{"id":55143,"date":"2020-11-19T16:20:55","date_gmt":"2020-11-19T16:20:55","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=55143"},"modified":"2023-09-19T02:14:37","modified_gmt":"2023-09-19T02:14:37","slug":"ncert-solutions-for-class-10th-mathematics-chapter-2-polynomials","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-class-10th-mathematics-chapter-2-polynomials\/","title":{"rendered":"NCERT Solutions for Class 10th Mathematics: Chapter 2 Polynomials"},"content":{"rendered":"\n

Class 10: Mathematics Chapter 2 solutions. Complete Class 10 Mathematics Chapter 2 Notes.<\/p>\n\n\n\n

NCERT Solutions for Class 10th Mathematics: Chapter 2 Polynomials<\/h2>\n\n\n\n

NCERT 10th Mathematics Chapter 2, class 10 Mathematics Chapter 2 solutions<\/p>\n\n\n\n

Page No: 28<\/p>\n\n\n\n

Exercises 2.1<\/strong><\/p>\n\n\n\n

1. The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.<\/strong><\/p>\n\n\n\n

\"NCERT<\/a><\/figure>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) The number of zeroes is 0 as the graph does not cut the x<\/em>-axis at any point.
(ii) The number of zeroes is 1 as the graph intersects the x<\/em>-axis at only 1 point.
(iii) The number of zeroes is 3 as the graph intersects the x<\/em>-axis at 3 points.
(iv) The number of zeroes is 2 as the graph intersects the x<\/em>-axis at 2 points.
(v) The number of zeroes is 4 as the graph intersects the x<\/em>-axis at 4 points.
(vi) The number of zeroes is 3 as the graph intersects the x<\/em>-axis at 3 points.<\/p>\n\n\n\n

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
<\/strong>(i) x<\/em>2<\/sup> \u2013 2x<\/em> \u2013 8
(ii) 4s<\/em>2<\/sup> \u2013 4s<\/em> + 1
(iii) 6x<\/em>2<\/sup> \u2013 3 \u2013 7x<\/em>
(iv) 4u<\/em>2<\/sup> + 8u<\/em>
(v) t<\/em>2<\/sup> \u2013 15
(vi) 3x<\/em>2<\/sup> \u2013 x<\/em> \u2013 4<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) x<\/em>2<\/sup> \u2013 2x<\/em> \u2013 8
<\/strong>= (x<\/em> – 4) (x<\/em> + 2)
The value of x<\/em>2<\/sup> \u2013 2x<\/em> \u2013 8 is zero when x<\/em> – 4 = 0 or x<\/em> + 2 = 0, i.e., when x<\/em> = 4 or x<\/em> = -2
Therefore, the zeroes of x<\/em>2<\/sup> \u2013 2x<\/em> \u2013 8 are 4 and -2.<\/p>\n\n\n\n

Sum of zeroes = 4 + (-2) = 2 = -(-2)\/1 = -(Coefficient of x<\/em>)\/Coefficient of x<\/em>2<\/sup><\/p>\n\n\n\n

Product of zeroes = 4 \u00d7 (-2) = -8 = -8\/1 = Constant term\/Coefficient of x<\/em>2<\/sup><\/p>\n\n\n\n

(ii) 4s<\/em>2<\/sup> \u2013 4s<\/em> + 1
<\/strong>= (2s<\/em>-1)2<\/sup>
The value of 4s<\/em>2<\/sup> – 4s<\/em> + 1 is zero when 2s<\/em> – 1 = 0, i.e., s = 1\/2<\/p>\n\n\n\n

Therefore, the zeroes of 4s2<\/sup> – 4s + 1 are 1\/2 and 1\/2.<\/p>\n\n\n\n

Sum of zeroes = 1\/2 + 1\/2 = 1 = -(-4)\/4 = -(Coefficient of s)<\/em>\/Coefficient of s<\/em>2<\/sup>
Product of zeroes = 1\/2 \u00d7 1\/2 = 1\/4 = Constant term\/Coefficient of s<\/em>2<\/sup>.<\/p>\n\n\n\n

