49\/4) + (-77\/2) \u2013 14<\/p>\n\n\n\n

= -343\/4 + 245\/4 \u2013 77\/2 \u2013 14<\/p>\n\n\n\n

= (-343 + 245 + 154 \u2013 56)\/4<\/p>\n\n\n\n

= -399 + 399\/4<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Therefore, (2x + 7) is a factor of 2x3<\/sup> + 5x2<\/sup> \u2013 11x \u2013 14<\/p>\n\n\n\n

Then, dividing f(x) by (2x + 1), we get<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, 2x3<\/sup> + 5x2<\/sup> \u2013 11x \u2013 14 = (2x + 7) (x2<\/sup> \u2013 x \u2013 2)<\/p>\n\n\n\n

= (2x + 7) (x2<\/sup> \u2013 2x + x \u2013 2)<\/p>\n\n\n\n

= (2x + 7) (x(x \u2013 2) + 1 (x \u2013 2))<\/p>\n\n\n\n

= (x + 1) (x \u2013 2) (2x + 7)<\/p>\n\n\n\n

14. Use factor theorem to factorize the following polynomials completely.<\/strong><\/p>\n\n\n\n

(i) x3<\/sup> + 2x2<\/sup> \u2013 5x \u2013 6<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume x = -1,<\/p>\n\n\n\n

Given, f(x) = x3<\/sup> + 2x2<\/sup> \u2013 5x \u2013 6<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(-1) = (-1)3<\/sup> + 2(-1)2<\/sup> \u2013 5(-1) \u2013 6<\/p>\n\n\n\n

= -1 +2 (1) + 5 \u2013 6<\/p>\n\n\n\n

= -1 +2 + 5 \u2013 6<\/p>\n\n\n\n

= -7 + 7<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Then, dividing f(x) by (x + 1), we get<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, x3<\/sup> + 2x2<\/sup> \u2013 5x \u2013 6 = (x + 1) (x2<\/sup> + 3x \u2013 2x \u2013 6)<\/p>\n\n\n\n

= (x + 1) (x(x + 3) \u2013 2(x + 3))<\/p>\n\n\n\n

= (x + 1) (x \u2013 2) (x + 3)<\/p>\n\n\n\n

(ii) x3<\/sup> \u2013 13x \u2013 12<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume x = -1,<\/p>\n\n\n\n

Given, f(x) = x3<\/sup> \u2013 13x \u2013 12<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(-1) = (-1)3<\/sup> \u2013 13(-1) \u2013 12<\/p>\n\n\n\n

= -1 + 13 \u2013 12<\/p>\n\n\n\n

= \u2013 13 + 13<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Then, dividing f(x) by (x + 1), we get<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, x3<\/sup> \u2013 13x \u2013 12 = (x + 1) (x2<\/sup> \u2013 x \u2013 12)<\/p>\n\n\n\n

= (x + 1) (x2<\/sup> \u2013 4x + 3x \u2013 12)<\/p>\n\n\n\n

= (x + 1) (x(x \u2013 4)) + 3(x \u2013 4))<\/p>\n\n\n\n

= (x + 1) (x + 3) (x \u2013 4)<\/p>\n\n\n\n

15. Use the remainder theorem to factorize the following expression.<\/strong><\/p>\n\n\n\n

(i) 2x3<\/sup> + x2<\/sup> \u2013 13x + 6<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume x = 2,<\/p>\n\n\n\n

Then, f(x) = 2x3<\/sup> + x2<\/sup> \u2013 13x + 6<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(2) = (2 \u00d7 23<\/sup>) + 22<\/sup> <\/sub>\u2013 13 \u00d7 2 + 6<\/p>\n\n\n\n

= (2 \u00d7 8) + 4 \u2013 26 + 6<\/p>\n\n\n\n

= 16 + 4 \u2013 26 + 6<\/p>\n\n\n\n

= 26 \u2013 26<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Then, dividing f(x) by (x \u2013 2), we get<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, 2x3<\/sup> + x2<\/sup> \u2013 13x + 6 = (x \u2013 2) (2x2<\/sup> + 5x \u2013 3)<\/p>\n\n\n\n

= (x \u2013 2)(2x2<\/sup> + 6x \u2013 x \u2013 3)<\/p>\n\n\n\n

= (x \u2013 2) (2x(x + 3) \u2013 1 (x + 3))<\/p>\n\n\n\n

= (x \u2013 2) (x + 3) (2x \u2013 1)<\/p>\n\n\n\n

(ii) 3x2<\/sup> + 2x2<\/sup> \u2013 19x + 6<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Given, f(x) = 3x3<\/sup> + 2x2<\/sup> \u2013 19x + 6<\/p>\n\n\n\n

Let us assume x = 1<\/p>\n\n\n\n

Then, f(1) = 3(1)3<\/sup> + 2(1)2<\/sup> \u2013 (19 \u00d7 1) + 6<\/p>\n\n\n\n

= 3 + 2 \u2013 19 + 6<\/p>\n\n\n\n

= 11 \u2013 19<\/p>\n\n\n\n

= \u2013 8<\/p>\n\n\n\n

So, \u2013 8 \u2260 0<\/p>\n\n\n\n

Let us assume x = -1<\/p>\n\n\n\n

Then, f(-1) = 3(-1)3<\/sup> + 2(-1)2<\/sup> \u2013 (19 \u00d7 (-1)) + 6<\/p>\n\n\n\n

= \u2013 3 + 2 + 19 + 6<\/p>\n\n\n\n

= \u2013 3 + 27<\/p>\n\n\n\n

= 24<\/p>\n\n\n\n

So, 24 \u2260 0<\/p>\n\n\n\n

Now, assume x = 2<\/p>\n\n\n\n

Then, f(2) = 3(2)3<\/sup> + 2(2)2<\/sup> \u2013 (19 \u00d7 (2)) + 6<\/p>\n\n\n\n

= 24 + 8 \u2013 38 + 6<\/p>\n\n\n\n

= 38 \u2013 38<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

So, 0 = 0<\/p>\n\n\n\n

Therefore, (x \u2013 2) is a factor of f(x).<\/p>\n\n\n\n

f(x) = 3x3<\/sup> + 2x2<\/sup> \u2013 19x + 6<\/p>\n\n\n\n

= 3x3<\/sup> \u2013 6x2<\/sup> + 8x2<\/sup> \u2013 16x \u2013 3x + 6<\/p>\n\n\n\n

= 3x2<\/sup> (x \u2013 2) + 8x (x \u2013 2) \u2013 3(x \u2013 2)<\/p>\n\n\n\n

= (x \u2013 2) (3x2<\/sup> + 8x \u2013 3)<\/p>\n\n\n\n

= (x \u2013 2) (3x2<\/sup> + 9x \u2013 x \u2013 3)<\/p>\n\n\n\n

= (x \u2013 2) (3x(x + 3) \u2013 1(x + 3)<\/p>\n\n\n\n

= (x \u2013 2) (x + 3) (3x \u2013 1)<\/p>\n\n\n\n

(iii) 2x3<\/sup> + 3x2<\/sup> \u2013 9x \u2013 10<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Given, f(x) = 2x3<\/sup> + 3x2<\/sup> \u2013 9x \u2013 10<\/p>\n\n\n\n

Let us assume, x = -1<\/p>\n\n\n\n

= 2(-1)3<\/sup> + 3(-1)2<\/sup> \u2013 9 (-1) \u2013 10<\/p>\n\n\n\n

= -2 + 3 + 9 \u2013 10<\/p>\n\n\n\n

= 12 \u2013 12<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Therefore, (x + 1) is the factor of 2x3<\/sup> + 3x2<\/sup> \u2013 9x \u2013 10<\/p>\n\n\n\n

Then, dividing f(x) by (x + 1), we get<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, 2x3<\/sup> + 3x2<\/sup> \u2013 9x \u2013 10 = 2x2<\/sup> + 5x \u2013 4x \u2013 10<\/p>\n\n\n\n