(iii) 6x<\/em>2<\/sup> \u2013 3 \u2013 7x<\/em>
<\/strong>= <\/em>6x<\/em>2 <\/sup>\u2013 7x <\/em>\u2013 3
= (3x<\/em> + 1) (2x<\/em> – 3)
The value of 6x<\/em>2<\/sup> – 3 – 7x<\/em> is zero when 3x<\/em> + 1 = 0 or 2x<\/em> – 3 = 0, i.e., x<\/em> = -1\/3 or x<\/em> = 3\/2<\/p>\n\n\n\n

Therefore, the zeroes of 6x<\/em>2<\/sup> – 3 – 7x<\/em> are -1\/3 and 3\/2.<\/p>\n\n\n\n

Sum of zeroes = -1\/3 + 3\/2 = 7\/6 = -(-7)\/6 = -(Coefficient of x<\/em>)\/Coefficient of x<\/em>2<\/sup>
Product of zeroes = -1\/3 \u00d7 3\/2 = -1\/2 = -3\/6 = Constant term\/Coefficient of x<\/em>2<\/sup>.<\/p>\n\n\n\n

(iv) 4u<\/em>2<\/sup> + 8u<\/em>
<\/strong>= <\/em>4u<\/em>2<\/sup> + 8u + <\/em>0
= 4u<\/em>(u<\/em> + 2)
The value of 4u<\/em>2<\/sup> + 8u<\/em> is zero when 4u<\/em> = 0 or u<\/em> + 2 = 0, i.e., u<\/em> = 0 or u<\/em> = – 2
Therefore, the zeroes of 4u<\/em>2<\/sup> + 8u<\/em> are 0 and – 2.
Sum of zeroes = 0 + (-2) = -2 = -(8)\/4 = -(Coefficient of u<\/em>)\/Coefficient of u<\/em>2<\/sup>
Product of zeroes = 0 \u00d7 (-2) = 0 = 0\/4 = Constant term\/Coefficient of u<\/em>2<\/sup>.<\/p>\n\n\n\n

(v) t<\/em>2<\/sup> \u2013 15
<\/strong>= t<\/em>2 <\/sup>– 0.t<\/em> – 15
= (t <\/em>– \u221a15) (t<\/em> + \u221a15)
The value of t<\/em>2<\/sup> – 15 is zero when t<\/em> – \u221a15 = 0 or t<\/em> + \u221a15 = 0, i.e., when t<\/em> = \u221a15 or t <\/em>= -\u221a15
Therefore, the zeroes of t<\/em>2<\/sup> – 15 are \u221a15 and -\u221a15.Sum of zeroes = \u221a15 + -\u221a15 = 0 = -0\/1 = -(Coefficient of t<\/em>)\/Coefficient of t<\/em>2<\/sup>
Product of zeroes = (\u221a15) (-\u221a15) = -15 = -15\/1 = Constant term\/Coefficient of u<\/em>2<\/sup>.<\/p>\n\n\n\n

(vi) 3x<\/em>2<\/sup> \u2013 x<\/em> \u2013 4
<\/strong>= (3x<\/em> – 4) (x<\/em> + 1)
The value of 3x<\/em>2<\/sup> \u2013 x<\/em> \u2013 4 is zero when 3x<\/em> – 4 = 0 and x<\/em> + 1 = 0,i.e., when x<\/em> = 4\/3 or x<\/em> = -1
Therefore, the zeroes of 3x<\/em>2<\/sup> \u2013 x<\/em> \u2013 4 are 4\/3 and -1.<\/p>\n\n\n\n

Sum of zeroes = 4\/3 + (-1) = 1\/3 = -(-1)\/3 = -(Coefficient of x<\/em>)\/Coefficient of x<\/em>2<\/sup><\/p>\n\n\n\n

Product of zeroes = 4\/3 \u00d7 (-1) = -4\/3 = Constant term\/Coefficient of x<\/em>2<\/sup>.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 2, class 10 Mathematics Chapter 2 solutions<\/p>\n\n\n\n

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.<\/strong><\/p>\n\n\n\n

(i) 1\/4 , -1
(ii) \u221a2 , 1\/3
(iii) 0, \u221a5
(iv) 1,1 
(v) -1\/4 ,1\/4
(vi) 4,1<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 1\/4 , -1
<\/strong>Let the polynomial be ax<\/em>2<\/sup> + bx<\/em> + c<\/em>, and its zeroes be \u03b1 and \u00df
\u03b1 + \u00df = 1\/4 = –b<\/em>\/a<\/em>
\u03b1\u00df = -1 = -4\/4 = c<\/em>\/a<\/em>
If a<\/em> = 4, then b<\/em> = -1, c<\/em> = -4
Therefore, the quadratic polynomial is 4x<\/em>2<\/sup> – x<\/em> -4.<\/p>\n\n\n\n

(ii) \u221a2 , 1\/3
<\/strong>Let the polynomial be ax<\/em>2<\/sup> + bx<\/em> + c<\/em>, and its zeroes be \u03b1 and \u00df
\u03b1 + \u00df = \u221a2 = 3\u221a2\/3 = –b<\/em>\/a<\/em>
\u03b1\u00df = 1\/3 = c<\/em>\/a<\/em>
If a<\/em> = 3, then b<\/em> = -3\u221a2, c<\/em> = 1
Therefore, the quadratic polynomial is 3x<\/em>2<\/sup> -3\u221a2x<\/em> +1.<\/p>\n\n\n\n

(iii) 0, \u221a5
<\/strong>Let the polynomial be ax<\/em>2<\/sup> + bx<\/em> + c<\/em>, and its zeroes be \u03b1 and \u00df
\u03b1 + \u00df = 0 = 0\/1 = –b<\/em>\/a<\/em>
\u03b1\u00df = \u221a5 = \u221a5\/1 = c<\/em>\/a<\/em>
If a<\/em> = 1, then b<\/em> = 0, c<\/em> = \u221a5
Therefore, the quadratic polynomial is x<\/em>2<\/sup> + \u221a5.<\/p>\n\n\n\n

(iv) 1, 1
<\/strong>Let the polynomial be ax<\/em>2<\/sup> + bx<\/em> + c<\/em>, and its zeroes be \u03b1 and \u00df
\u03b1 + \u00df = 1 = 1\/1 = –b<\/em>\/a<\/em>
\u03b1\u00df = 1 = 1\/1 = c<\/em>\/a<\/em>
If a<\/em> = 1, then b<\/em> = -1, c<\/em> = 1
Therefore, the quadratic polynomial is x<\/em>2<\/sup> – x<\/em> +1.<\/p>\n\n\n\n

(v) -1\/4 ,1\/4
<\/strong>Let the polynomial be ax<\/em>2<\/sup> + bx<\/em> + c<\/em>, and its zeroes be \u03b1 and \u00df
\u03b1 + \u00df = -1\/4 = –b<\/em>\/a<\/em>
\u03b1\u00df = 1\/4 = c<\/em>\/a<\/em>
If a<\/em> = 4, then b<\/em> = 1, c<\/em> = 1
Therefore, the quadratic polynomial is 4x<\/em>2<\/sup> + x<\/em>+1<\/p>\n\n\n\n

.(vi) 4,1
Let the polynomial be ax<\/em>2<\/sup> + bx<\/em> + c<\/em>, and its zeroes be \u03b1 and \u00df
\u03b1 + \u00df = 4 = 4\/1 = –b<\/em>\/a<\/em>
\u03b1\u00df = 1 = 1\/1 = c<\/em>\/a<\/em>
If a<\/em> = 1, then b<\/em> = -4, c<\/em> = 1
Therefore, the quadratic polynomial is x<\/em>2<\/sup> – 4x<\/em>+1<\/p>\n\n\n\n