= x(2x + 5) \u2013 2 (2x + 5) \u2013 (2x + 5) (x \u2013 2)<\/p>\n\n\n\n

Hence the factors are (x + 1) (x \u2013 2) (2x + 5)<\/p>\n\n\n\n

(iv) x3<\/sup> + 10x2<\/sup> \u2013 37x + 26<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Given, f(x) = x3<\/sup> + 10x2<\/sup> \u2013 37x + 26<\/p>\n\n\n\n

Let us assume, x = 1<\/p>\n\n\n\n

Then, f(1) = 13<\/sup> + 10(1)2<\/sup> \u2013 37 (1) + 26<\/p>\n\n\n\n

= 1 + 10 \u2013 37 + 26<\/p>\n\n\n\n

= 37 \u2013 37<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Therefore, x \u2013 1is a factor of x3<\/sup> + 10x2<\/sup> \u2013 37x + 26<\/p>\n\n\n\n

Then, dividing f(x) by (x \u2013 1), we get<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, x3<\/sup> + 10x2<\/sup> \u2013 37x + 26 = (x \u2013 1) (x2<\/sup> + 11x \u2013 26)<\/p>\n\n\n\n

= (x \u2013 1) (x2<\/sup> + 13x \u2013 2x \u2013 26)<\/p>\n\n\n\n

= (x \u2013 1) (x (x + 13) \u2013 2(x + 13))<\/p>\n\n\n\n

= (x \u2013 1) ((x \u2013 2) (x + 13))<\/p>\n\n\n\n

16. If (2x + 1) is a factor of 6x3<\/sup> + 5x2<\/sup> + ax \u2013 2 find the value of a.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume 2x + 1 = 0<\/p>\n\n\n\n

Then, 2x = \u2013 1<\/p>\n\n\n\n

X = -\u00bd<\/p>\n\n\n\n

Given, f(x) = 6x3<\/sup> + 5x2<\/sup> + ax \u2013 2<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f (-\u00bd) = 6 (-\u00bd)3<\/sup> + 5 (-\u00bd)2<\/sup> + a (-\u00bd) \u2013 2<\/p>\n\n\n\n

= 6 (-1\/8) + 5 (\u00bc) \u2013 \u00bda \u2013 2<\/p>\n\n\n\n

= -3\/4 + 5\/4 \u2013 a\/2 \u2013 2<\/p>\n\n\n\n

= (-3 + 4 \u2013 2a \u2013 8)\/4<\/p>\n\n\n\n

= (-6 \u2013 2a)\/4<\/p>\n\n\n\n

From the question, (2x + 1) is a factor of 6x3<\/sup> + 5x2<\/sup> + ax \u2013 2<\/p>\n\n\n\n

Then, remainder is 0.<\/p>\n\n\n\n

So, (-6 \u2013 2a)\/4 = 0<\/p>\n\n\n\n

-6 \u2013 2a = 4 \u00d7 0<\/p>\n\n\n\n

\u2013 6 \u2013 2a = 0<\/p>\n\n\n\n

-2a = 6<\/p>\n\n\n\n

a = -6\/2<\/p>\n\n\n\n

a = \u2013 3<\/p>\n\n\n\n

Therefore, the value of a is \u2013 3.<\/p>\n\n\n\n

17. If (3x \u2013 2) is a factor of 3x3<\/sup> \u2013 kx2<\/sup> + 21x \u2013 10, find the value of k.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume 3x \u2013 2 = 0<\/p>\n\n\n\n

Then, 3x = 2<\/p>\n\n\n\n

X = 2\/3<\/p>\n\n\n\n

Given, f(x) = 3x3<\/sup> \u2013 kx2<\/sup> + 21x \u2013 10<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f (2\/3) = 3 (2\/3)3<\/sup> \u2013 k (2\/3)2<\/sup> + 21 (2\/3) \u2013 10<\/p>\n\n\n\n

= 3 (8\/27) \u2013 k (4\/9) + 14 \u2013 10<\/p>\n\n\n\n

= 8\/9 \u2013 4k\/9 + 14 \u2013 10<\/p>\n\n\n\n

= 8\/9 \u2013 4k\/9 + 4<\/p>\n\n\n\n

= (8 \u2013 4k + 36)\/9<\/p>\n\n\n\n

= (44 \u2013 4k)\/9<\/p>\n\n\n\n

From the question, (3x \u2013 2) is a factor of 3x3<\/sup> \u2013 kx2<\/sup> + 21x \u2013 10<\/p>\n\n\n\n

Then, remainder is 0<\/p>\n\n\n\n

So, (44 \u2013 4k)\/9 = 0<\/p>\n\n\n\n

44 \u2013 4k = 0 \u00d7 9<\/p>\n\n\n\n

44 = 4k<\/p>\n\n\n\n

K = 44\/4<\/p>\n\n\n\n

K = 11<\/p>\n\n\n\n

18. If (x \u2013 2) is a factor of 2x3<\/sup> \u2013 x2<\/sup> + px \u2013 2, then<\/strong> (i) find the value of p.<\/strong> (ii) with this value of p, factorize the above expression completely.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume x -2 = 0<\/p>\n\n\n\n

Then, x = 2<\/p>\n\n\n\n

Given, f(x) = 2x3<\/sup> \u2013 x2<\/sup> + px \u2013 2<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(2) = (2 \u00d7 23<\/sup>) \u2013 22<\/sup> + (p \u00d7 2) \u2013 2<\/p>\n\n\n\n

= (2 \u00d7 8) \u2013 4 + 2p \u2013 2<\/p>\n\n\n\n

= 16 \u2013 4 + 2p \u2013 2<\/p>\n\n\n\n

= 16 \u2013 6 + 2p<\/p>\n\n\n\n

= 10 + 2p<\/p>\n\n\n\n

From the question, (x \u2013 2) is a factor of 2x3<\/sup> \u2013 x2<\/sup> + px \u2013 2<\/p>\n\n\n\n

Then, remainder is 0.<\/p>\n\n\n\n

10 + 2p = 0<\/p>\n\n\n\n

2p = \u2013 10<\/p>\n\n\n\n

P = -10\/2<\/p>\n\n\n\n

P = -5<\/p>\n\n\n\n

So, (x \u2013 2) is a factor of 2x3<\/sup> \u2013 x2<\/sup> + 5x \u2013 2<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, 2x3<\/sup> \u2013 x2<\/sup> + 5x \u2013 2 = (x \u2013 2) (2x2<\/sup> + 3x + 1)<\/p>\n\n\n\n

= (x \u2013 2) (2x2<\/sup> + 2x + x + 1)<\/p>\n\n\n\n

= (x \u2013 2) (2x(x + 1) + 1(x + 1))<\/p>\n\n\n\n

= (x + 1) (x \u2013 2) (2x + 1)<\/p>\n\n\n\n

19. What number should be subtracted from 2x3<\/sup> \u2013 5x2<\/sup> + 5x so that the resulting polynomial has 2x \u2013 3 as a factor?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume the number to be subtracted from 2x3<\/sup> \u2013 5x2<\/sup> + 5x be p.<\/p>\n\n\n\n

Then, f(x) = 2x3<\/sup> \u2013 5x2<\/sup> + 5x \u2013 p<\/p>\n\n\n\n

Given, 2x \u2013 3 = 0<\/p>\n\n\n\n

x = 3\/2<\/p>\n\n\n\n

f(3\/2) = 0<\/p>\n\n\n\n

So, f(3\/2) = 2(3\/2)3<\/sup> \u2013 5(3\/2)2<\/sup> + 5(3\/2) \u2013 p = 0<\/p>\n\n\n\n

2(27\/8) \u2013 5(9\/4) + 15\/2 \u2013 p = 0<\/p>\n\n\n\n

27\/4 \u2013 45\/4 + 15\/2 \u2013 p = 0 [multiply by 4 for all numerator]<\/p>\n\n\n\n