<\/p>

Exercise 2.3<\/h4>

<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 2, class 10 Mathematics Chapter 2 solutions<\/p>\n\n\n\n

1. Divide the polynomial p<\/em>(x<\/em>) by the polynomial g<\/em>(x<\/em>) and find the quotient and remainder in each of the following:<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) p<\/em>(x<\/em>) = x<\/em>3<\/sup> \u2013 3x<\/em>2<\/sup> + 5x<\/em> \u2013 3, g<\/em>(x<\/em>) = x<\/em>2<\/sup> \u2013 2<\/strong><\/p>\n\n\n\n

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

Quotient = x<\/em>-3 and remainder 7x<\/em> – 9<\/p>\n\n\n\n

(ii) p<\/em>(x<\/em>) = x<\/em>4<\/sup> \u2013 3x<\/em>2<\/sup> + 4x + 5, g<\/em>(x<\/em>) = x<\/em>2<\/sup> + 1 \u2013 x<\/em><\/strong><\/p>\n\n\n\n

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

Quotient = x<\/em>2<\/sup> + x <\/em>– 3 and remainder 8<\/p>\n\n\n\n

(iii) p<\/em>(x<\/em>) = x<\/em>4<\/sup> \u2013 5x<\/em> + 6, g<\/em>(x<\/em>) = 2 \u2013 x<\/em>2<\/sup><\/strong><\/p>\n\n\n\n

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

Quotient = –x<\/em>2<\/sup> -2 and remainder -5x<\/em> +10<\/p>\n\n\n\n

2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) t<\/em>2<\/sup> \u2013 3,  2t<\/em>4<\/sup> + 3t<\/em>3<\/sup> \u2013 2t<\/em>2<\/sup> \u2013 9t<\/em> \u2013 12<\/strong><\/p>\n\n\n\n

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

t<\/em>2<\/sup> \u2013 3 exactly divides  2t<\/em>4<\/sup> + 3t<\/em>3<\/sup> \u2013 2t<\/em>2<\/sup> \u2013 9t<\/em> \u2013 12 leaving no remainder. Hence, it is a factor of  2t<\/em>4<\/sup> + 3t<\/em>3<\/sup> \u2013 2t<\/em>2<\/sup> \u2013 9t<\/em> \u2013 12.<\/p>\n\n\n\n

(ii) x<\/em>2<\/sup> + 3x<\/em> + 1, 3x<\/em>4<\/sup> + 5x<\/em>3<\/sup> \u2013 7x<\/em>2<\/sup> + 2x<\/em> + 2<\/strong><\/p>\n\n\n\n

\"NCERT<\/a><\/figure>\n\n\n\n

x<\/em>2<\/sup> + 3x<\/em> + 1 exactly divides 3x<\/em>4<\/sup> + 5x<\/em>3<\/sup> \u2013 7x<\/em>2<\/sup> + 2x<\/em> + 2 leaving no remainder. Hence, it is factor of 3x<\/em>4<\/sup> + 5x<\/em>3<\/sup> \u2013 7x<\/em>2<\/sup> + 2x<\/em> + 2.<\/p>\n\n\n\n

(iii) x<\/em>3<\/sup> \u2013 3x<\/em> + 1, x<\/em>5<\/sup> \u2013 4x<\/em>3<\/sup> + x<\/em>2<\/sup> + 3x<\/em> + 1<\/strong><\/p>\n\n\n\n

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

x<\/em>3<\/sup> \u2013 3x<\/em> + 1 didn’t divides exactly x<\/em>5<\/sup> \u2013 4x<\/em>3<\/sup> + x<\/em>2<\/sup> + 3x<\/em> + 1 and leaves 2 as remainder. Hence, it not a factor of x<\/em>5<\/sup> \u2013 4x<\/em>3<\/sup> + x<\/em>2<\/sup> + 3x<\/em> + 1.<\/p>\n\n\n\n

3. Obtain all other zeroes of 3x<\/em>4<\/sup> + 6x<\/em>3<\/sup> \u2013 2x<\/em>2<\/sup> \u2013 10x<\/em> \u2013 5, if two of its zeroes are \u221a(5\/3) and – \u221a(5\/3).<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

p<\/em>(x<\/em>) = 3x<\/em>4<\/sup> + 6x<\/em>3<\/sup> \u2013 2x<\/em>2<\/sup> \u2013 10x<\/em> \u2013 5
Since the two zeroes are \u221a(5\/3) and – \u221a(5\/3).<\/p>\n\n\n\n

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

We factorize x<\/em>2<\/sup> + 2x <\/em>+ 1
= (x <\/em>+ 1)2<\/sup>
Therefore, its zero is given by x<\/em> + 1 = 0
x<\/em> = -1
As it has the term (x <\/em>+ 1)2<\/sup> , therefore, there will be 2 zeroes at x<\/em> = – 1.
Hence, the zeroes of the given polynomial are \u221a(5\/3) and – \u221a(5\/3), – 1 and – 1.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 2, class 10 Mathematics Chapter 2 solutions<\/p>\n\n\n\n

4.  On dividing x<\/em>3<\/sup> – 3x<\/em>2<\/sup> + x<\/em> + 2 by a polynomial g<\/em>(x<\/em>), the quotient and remainder were x<\/em> – 2 and -2x<\/em> + 4, respectively. Find g<\/em>(x<\/em>).<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Here in the given question,
Dividend = x<\/em>3<\/sup> – 3x<\/em>2<\/sup> + x<\/em> + 2
Quotient = x<\/em> – 2
Remainder = -2x<\/em> + 4
Divisor = g<\/em>(x<\/em>)
We know that,
Dividend = Quotient \u00d7 Divisor + Remainder
\u21d2 x<\/em>3<\/sup> – 3x<\/em>2<\/sup> + x<\/em> + 2 = (x<\/em> – 2) \u00d7 g<\/em>(x<\/em>) + (-2x<\/em> + 4)\u21d2 x<\/em>3<\/sup> – 3x<\/em>2<\/sup> + x<\/em> + 2 – (-2x<\/em> + 4) = (x<\/em> – 2) \u00d7 g<\/em>(x<\/em>)
\u21d2 x<\/em>3<\/sup> – 3x<\/em>2<\/sup> + 3x<\/em> – 2 = (x<\/em> – 2) \u00d7 g<\/em>(x<\/em>)
\u21d2 g<\/em>(x<\/em>) =  (x<\/em>3<\/sup> – 3x<\/em>2<\/sup> + 3x<\/em> – 2)\/(x<\/em> – 2)<\/p>\n\n\n\n

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

\u2234 g<\/em>(x<\/em>) = (x<\/em>2<\/sup> – x<\/em> + 1)<\/p>\n\n\n\n

5.Give examples of polynomial p<\/em>(x<\/em>), g<\/em>(x<\/em>), q<\/em>(x<\/em>) and r<\/em>(x<\/em>), which satisfy the division algorithm and<\/strong>
(i) deg p<\/em>(x<\/em>) = deg q<\/em>(x<\/em>)
(ii) deg q<\/em>(x<\/em>) = deg r<\/em>(x<\/em>)
(iii) deg r<\/em>(x<\/em>) = 0<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) Let us assume the division of 6x<\/em>2<\/sup> + 2x<\/em> + 2 by 2
Here, p<\/em>(x<\/em>) = 6x<\/em>2<\/sup> + 2x<\/em> + 2
g<\/em>(x<\/em>) = 2
q<\/em>(x<\/em>) = 3x<\/em>2<\/sup> + x<\/em> + 1
r<\/em>(x<\/em>) = 0
Degree of p<\/em>(x<\/em>) and q<\/em>(x<\/em>) is same i.e. 2.
Checking for division algorithm,
p<\/em>(x<\/em>) = g<\/em>(x<\/em>) \u00d7 q<\/em>(x<\/em>) + r<\/em>(x<\/em>)
Or, 6x<\/em>2<\/sup> + 2x<\/em> + 2 = 2x<\/em> (3x<\/em>2<\/sup> + x <\/em>+ 1)
Hence, division algorithm is satisfied.<\/p>\n\n\n\n