<\/strong>27 \u2013 45 + 30 \u2013 4p = 0<\/p>\n\n\n\n

57 \u2013 45 \u2013 4p = 0<\/p>\n\n\n\n

12 \u2013 4p = 0<\/p>\n\n\n\n

P = 12\/4<\/p>\n\n\n\n

P = 3<\/p>\n\n\n\n

Therefore, 3 is the number should be subtracted from 2x3<\/sup> \u2013 5x2<\/sup> + 5x.<\/p>\n\n\n\n

20. (i) Find the value of the constants a and b, if (x \u2013 2) and (x + 3) are both factors of the expression x3<\/sup> + ax2<\/sup> + bx \u2013 12.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume x \u2013 2 = 0<\/p>\n\n\n\n

Then, x = 2<\/p>\n\n\n\n

Given, f(x) = x3<\/sup> + ax2<\/sup> + bx \u2013 12<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(2) = 23<\/sup> + a(2)2<\/sup> + b(2) \u2013 12<\/p>\n\n\n\n

= 8 + 4a + 2b \u2013 12<\/p>\n\n\n\n

= 4a + 2b \u2013 4<\/p>\n\n\n\n

From the question, (x \u2013 2) is a factor of x3<\/sup> + ax2<\/sup> + bx \u2013 12.<\/p>\n\n\n\n

So, 4a + 2b \u2013 4 = 0<\/p>\n\n\n\n

4a + 2b = 4<\/p>\n\n\n\n

By dividing both the side by 2 we get,<\/p>\n\n\n\n

2a + b = 2 \u2026 [equation (i)]<\/p>\n\n\n\n

Now, assume x + 3 = 0<\/p>\n\n\n\n

Then, x = -3<\/p>\n\n\n\n

Given, f(x) = x3<\/sup> + ax2<\/sup> + bx \u2013 12<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(-3) = (-3)3<\/sup> + a(-3)2<\/sup> + b(-3) \u2013 12<\/p>\n\n\n\n

= -27 + 9a \u2013 3b \u2013 12<\/p>\n\n\n\n

= 9a \u2013 3b \u2013 39<\/p>\n\n\n\n

From the question, (x \u2013 3) is a factor of x3<\/sup> + ax2<\/sup> + bx \u2013 12.<\/p>\n\n\n\n

So, 9a \u2013 3b \u2013 39 = 0<\/p>\n\n\n\n

9a \u2013 3b = 39<\/p>\n\n\n\n

By dividing both the side by 3 we get,<\/p>\n\n\n\n

3a \u2013 b = 13 \u2026 [equation (ii)]<\/p>\n\n\n\n

Now, adding both equation (i) and equation (ii) we get,<\/p>\n\n\n\n

(2a + b) + (3a \u2013 b) = 2 + 13<\/p>\n\n\n\n

2a + 3a + b \u2013 b = 15<\/p>\n\n\n\n

5a = 15<\/p>\n\n\n\n

a = 15\/5<\/p>\n\n\n\n

a = 3<\/p>\n\n\n\n

Consider the equation (i) to find out \u2018b\u2019.<\/p>\n\n\n\n

2a + b = 2<\/p>\n\n\n\n

2(3) + b = 2<\/p>\n\n\n\n

6 + b = 2<\/p>\n\n\n\n

b = 2 \u2013 6<\/p>\n\n\n\n

b = -4<\/p>\n\n\n\n

(ii) If (x + 2) and (x + 3) are factors of x3<\/sup> + ax + b, Find the values of a and b.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume x + 2 = 0<\/p>\n\n\n\n

Then, x = -2<\/p>\n\n\n\n

Given, f(x) = x3<\/sup> + ax + b<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(-2) = (-2)3<\/sup> + a(-2) + b<\/p>\n\n\n\n

= -8 \u2013 2a + b<\/p>\n\n\n\n

From the question, (x + 2) is a factor of x3<\/sup> + ax + b.<\/p>\n\n\n\n

Therefore, remainder is 0.<\/p>\n\n\n\n

f(x) = 0<\/p>\n\n\n\n

\u2013 8 \u2013 2a + b = 0<\/p>\n\n\n\n

2a \u2013 b = \u2013 8 \u2026 [equation (i)]<\/p>\n\n\n\n

Let us assume x + 3 = 0<\/p>\n\n\n\n

Then, x = -3<\/p>\n\n\n\n

Given, f(x) = x3<\/sup> + ax + b<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(-2) = (-3)3<\/sup> + a(-3) + b<\/p>\n\n\n\n

= -27 \u2013 3a + b<\/p>\n\n\n\n

From the question, (x + 3) is a factor of x3<\/sup> + ax + b.<\/p>\n\n\n\n

Therefore, remainder is 0.<\/p>\n\n\n\n

f(x) = 0<\/p>\n\n\n\n

\u2013 27 \u2013 3a + b = 0<\/p>\n\n\n\n

3a \u2013 b = \u2013 27 \u2026 [equation (i)]<\/p>\n\n\n\n

Now, subtracting both equation (i) and equation (ii) we get,<\/p>\n\n\n\n

(2a \u2013 b) \u2013 (3a \u2013 b) = -8 \u2013 (-27)<\/p>\n\n\n\n

2a \u2013 3a \u2013 b + b = \u2013 8 + 27<\/p>\n\n\n\n

-a = 19<\/p>\n\n\n\n

a = -19<\/p>\n\n\n\n

Consider the equation (i) to find out \u2018b\u2019.<\/p>\n\n\n\n

2a \u2013 b = \u2013 8<\/p>\n\n\n\n

2(-19) \u2013 b = -8<\/p>\n\n\n\n

-38 \u2013 b = \u2013 8<\/p>\n\n\n\n

b = -38 +8<\/p>\n\n\n\n

b = -30<\/p>\n\n\n\n

21. If (x + 2) and (x \u2013 3) are factors of x3<\/sup> + ax + b, find the values of a and b. With these values of a and b, factorize the given expression.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume x + 2 = 0<\/p>\n\n\n\n

Then, x = -2<\/p>\n\n\n\n

Given, f(x) = x3<\/sup> + ax + b<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(-2) = (-2)3<\/sup> + a(-2) + b<\/p>\n\n\n\n

= -8 \u2013 2a + b<\/p>\n\n\n\n

From the question, (x + 2) is a factor of x3<\/sup> + ax + b.<\/p>\n\n\n\n

Therefore, remainder is 0.<\/p>\n\n\n\n

f(x) = 0<\/p>\n\n\n\n

\u2013 8 \u2013 2a + b = 0<\/p>\n\n\n\n

2a \u2013 b = \u2013 8 \u2026 [equation (i)]<\/p>\n\n\n\n

Now, assume x \u2013 3 = 0<\/p>\n\n\n\n

Then, x = 3<\/p>\n\n\n\n

Given, f(x) = x3<\/sup> + ax + b<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(3) = (3)3<\/sup> + a(3) + b<\/p>\n\n\n\n

= 27 + 3a + b<\/p>\n\n\n\n

From the question, (x \u2013 3) is a factor of x3<\/sup> + ax + b.<\/p>\n\n\n\n

Therefore, remainder is 0.<\/p>\n\n\n\n

f(x) = 0<\/p>\n\n\n\n

27 + 3a + b = 0<\/p>\n\n\n\n

3a + b = \u2013 27 \u2026 [equation (ii)]<\/p>\n\n\n\n

Now, adding both equation (i) and equation (ii) we get,<\/p>\n\n\n\n

(2a \u2013 b) + (3a + b) = \u2013 8 \u2013 27<\/p>\n\n\n\n

2a \u2013 b + 3a + b = -35<\/p>\n\n\n\n

5a = -35<\/p>\n\n\n\n

a = -35\/5<\/p>\n\n\n\n

a = -7<\/p>\n\n\n\n

Consider the equation (i) to find out \u2018b\u2019.<\/p>\n\n\n\n

2a \u2013 b = \u2013 8<\/p>\n\n\n\n

2(-7) \u2013 b = -8<\/p>\n\n\n\n

-14 \u2013 b = -8<\/p>\n\n\n\n

b = \u2013 14 + 8<\/p>\n\n\n\n

b = -6<\/p>\n\n\n\n

Therefore, value of a = -7 and b = -6.<\/p>\n\n\n\n

Then, f(x) = x3<\/sup> \u2013 7x \u2013 6<\/p>\n\n\n\n

(x + 2) (x \u2013 3)<\/p>\n\n\n\n

= x(x \u2013 3) + 2(x \u2013 3)<\/p>\n\n\n\n

= x2<\/sup> \u2013 3x + 2x \u2013 6<\/p>\n\n\n\n

= x2<\/sup> \u2013 x \u2013 6<\/p>\n\n\n\n

Dividing f(x) by x2<\/sup> \u2013 x \u2013 6 we get,<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, x3<\/sup> \u2013 7x \u2013 6 = (x + 1) (x + 2) (x \u2013 3)<\/p>\n\n\n\n

22. (x \u2013 2) is a factor of the expression x3<\/sup> + ax2<\/sup> + bx + 6. When this expression is divided by (x \u2013 3), it leaves the remainder 3. Find the values of a and b.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question it is given that, (x \u2013 2) is a factor of the expression x3<\/sup> + ax2<\/sup> + bx + 6<\/p>\n\n\n\n

Then, f(x) = x3<\/sup> + ax2<\/sup> + bx + 6 \u2026 [equation (i)]<\/p>\n\n\n\n

Let assume x \u2013 2 = 0<\/p>\n\n\n\n

Then, x = 2<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(2) = 23<\/sup> <\/sub>+ a(2)2<\/sup> + 2b + 6<\/p>\n\n\n\n

= 8 + 4a + 2b + 6<\/p>\n\n\n\n

= 14 + 4a + 2b<\/p>\n\n\n\n

By dividing the numbers by 2 we get,<\/p>\n\n\n\n

= 7 + 2a + b<\/p>\n\n\n\n

From the question, (x \u2013 2) is a factor of the expression x3<\/sup> + ax2<\/sup> + bx + 6.<\/p>\n\n\n\n

So, remainder is 0.<\/p>\n\n\n\n

f(x) = 0<\/p>\n\n\n\n

7 + 2a + b = 0<\/p>\n\n\n\n

2a + b = -7 \u2026 [equation (ii)]<\/p>\n\n\n\n

Now, expression is divided by (x \u2013 3), it leaves the remainder 3.<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So, remainder = 33 + 9a + 3b = 3<\/p>\n\n\n\n

9a + 3b = 3 \u2013 33<\/p>\n\n\n\n

9a + 3b = -30<\/p>\n\n\n\n

By dividing the numbers by 3 we get,<\/p>\n\n\n\n

= 3a + b = \u2013 10 \u2026 [equation (iii)]<\/p>\n\n\n\n

Now, subtracting equation (iii) from equation (ii) we get,<\/p>\n\n\n\n

(3a + b) \u2013 (2a + b) = \u2013 10 \u2013 (-7)<\/p>\n\n\n\n

3a \u2013 2a + b \u2013 b = \u2013 10 + 7<\/p>\n\n\n\n

a = -3<\/p>\n\n\n\n

Consider the equation (ii) to find out \u2018b\u2019.<\/p>\n\n\n\n

2a + b = \u2013 7<\/p>\n\n\n\n

2(-3) + b = \u2013 7<\/p>\n\n\n\n

-6 + b = \u2013 7<\/p>\n\n\n\n

b = \u2013 7 + 6<\/p>\n\n\n\n

b = \u2013 1<\/p>\n\n\n\n

23. If (x \u2013 2) is a factor of the expression 2x3<\/sup> + ax2<\/sup> + bx \u2013 14 and when the expression is divided by (x \u2013 3), it leaves a remainder 52, find the values of a and b.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question it is given that, (x \u2013 2) is a factor of the expression 2x3<\/sup> + ax2<\/sup> + bx \u2013 14<\/p>\n\n\n\n

Then, f(x) = 2x3<\/sup> + ax2<\/sup> + bx \u2013 14 \u2026 [equation (i)]<\/p>\n\n\n\n

Let assume x \u2013 2 = 0<\/p>\n\n\n\n

Then, x = 2<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(2) = 2(2)3<\/sup> <\/sub>+ a(2)2<\/sup> + 2b \u2013 14<\/p>\n\n\n\n

= 16 + 4a + 2b \u2013 14<\/p>\n\n\n\n

= 2 + 4a + 2b<\/p>\n\n\n\n

By dividing the numbers by 2 we get,<\/p>\n\n\n\n

= 1 + 2a + b<\/p>\n\n\n\n

From the question, (x \u2013 2) is a factor of the expression 2x3<\/sup> + ax2<\/sup> + bx \u2013 14.<\/p>\n\n\n\n

So, remainder is 0.<\/p>\n\n\n\n

f(x) = 0<\/p>\n\n\n\n

1 + 2a + b = 0<\/p>\n\n\n\n

2a + b = -1 \u2026 [equation (ii)]<\/p>\n\n\n\n

Now, expression is divided by (x \u2013 3), it leaves the remainder 52.<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So, remainder = 9a + 3b + 40 = 52<\/p>\n\n\n\n

9a + 3b = 52 \u2013 40<\/p>\n\n\n\n

9a + 3b = 12<\/p>\n\n\n\n

By dividing the numbers by 3 we get,<\/p>\n\n\n\n

= 3a + b = 4 \u2026 [equation (iii)]<\/p>\n\n\n\n

Now, subtracting equation (iii) from equation (ii) we get,<\/p>\n\n\n\n

(3a + b) \u2013 (2a + b) = 4 \u2013 (-1)<\/p>\n\n\n\n

3a \u2013 2a + b \u2013 b = 4 + 1<\/p>\n\n\n\n

a = 5<\/p>\n\n\n\n

a = 5<\/p>\n\n\n\n

Consider the equation (ii) to find out \u2018b\u2019.<\/p>\n\n\n\n

2a + b = \u2013 1<\/p>\n\n\n\n

2(5) + b = \u2013 1<\/p>\n\n\n\n

10 + b = \u2013 1<\/p>\n\n\n\n

b = \u2013 1 \u2013 10<\/p>\n\n\n\n

b = \u2013 11<\/p>\n\n\n\n

24. If ax3<\/sup> + 3x2<\/sup> + bx \u2013 3 has a factor (2x + 3) and leaves remainder \u2013 3 when divided by (x + 2), find the values of a and b. With these values of a and b, factorize the given expression.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume, 2x + 3 = 0<\/p>\n\n\n\n

Then, 2x = -3<\/p>\n\n\n\n

x = -3\/2<\/p>\n\n\n\n

Given, f(x) = ax3<\/sup> + 3x2<\/sup> + bx \u2013 3<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(-3\/2) = a(-3\/2)3<\/sup> + 3(-3\/2)2<\/sup> + b(-3\/2) \u2013 3<\/p>\n\n\n\n

= a(-27\/8) + 3(9\/4) \u2013 3b\/2 \u2013 3<\/p>\n\n\n\n

= -27a\/8 + 27\/4 \u2013 3b\/2 \u2013 3<\/p>\n\n\n\n

From the question it is given that, <\/strong>ax3<\/sup> + 3x2<\/sup> + bx \u2013 3 has a factor (2x + 3).<\/strong><\/p>\n\n\n\n

So, remainder is 0.<\/p>\n\n\n\n

-27a\/8 + 27\/4 \u2013 3b\/2 \u2013 3 = 0<\/p>\n\n\n\n

-27a + 54 \u2013 12b \u2013 24 = 0<\/p>\n\n\n\n

-27a \u2013 12b = -30<\/p>\n\n\n\n

By dividing the numbers by \u2013 3 we get,<\/p>\n\n\n\n

9a + 4b = 10 [equation (i)]<\/p>\n\n\n\n

Now, let us assume x + 2 = 0<\/p>\n\n\n\n

Then, x = -2<\/p>\n\n\n\n

Given, f(x) = ax3<\/sup> + 3x2<\/sup> + bx \u2013 3<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(2) = a(-2)3<\/sup> + 3(-2)2<\/sup> + b(-2) \u2013 3<\/p>\n\n\n\n