(ii) Let us assume the division of x<\/em>3<\/sup>+ x<\/em> by x<\/em>2<\/sup>,
Here, p<\/em>(x<\/em>) = x<\/em>3<\/sup> + x<\/em>
g<\/em>(x<\/em>) = x<\/em>2<\/sup>
q<\/em>(x<\/em>) = x<\/em> and r<\/em>(x<\/em>) = x<\/em>
Clearly, the degree of q<\/em>(x<\/em>) and r<\/em>(x<\/em>) is the same i.e., 1.
Checking for division algorithm,
p<\/em>(x<\/em>) = g<\/em>(x<\/em>) \u00d7 q<\/em>(x<\/em>) + r<\/em>(x<\/em>)
x<\/em>3<\/sup> + x<\/em> = (x<\/em>2<\/sup> ) \u00d7 x<\/em> + x<\/em>
x<\/em>3<\/sup> + x<\/em> = x<\/em>3<\/sup> + x<\/em>
Thus, the division algorithm is satisfied.<\/p>\n\n\n\n

(iii) Let us assume the division of x<\/em>3<\/sup>+ 1 by x<\/em>2<\/sup>.
Here, p<\/em>(x<\/em>) = x<\/em>3<\/sup> + 1
g(x) = x2<\/sup>
q<\/em>(x<\/em>) = x<\/em> and r<\/em>(x<\/em>) = 1
Clearly, the degree of r<\/em>(x<\/em>) is 0.
Checking for division algorithm,
p<\/em>(x<\/em>) = g<\/em>(x<\/em>) \u00d7 q<\/em>(x<\/em>) + r<\/em>(x<\/em>)
x<\/em>3<\/sup> + 1 = (x<\/em>2<\/sup> ) \u00d7 x <\/em>+ 1
x<\/em>3<\/sup> + 1 = x<\/em>3<\/sup> + 1
Thus, the division algorithm is satisfied.<\/p>\n\n\n\n

<\/p>

Exercise 2.4 (Optional)<\/h5>

1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:<\/strong>
(i) 2x<\/em>3<\/sup> + x<\/em>2<\/sup> – 5x <\/em>+ 2; 1\/2, 1, -2
(ii) x<\/em>3<\/sup> – 4x<\/em>2<\/sup> + 5x –<\/em> 2; 2, 1, 1

<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) p<\/em>(x<\/em>) = 2x<\/em>3<\/sup> + x<\/em>2<\/sup> – 5x <\/em>+ 2
Now for zeroes, putting the given value in x.<\/p>\n\n\n\n

p<\/em>(1\/2<\/em>) = 2(1\/2)3<\/sup> + (1\/2)<\/em>2<\/sup> – 5(1\/2)+ 2
= (2\u00d71\/8) + 1\/4 – 5\/2 + 2
= 1\/4 + 1\/4 – 5\/2 + 2
= 1\/2 – 5\/2 + 2 = 0<\/p>\n\n\n\n

p<\/em>(1<\/em>) = 2(1)3<\/sup> + (1)<\/em>2<\/sup> – 5(1)+ 2
= (2\u00d71) + 1 – 5 + 2
= 2 + 1 – 5 + 2 = 0<\/p>\n\n\n\n

p<\/em>(-2<\/em>) = 2(-2)3<\/sup> + (-2)2<\/sup> – 5(-2)+ 2
= (2 \u00d7 -8) + 4 + 10 + 2
= -16 + 16 = 0<\/p>\n\n\n\n