= -8a + 12 \u2013 2b \u2013 3<\/p>\n\n\n\n

= -8a \u2013 2b + 9<\/p>\n\n\n\n

Leaves the remainder -3<\/p>\n\n\n\n

So, -8a \u2013 2b + 9 = -3<\/p>\n\n\n\n

-8a \u2013 2b = -3 \u2013 9<\/p>\n\n\n\n

-8a \u2013 2b = -12<\/p>\n\n\n\n

By dividing both sides by -2 we get,<\/p>\n\n\n\n

4a + b = 6 [equation (ii)]<\/p>\n\n\n\n

By multiplying equation (ii) by 4,<\/p>\n\n\n\n

16a + 4b = 24<\/p>\n\n\n\n

Now, subtracting equation (ii) from equation (i) we get,<\/p>\n\n\n\n

(16a + 4b) \u2013 (9a + 4b) = 24 \u2013 10<\/p>\n\n\n\n

16a \u2013 9a + 4b \u2013 4b = 14<\/p>\n\n\n\n

7a = 14<\/p>\n\n\n\n

a = 14\/7<\/p>\n\n\n\n

a = 2<\/p>\n\n\n\n

Consider the equation (i) to find out \u2018b\u2019.<\/p>\n\n\n\n

9a + 4b = 10<\/p>\n\n\n\n

9(2) + 4b = 10<\/p>\n\n\n\n

18 + 4b = 10<\/p>\n\n\n\n

4b = 10 \u2013 18<\/p>\n\n\n\n

4b = -8<\/p>\n\n\n\n

b = -8\/4<\/p>\n\n\n\n

b = -2<\/p>\n\n\n\n

Therefore, f(x) = ax3<\/sup> + 3x2<\/sup> + bx \u2013 3<\/p>\n\n\n\n

= 2x3<\/sup> + 3x2<\/sup> \u2013 2x \u2013 3<\/p>\n\n\n\n

Given, 2x + 3 is a factor of f(x)<\/p>\n\n\n\n

So, divide f(x) by 2x + 3<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, 2x3<\/sup> + 3x2<\/sup> \u2013 2x \u2013 3 = (2x + 3) (x2<\/sup> \u2013 1)<\/p>\n\n\n\n

= (2x + 3) (x + 1) (x \u2013 1)<\/p>\n\n\n\n

25. Given f(x) = ax2<\/sup> + bx + 2 and g(x) = bx2<\/sup> + ax + 1. If x \u2013 2 is a factor of f(x) but leaves the remainder \u2013 15 when it divides g(x), find the values of a and b. With these values of a and b, factorize the expression. f(x) + g(x) + 4x2<\/sup> + 7x.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question it is given that, f(x) = ax2<\/sup> + bx + 2 and g(x) = bx2<\/sup> + ax + 1 and x \u2013 2 is a factor of f(x),<\/p>\n\n\n\n

So, x = 2<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(2) = 0<\/p>\n\n\n\n

a(2)2<\/sup> + b(2) + 2 = 0<\/p>\n\n\n\n

4a + 2b + 2 = 0<\/p>\n\n\n\n

By dividing both sides by 2 we get,<\/p>\n\n\n\n

2a + b + 1 = 0 \u2026 [equation (i)]<\/p>\n\n\n\n

Given, g(x) divide by (x \u2013 2), leaves remainder \u2013 15<\/p>\n\n\n\n

g(x) = bx2<\/sup> + ax + 1<\/p>\n\n\n\n

So, g(2) = -15<\/p>\n\n\n\n

b(2)2<\/sup> + 2a + 1 = -15<\/p>\n\n\n\n

4b + 2a + 1 + 15 = 0<\/p>\n\n\n\n

4b + 2a + 16 = 0<\/p>\n\n\n\n

By dividing both sides by 2 we get,<\/p>\n\n\n\n

2b + a + 8 = 0 \u2026 [equation (ii)]<\/p>\n\n\n\n

Now, subtracting equation (ii) from equation (i) multiplied by 2,<\/p>\n\n\n\n

(4a + 2b + 2) \u2013 (a + 2b + 8) = 0 \u2013 0<\/p>\n\n\n\n

4a \u2013 a + 2b \u2013 2b + 2 \u2013 8 = 0<\/p>\n\n\n\n

3a \u2013 6 = 0<\/p>\n\n\n\n

3a = 6<\/p>\n\n\n\n

a = 6\/3<\/p>\n\n\n\n

a = 2<\/p>\n\n\n\n

Consider the equation (i) to find out \u2018b\u2019.<\/p>\n\n\n\n

2a + b + 1 = 0<\/p>\n\n\n\n

2(2) + b = \u2013 1<\/p>\n\n\n\n

4 + b = \u2013 1<\/p>\n\n\n\n

b = \u2013 1 \u2013 4<\/p>\n\n\n\n

b = \u2013 5<\/p>\n\n\n\n

Now, f(x) = ax2<\/sup> + bx + 2 = 2x2<\/sup> \u2013 5x + 2<\/p>\n\n\n\n

g(x) = bx2<\/sup> + ax + 1 = -5x2<\/sup> + 2x + 1<\/p>\n\n\n\n

then, f(x) + g(x) + 4x2<\/sup> + 7x<\/p>\n\n\n\n

= 2x2<\/sup> \u2013 5x + 2 \u2013 5x2<\/sup> + 2x + 1 + 4x2<\/sup> + 7x<\/p>\n\n\n\n

= x2<\/sup> + 4x + 3<\/p>\n\n\n\n

= x2<\/sup> + 3x + x + 3<\/p>\n\n\n\n

= x(x + 3) + 1(x + 3)<\/p>\n\n\n\n

= (x + 1) (x + 3)<\/p>\n\n\n\n

Chapter Test<\/p>\n\n\n\n

1. Find the remainder when 2x3<\/sup> \u2013 3x2<\/sup> + 4x + 7 is divided by<\/strong> <\/p>\n\n\n\n

(i) x \u2013 2<\/strong> <\/p>\n\n\n\n

(ii) x + 3<\/strong> <\/p>\n\n\n\n

(iii) 2x + 1<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question it is given that, f(x) = 2x3<\/sup> \u2013 3x2<\/sup> + 4x + 7<\/p>\n\n\n\n

(i) Consider x -2<\/p>\n\n\n\n

let us assume x \u2013 2 = 0<\/p>\n\n\n\n

Then, x = 2<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(2) = 2(2)3<\/sup> \u2013 3(2)2<\/sup> + 4(2) + 7<\/p>\n\n\n\n

= 16 \u2013 12 + 8 + 7<\/p>\n\n\n\n

= 31 -12<\/p>\n\n\n\n

= 19<\/p>\n\n\n\n

Therefore, the remainder is 19<\/p>\n\n\n\n

(ii) consider x + 3<\/p>\n\n\n\n

let us assume x + 3 = 0<\/p>\n\n\n\n

Then, x = -3<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(2) = 2(-3)3<\/sup> \u2013 3(-3)2<\/sup> + 4(-3) + 7<\/p>\n\n\n\n

= 2(-27) \u2013 3(9) \u2013 12 + 7<\/p>\n\n\n\n

= \u2013 54 \u2013 27 \u2013 12 + 7<\/p>\n\n\n\n

= \u2013 93 + 7<\/p>\n\n\n\n

= \u2013 86<\/p>\n\n\n\n

Therefore, remainder is -86.<\/p>\n\n\n\n

(iii) consider 2x + 1<\/p>\n\n\n\n

Let us assume, 2x + 1 = 0<\/p>\n\n\n\n

Then, 2x = -1<\/p>\n\n\n\n

X = -\u00bd<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f (-\u00bd) = 2 (-\u00bd)3<\/sup> \u2013 3(-\u00bd)2<\/sup> + 4 (-\u00bd) + 7<\/p>\n\n\n\n