Thus, 1\/2, 1 and -2 are the zeroes of the given polynomial.<\/p>\n\n\n\n

Comparing the given polynomial with ax<\/em>3<\/sup> + bx<\/em>2<\/sup> + cx <\/em>+ d, we get a=2, b=1, c=-5, d=2
Also, \u03b1=1\/2, \u03b2=1 and \u03b3=-2
Now,
-b\/a = \u03b1+\u03b2+\u03b3
\u21d2 1\/2 = 1\/2 + 1 – 2
\u21d2 1\/2 = 1\/2

c\/a = <\/p>\n\n\n\n

\u03b1\u03b2+\u03b2\u03b3+\u03b3\u03b1
\u21d2 -5\/2 = (1\/2 \u00d7 1) + (1 \u00d7 -2) + (-2 \u00d7 1\/2)
\u21d2 -5\/2 = 1\/2 – 2 – 1
\u21d2 -5\/2 = -5\/2<\/p>\n\n\n\n

-d\/a = \u03b1\u03b2\u03b3
\u21d2 -2\/2 = (1\/2 \u00d7 1 \u00d7 -2)
\u21d2 -1 = 1<\/p>\n\n\n\n

Thus, the relationship between zeroes and the coefficients are verified.<\/p>\n\n\n\n

(ii)  p<\/em>(x<\/em>) = x<\/em>3<\/sup> – 4x<\/em>2<\/sup> + 5x –<\/em> 2
Now for zeroes, putting the given value in x.<\/p>\n\n\n\n

p<\/em>(2<\/em>) = 23<\/sup> – 4(2)<\/em>2<\/sup> + 5(2)- 2
= 8 – 16 + 10 – 2
= 0<\/p>\n\n\n\n

p<\/em>(1<\/em>) = 13<\/sup> – 4(1)<\/em>2<\/sup> + 5(1)- 2
= 1 – 4 + 5 – 2
= 0<\/p>\n\n\n\n

p<\/em>(1<\/em>) = 13<\/sup> – 4(1)<\/em>2<\/sup> + 5(1)- 2
= 1 – 4 + 5 – 2
= 0<\/p>\n\n\n\n

Thus, 2, 1 and 1 are the zeroes of the given polynomial.<\/p>\n\n\n\n

Comparing the given polynomial with ax<\/em>3<\/sup> + bx<\/em>2<\/sup> + cx <\/em>+ d, we get a=1, b=-4, c=5, d=-2
Also, \u03b1=2, \u03b2=1 and \u03b3=1
Now,
-b\/a = \u03b1+\u03b2+\u03b3
\u21d2 4\/1 = 2 + 1 + 1
\u21d2 4 = 4

c\/a = <\/p>\n\n\n\n

\u03b1\u03b2+\u03b2\u03b3+\u03b3\u03b1
\u21d2 5\/1 = (2 \u00d7 1) + (1 \u00d7 1) + (1 \u00d7 2)
\u21d2 5 = 2 + 1 + 2
\u21d2 5 = 5<\/p>\n\n\n\n

-d\/a = \u03b1\u03b2\u03b3
\u21d2 2\/1 = (2 \u00d7 1 \u00d7 1)
\u21d2 2 = 2<\/p>\n\n\n\n

Thus, the relationship between zeroes and the coefficients are verified.