= 2(-1\/8) \u2013 3 (\u00bc) + 4 (-\u00bd) + 7<\/p>\n\n\n\n

= -\u00bc \u2013 \u00be \u2013 2 + 7<\/p>\n\n\n\n

= -1 \u2013 2 + 7<\/p>\n\n\n\n

= 4<\/p>\n\n\n\n

Therefore, remainder is 4.<\/p>\n\n\n\n

2. When 2x3<\/sup> \u2013 9x2<\/sup> + 10x \u2013 p is divided by (x + 1), the remainder is \u2013 24. Find the value of p.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume x + 1 = 0<\/p>\n\n\n\n

Then, x = -1<\/p>\n\n\n\n

Given, f(x) = 2x3<\/sup> \u2013 9x2<\/sup> + 10x \u2013 p<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(-1) = 2(-1)3<\/sup> \u2013 9(-1)2<\/sup> + 10(-1) \u2013 p<\/p>\n\n\n\n

= -2 \u2013 9 \u2013 10 + p<\/p>\n\n\n\n

= -21 + p<\/p>\n\n\n\n

From the question it is given that, the remainder is \u2013 24,<\/p>\n\n\n\n

So, -21 + p = -24<\/p>\n\n\n\n

p = \u2013 24 + 21<\/p>\n\n\n\n

p = -3<\/p>\n\n\n\n

So, f(x) = 2x3<\/sup> \u2013 9x2<\/sup> + 10x \u2013 (-3)<\/p>\n\n\n\n

= 2x3<\/sup> \u2013 9x2<\/sup> + 10x + 3<\/p>\n\n\n\n

Therefore, the value of p is 3.<\/p>\n\n\n\n

3. If (2x \u2013 3) is a factor of 6x2<\/sup> + x + a, find the value of a. With this value of a, factorise the given expression.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume 2x \u2013 3 = 0<\/p>\n\n\n\n

Then, 2x = 3<\/p>\n\n\n\n

X = 3\/2<\/p>\n\n\n\n

Given, f(x) = 6x2<\/sup> + x + a<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(3\/2) = 6(3\/2)2<\/sup> + (3\/2) + a<\/p>\n\n\n\n

= 6(9\/4) + (3\/2) + a<\/p>\n\n\n\n

= 3(9\/2) + (3\/2) + a<\/p>\n\n\n\n

= 27\/2 + 3\/2 + a<\/p>\n\n\n\n

= 30\/2 + a<\/p>\n\n\n\n

= 15 + a<\/p>\n\n\n\n

From the question, (2x \u2013 3) is a factor of 6x2<\/sup> + x + a.<\/p>\n\n\n\n

So, remainder is 0.<\/p>\n\n\n\n

Then, 15 + a = 0<\/p>\n\n\n\n

a = -15<\/p>\n\n\n\n

Therefore, f(x) = 6x2<\/sup> + x \u2013 15<\/p>\n\n\n\n

Dividing f(x) by 2x \u2013 3 we get,<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, 6x2<\/sup> + x \u2013 15 = (2x \u2013 3) (3x + 5)<\/p>\n\n\n\n

4. When 3x2<\/sup> \u2013 5x + p is divided by (x \u2013 2), the remainder is 3. Find the value of p. Also factorize the polynomial 3x2<\/sup> \u2013 5x + p \u2013 3.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume x \u2013 2 = 0<\/p>\n\n\n\n

Then, x = 2<\/p>\n\n\n\n

Given, f(x) = 3x2<\/sup> \u2013 5x + p<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

So, f(2) = 3(2)2<\/sup> \u2013 5(2) + p<\/p>\n\n\n\n

= 3(4) \u2013 10 + p<\/p>\n\n\n\n

= 12 \u2013 10 + p<\/p>\n\n\n\n

= 2 + p<\/p>\n\n\n\n

From the question it is given that, remainder is 3.<\/p>\n\n\n\n

So, 2 + p = 3<\/p>\n\n\n\n

p = 3 \u2013 2<\/p>\n\n\n\n

p = 1<\/p>\n\n\n\n

Therefore, f(x) = 3x2<\/sup> \u2013 5x + 1<\/p>\n\n\n\n

Consider the polynomial, 3x2<\/sup> \u2013 5x + p \u2013 3<\/p>\n\n\n\n

Now, substitute the value of p in polynomial,<\/p>\n\n\n\n

= 3x2<\/sup> \u2013 5x + 1 \u2013 3<\/p>\n\n\n\n

= 3x2<\/sup> \u2013 5x \u2013 2<\/p>\n\n\n\n

Now, by factorizing the polynomial 3x2<\/sup> \u2013 5x \u2013 2,<\/p>\n\n\n\n

Dividing 3x2<\/sup> \u2013 5x \u2013 2 by x \u2013 2 we get,<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, 3x2<\/sup> \u2013 5x \u2013 2 = (x \u2013 2) (3x + 1)<\/p>\n\n\n\n

5. Prove that (5x + 4) is a factor of 5x3<\/sup> + 4x2<\/sup> \u2013 5x \u2013 4. Hence factorize the given polynomial completely.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume (5x + 4) = 0<\/p>\n\n\n\n

Then, 5x = -4<\/p>\n\n\n\n

x = -4\/5<\/p>\n\n\n\n

Given, f(x) = 5x3<\/sup> + 4x2<\/sup> \u2013 5x \u2013 4<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

So, f(-4\/5) = 5(-4\/5)3<\/sup> + 4(-4\/5)2<\/sup> \u2013 5(-4\/5) \u2013 4<\/p>\n\n\n\n

= 5(-64\/125) + 4 (16\/25) + 4 \u2013 4<\/p>\n\n\n\n

= -64\/25 + 64\/25<\/p>\n\n\n\n

= (-64 + 64)\/25<\/p>\n\n\n\n

= 0\/25<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Hence, (5x + 4) is a factor of 5x3<\/sup> + 4x2<\/sup> \u2013 5x \u2013 4.<\/p>\n\n\n\n

So, dividing 5x3<\/sup> + 4x2<\/sup> \u2013 5x \u2013 4 by 5x + 4 we get,<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, 5x3<\/sup> + 4x2<\/sup> \u2013 5x \u2013 4 = (5x + 4) (x2<\/sup> \u2013 1)<\/p>\n\n\n\n

= (5x + 4) (x2<\/sup> \u2013 12<\/sup>)<\/p>\n\n\n\n

= (5x + 4) (x + 1) (x \u2013 1)<\/p>\n\n\n\n

6. Use factor theorem to factorize the following polynomials completely:<\/strong> <\/p>\n\n\n\n

(i) 4x3<\/sup> + 4x2<\/sup> \u2013 9x \u2013 9<\/strong> <\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume x = -1,<\/p>\n\n\n\n

Given, f(x) = 4x3<\/sup> + 4x2<\/sup> \u2013 9x \u2013 9 <\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(-1) = 4(-1)3<\/sup> + 4(-1)2<\/sup> \u2013 9(-1) \u2013 9<\/p>\n\n\n\n

= -4 + 4 + 9 \u2013 9<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Therefore, x + 1 is the factor of 4x3<\/sup> + 4x2<\/sup> \u2013 9x \u2013 9.<\/p>\n\n\n\n

Now, dividing 4x3<\/sup> + 4x2<\/sup> \u2013 9x \u2013 9 by x + 1 we get,<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, 4x3<\/sup> + 4x2<\/sup> \u2013 9x \u2013 9 = (x + 1) (4x2<\/sup> \u2013 9)<\/p>\n\n\n\n

= (x + 1) ((2x)2<\/sup> \u2013 (3)2<\/sup>)<\/p>\n\n\n\n

= (x + 1) (2x + 3) (2x \u2013 3)<\/p>\n\n\n\n

(ii) x3<\/sup> \u2013 19x \u2013 30<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume x = -2,<\/p>\n\n\n\n

Given, f(x) = x3<\/sup> \u2013 19x \u2013 30<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(-1) = (-2)3<\/sup> \u2013 19(-2) \u2013 30<\/p>\n\n\n\n

= -8 + 38 \u2013 30<\/p>\n\n\n\n

= -38 + 38<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Therefore, x + 2 is the factor of x3<\/sup> \u2013 19x \u2013 30.<\/p>\n\n\n\n

Now, dividing x3<\/sup> \u2013 19x \u2013 30 by x + 2 we get,<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, x3<\/sup> \u2013 19x \u2013 30 = (x + 2)(x2<\/sup> \u2013 2x \u2013 15)<\/p>\n\n\n\n

= (x + 2) (x2<\/sup> \u2013 5x + 3x \u2013 15)<\/p>\n\n\n\n

= (x + 2) (x \u2013 5) (x + 3)<\/p>\n\n\n\n

7. If x3<\/sup> \u2013 2x2<\/sup> + px + q has a factor (x + 2) and leaves a remainder 9, when divided by (x + 1), find the values of p and q. With these values of p and q, factorize the given polynomial completely.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question it is given that, (x + 2) is a factor of the expression x3<\/sup> \u2013 2x2<\/sup> + px + q<\/p>\n\n\n\n

Then, f(x) = x3<\/sup> \u2013 2x2<\/sup> + px + q<\/p>\n\n\n\n

Let assume x + 2 = 0<\/p>\n\n\n\n

Then, x = -2<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(-2) = (-2)3<\/sup> \u2013 2(-2)2<\/sup> + p(-2) + q<\/p>\n\n\n\n

= -8 \u2013 8 \u2013 2p + q<\/p>\n\n\n\n

= -16 \u2013 2p + q<\/p>\n\n\n\n

2p \u2013 q = \u2013 16 \u2026 [equation (i)]<\/p>\n\n\n\n

Now, consider (x + 1)<\/p>\n\n\n\n

Then, f(x) = x3<\/sup> \u2013 2x2<\/sup> + px + q<\/p>\n\n\n\n

Let assume x + 1 = 0<\/p>\n\n\n\n

Then, x = -1<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(-1) = (-1)3<\/sup> \u2013 2(-1)2<\/sup> + p(-1) + q<\/p>\n\n\n\n

= -1 \u2013 2 \u2013p + q<\/p>\n\n\n\n

= \u2013 3 \u2013 p + q<\/p>\n\n\n\n

Given, remainder is 9<\/p>\n\n\n\n

So, -3 \u2013 p + q = 9<\/p>\n\n\n\n

\u2013 p + q = 9 + 3<\/p>\n\n\n\n

-p + q = 12 \u2026 [equation (ii)]<\/p>\n\n\n\n

Now, adding equation (i) and equation (ii) we get,<\/p>\n\n\n\n

(2p \u2013 q) + (-p + q) = \u2013 16 + 12<\/p>\n\n\n\n

2p \u2013 q \u2013 p + q = -4<\/p>\n\n\n\n

P = -4<\/p>\n\n\n\n

Consider the equation (ii) to find out \u2018b\u2019.<\/p>\n\n\n\n

\u2013 p + q = 12<\/p>\n\n\n\n

-(-4) + q = 12<\/p>\n\n\n\n

4 + q = 12<\/p>\n\n\n\n

q = 12 \u2013 4<\/p>\n\n\n\n

q = 8<\/p>\n\n\n\n

Therefore, by substituting the value of p and q f(x) = x3<\/sup> \u2013 2x2<\/sup> \u2013 4x + 8<\/p>\n\n\n\n

Dividing f(x) be (x + 2) we get,<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

x3<\/sup> \u2013 2x2<\/sup> \u2013 4x + 8 = (x + 2) (x2<\/sup> \u2013 4x + 4)<\/p>\n\n\n\n

= (x + 2) (x2<\/sup> \u2013 2 \u00d7 x (-2) + 22<\/sup>)<\/p>\n\n\n\n

= (x + 2) (x \u2013 2)2<\/sup><\/p>\n\n\n\n

8. If (x + 3) and (x \u2013 4) are factors of x3<\/sup> + ax2<\/sup> \u2013 bx + 24, find the values of a and b: With these values of a and b, factorize the given expression.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume x + 3 = 0<\/p>\n\n\n\n

Then, x = -3<\/p>\n\n\n\n

Given, f(x) = x3<\/sup> + ax2<\/sup> \u2013 bx + 24<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(-3) = (-3)3<\/sup> + a(-3)2<\/sup> \u2013 b(-3) + 24<\/p>\n\n\n\n

= -27 + 9a + 3b + 24<\/p>\n\n\n\n

= 9a + 3b \u2013 3<\/p>\n\n\n\n

Dividing all terms by 3 we get,<\/p>\n\n\n\n

= 3a + b \u2013 1<\/p>\n\n\n\n

From the question, (x + 3) is a factor of x3<\/sup> + ax2<\/sup> \u2013 bx + 24.<\/p>\n\n\n\n

Therefore, remainder is 0.<\/p>\n\n\n\n

f(x) = 0<\/p>\n\n\n\n

3a + b \u2013 1 = 0<\/p>\n\n\n\n

3a + b = 1 \u2026 [equation (i)]<\/p>\n\n\n\n

Now, assume x \u2013 4 = 0<\/p>\n\n\n\n

Then, x = 4<\/p>\n\n\n\n

Given, f(x) = x3<\/sup> + ax2<\/sup> \u2013 bx + 24<\/p>\n\n\n\n

Now, substitute the value of x in f(x),<\/p>\n\n\n\n

f(4) = 43<\/sup> + a(4)2<\/sup> \u2013 b(4) + 24<\/p>\n\n\n\n

= 64 + 16a \u2013 4b + 24<\/p>\n\n\n\n

= 88 + 16a \u2013 4b<\/p>\n\n\n\n

Dividing all terms by 4 we get,<\/p>\n\n\n\n

= 22 + 4a \u2013 b<\/p>\n\n\n\n

From the question, (x \u2013 4) is a factor of x3<\/sup> + ax2<\/sup> \u2013 bx + 24.<\/p>\n\n\n\n

Therefore, remainder is 0.<\/p>\n\n\n\n

f(x) = 0<\/p>\n\n\n\n

22 + 4a \u2013 b = 0<\/p>\n\n\n\n

4a \u2013 b = \u2013 22 \u2026 [equation (ii)]<\/p>\n\n\n\n

Now, adding both equation (i) and equation (ii) we get,<\/p>\n\n\n\n

(3a + b) + (4a \u2013 b) = 1 \u2013 22<\/p>\n\n\n\n

3a + b + 4a \u2013 b = \u2013 21<\/p>\n\n\n\n

7a = \u2013 21<\/p>\n\n\n\n

a = -21\/7<\/p>\n\n\n\n

a = -3<\/p>\n\n\n\n

Consider the equation (i) to find out \u2018b\u2019.<\/p>\n\n\n\n

3a + b = 1<\/p>\n\n\n\n

3(-3) + b = 1<\/p>\n\n\n\n

-9 + b = 1<\/p>\n\n\n\n

b = 1 + 9<\/p>\n\n\n\n

b = 10<\/p>\n\n\n\n

Therefore, value of a = -3 and b = 10.<\/p>\n\n\n\n

Then, by substituting the value of a and b f(x) = x3<\/sup> \u2013 3x2<\/sup> \u2013 10x + 24<\/p>\n\n\n\n

(x + 3) (x \u2013 4)<\/p>\n\n\n\n

= x(x \u2013 4) + 3(x \u2013 4)<\/p>\n\n\n\n

= x2<\/sup> \u2013 4x + 3x \u2013 12<\/p>\n\n\n\n

= x2<\/sup> \u2013 x \u2013 12<\/p>\n\n\n\n

Dividing f(x) by x2<\/sup> \u2013 x \u2013 12 we get,<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, x3<\/sup> \u2013 3x2<\/sup> \u2013 10x + 24 = (x2<\/sup> \u2013 x \u2013 12) (x \u2013 2)<\/p>\n\n\n\n

= (x + 3) (x \u2013 4) (x \u2013 2)<\/p>\n\n\n\n

9. If (2x + 1) is a factor of both the expressions 2x2<\/sup> \u2013 5x + p and 2x2<\/sup> + 5x + q, find the value of p and q. Hence find the other factors of both the polynomials.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume 2x + 1 = 0<\/p>\n\n\n\n

Then, 2x = -1<\/p>\n\n\n\n

x = -\u00bd<\/p>\n\n\n\n

Given, p(x) = 2x2<\/sup> \u2013 5x + p<\/p>\n\n\n\n

Now, substitute the value of x in p(x),<\/p>\n\n\n\n

p (-\u00bd) = 2 (-\u00bd)2<\/sup> \u2013 5(-\u00bd) + p<\/p>\n\n\n\n

= 2(1\/4) + 5\/2 + p<\/p>\n\n\n\n

= \u00bd + 5\/2 + p<\/p>\n\n\n\n

= 6\/2 + p<\/p>\n\n\n\n

= 3 + p<\/p>\n\n\n\n

From the question it is given that, (2x + 1) is a factor of both the expressions 2x2<\/sup> \u2013 5x + p<\/p>\n\n\n\n

So, remainder is 0.<\/p>\n\n\n\n

Then, 3 + p = 0<\/p>\n\n\n\n

p = \u2013 3<\/p>\n\n\n\n

Now consider q(x) = 2x2<\/sup> + 5x + q<\/p>\n\n\n\n

Substitute the value of x in q(x)<\/p>\n\n\n\n

q (-\u00bd) = 2 (-\u00bd)2<\/sup> + 5(-\u00bd) + q<\/p>\n\n\n\n

= 2(1\/4) \u2013 5\/2 + q<\/p>\n\n\n\n

= \u00bd \u2013 5\/2 + q<\/p>\n\n\n\n

= (1 \u2013 5)\/2 + q<\/p>\n\n\n\n

= -4\/2 + q<\/p>\n\n\n\n

= q \u2013 2<\/p>\n\n\n\n

From the question it is given that, (2x + 1) is a factor of both the expressions 2x2<\/sup> + 5x + q<\/p>\n\n\n\n

So, remainder is 0.<\/p>\n\n\n\n

q \u2013 2 = 0<\/p>\n\n\n\n

q = 2<\/p>\n\n\n\n

Therefore, p = \u2013 3 and q = 2<\/p>\n\n\n\n

P(x) = 2x2<\/sup> \u2013 5x \u2013 3<\/p>\n\n\n\n

q(x) = 2x2<\/sup> + 5x + 2<\/p>\n\n\n\n

Then, divide p(x) by 2x + 1<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, 2x2<\/sup> \u2013 5x \u2013 3 = (2x + 1) (x \u2013 3)<\/p>\n\n\n\n

Now, divide q(x) by 2x + 1<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, 2x2<\/sup> + 5x + 2 = (2x + 1) (x + 2)<\/p>\n\n\n\n

10. If a polynomial f(x)= x4<\/sup>-2x3<\/sup> + 3x2<\/sup>\u2013 ax + b leaves reminder 5 and 19 when divided by (x \u2013 1) and (x + 1) respectively, Find the values of a and b. Hence determined the reminder when f(x) is divided by (x-2).<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question it is given that,<\/p>\n\n\n\n

f(x) = x4<\/sup> \u2013 2x3<\/sup> +3x2<\/sup> \u2013 ax + b<\/p>\n\n\n\n

Factor (x \u2013 1) leaves remainder 5,<\/p>\n\n\n\n

Factor (x + 1) leaves remainder 19,<\/p>\n\n\n\n

Where x = 1 and x = \u2013 1<\/p>\n\n\n\n

f(-1) = (-1)4<\/sup> \u2013 2(-1)3<\/sup> + 3(-1)2<\/sup> \u2013 a(-1) + b = 19<\/p>\n\n\n\n

1 \u2013 2(-1) + 3(1) \u2013 a(-1) + b = 19<\/p>\n\n\n\n

1 + 2 + 3 + a + b = 19<\/p>\n\n\n\n

6 + a + b = 19<\/p>\n\n\n\n

a + b = 19 \u2013 6<\/p>\n\n\n\n

a + b = 13 \u2026 [equation (i)]<\/p>\n\n\n\n

f(1) = (1)4<\/sup> \u2013 2(1)3<\/sup> + 3(1)2<\/sup> \u2013 a(1) + b = 5<\/p>\n\n\n\n

1 \u2013 2(1) + 3(1) \u2013 a(1) + b = 5<\/p>\n\n\n\n

1 \u2013 2 + 3 \u2013 a + b = 5<\/p>\n\n\n\n

2 \u2013 a + b = 5<\/p>\n\n\n\n

\u2013 a + b = 5 \u2013 2<\/p>\n\n\n\n

\u2013 a + b = 3 \u2026 [equation (ii)]<\/p>\n\n\n\n

Now, subtracting equation (ii) from equation (i) we get,<\/p>\n\n\n\n

(a + b) \u2013 (- a + b) = 13 \u2013 3<\/p>\n\n\n\n

a + b + a \u2013 b = 10<\/p>\n\n\n\n

2a = 10<\/p>\n\n\n\n

a = 10\/2<\/p>\n\n\n\n

a = 5<\/p>\n\n\n\n

To find out the value of b, substitute the value of a in equation (i) we get,<\/p>\n\n\n\n

a + b = 13<\/p>\n\n\n\n

5 + b = 13<\/p>\n\n\n\n

b = 13 \u2013 5<\/p>\n\n\n\n

b = 8<\/p>\n\n\n\n

Therefore, value of a = 5 and b = 8<\/p>\n\n\n\n

11. When a polynomial f(x) is divided by (x \u2013 1), the remainder is 5 and when it is, divided by (x \u2013 2), the remainder is 7. Find the remainder when it is divided by (x \u2013 1) (x \u2013 2).<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question it is given that,<\/p>\n\n\n\n

Polynomial f(x) is divided by (x \u2013 1),<\/p>\n\n\n\n

Remainder = 5<\/p>\n\n\n\n

Let us assume x \u2013 1 = 0<\/p>\n\n\n\n

x = 1<\/p>\n\n\n\n

f(1) = 5<\/p>\n\n\n\n

and the divided be (x \u2013 2), remainder = 7<\/p>\n\n\n\n

let us assume x \u2013 2 = 0<\/p>\n\n\n\n

x = 2<\/p>\n\n\n\n

Therefore, f(2) = 7<\/p>\n\n\n\n

So, f(x) = (x \u2013 1) (x \u2013 2) q(x) + ax + b<\/p>\n\n\n\n

Where, q(x) is the quotient and ax + b is remainder,<\/p>\n\n\n\n

Now put x = 1, we get,<\/p>\n\n\n\n

f(1) = (1 \u2013 1)(1 \u2013 2)q(1) + (a \u00d7 1) + b<\/p>\n\n\n\n

a + b = 5 \u2026 [equation (i)]<\/p>\n\n\n\n

x = 2,<\/p>\n\n\n\n

f(2) = (2 \u2013 1)(2 \u2013 2)q(2) + (a \u00d7 2) + b<\/p>\n\n\n\n

2a + b = 7 \u2026 [equation (ii)]<\/p>\n\n\n\n

Now subtracting equation (i) from equation (ii) we get,<\/p>\n\n\n\n

(2a + b) \u2013 (a + b) = 7 \u2013 5<\/p>\n\n\n\n

2a + b \u2013 a \u2013 b = 2<\/p>\n\n\n\n

a = 2<\/p>\n\n\n\n

To find out the value of b, substitute the value of a in equation (i) we get,<\/p>\n\n\n\n

a + b = 5<\/p>\n\n\n\n

2 + b = 5<\/p>\n\n\n\n

b = 5 \u2013 2<\/p>\n\n\n\n

b = 3<\/p>\n\n\n\n

Therefore, the remainder = ax + b = 2x + 3<\/p>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

ML Aggarwal Solutions for Class 10 Maths Chapter 1- Goods and Service Tax (GST)<\/p>\n\n\n\n

Download PDF<\/strong>: ML Aggarwal Solutions for Class 10 Maths Chapter 1- Goods and Service Tax (GST) PDF<\/a><\/p>\n\n\n\n

Chapterwise ML Aggarwal Solutions for Class 10 Maths :<\/strong><\/h2>\n\n\n\n