NCERT 10th Mathematics Chapter 2, class 10 Mathematics Chapter 2 solutions<\/p>\n\n\n\n

2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, \u20137, \u201314 respectively.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Let the polynomial be ax<\/em>3<\/sup> + bx<\/em>2 <\/sup>+ cx + d <\/em>and the zeroes be \u03b1, \u03b2 and \u03b3
Then, \u03b1 + \u03b2 + \u03b3 = -(-2)\/1 = 2 = -b\/a
\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = -7 = -7\/1 = c\/a
\u03b1\u03b2\u03b3 = -14 = -14\/1 = -d\/a<\/p>\n\n\n\n

\u2234 a = 1, b = -2, c = -7 and d = 14
So, one cubic polynomial which satisfy the given conditions will be x<\/em>3<\/sup> – 2x<\/em>2  <\/sup>– 7x + 14<\/em><\/p>\n\n\n\n

3. If the zeroes of the polynomial x<\/em>3<\/sup> \u2013 3x<\/em>2<\/sup> + x<\/em> + 1 are a\u2013b, a, a+b, find a and b.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Since, (a – b), a, (a + b) are the zeroes of the polynomial x<\/em>3<\/sup> \u2013 3x<\/em>2<\/sup> + x<\/em> + 1.
Therefore, sum of the zeroes = (a – b) + a + (a + b) = -(-3)\/1 = 3<\/p>\n\n\n\n

\u21d2 3a = 3 \u21d2 a =1<\/p>\n\n\n\n

\u2234 Sum of the products of is zeroes taken two at a time = a(a – b) + a(a + b) + (a + b) (a – b) =1\/1 = 1
a2<\/sup> – ab + a2<\/sup> + ab + a2<\/sup> – b2 <\/sup>= 1
\u21d2 3a2<\/sup> – b2<\/sup> =1<\/p>\n\n\n\n

Putting the value of a,<\/p>\n\n\n\n

\u21d2 3(1)2<\/sup> – b2<\/sup> = 1
\u21d2 3 – b2<\/sup> = 1
\u21d2 b2<\/sup> = 2
\u21d2 b = \u00b1\u221a2
Hence, a = 1 and b = \u00b1\u221a2

4. If two zeroes of the polynomial x4<\/sup> \u2013 6x3<\/sup> \u2013 26x2<\/sup> + 138x \u2013 35 are  2\u00b1\u221a3,  find other zeroes.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

2+\u221a3 and 2-\u221a3 are two zeroes of the polynomial p(x) = x4<\/sup> \u2013 6x3<\/sup> \u2013 26x2<\/sup> + 138x \u2013 35.
Let x = 2\u00b1\u221a3
So, x-2 = \u00b1\u221a3
On squaring, we get x2<\/sup> – 4x + 4 = 3,
\u21d2 x2<\/sup> – 4x + 1= 0<\/p>\n\n\n\n

Now, dividing p(x) by x2<\/sup> – 4x + 1<\/p>\n\n\n\n

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

\u2234 p(x) = x4<\/sup> – 6x3<\/sup> – 26x2<\/sup> + 138x – 35
= (x2<\/sup> – 4x + 1) (x2<\/sup> – 2x – 35)
= (x2<\/sup> – 4x + 1) (x2<\/sup> – 7x + 5x – 35)
= (x2<\/sup> – 4x + 1) [x(x – 7) + 5 (x – 7)]
= (x2<\/sup> – 4x + 1) (x + 5) (x – 7)<\/p>\n\n\n\n

\u2234 (x + 5) and (x – 7) are other factors of p(x).
\u2234 – 5 and 7 are other zeroes of the given polynomial.<\/p>\n\n\n\n

5. If the polynomial x4<\/sup> \u2013 6x3<\/sup> + 16x2<\/sup> \u2013 25x + 10 is divided by another polynomial x2<\/sup> \u2013 2x + k, the remainder comes out to be x + a, find k and a.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

On dividing x4<\/sup> \u2013 6x3<\/sup> + 16x2<\/sup> \u2013 25x + 10 by x2<\/sup> \u2013 2x + k<\/p>\n\n\n\n

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

\u2234 Remainder = (2k – 9)x – (8 – k)k + 10 <\/p>\n\n\n\n

But the remainder is given as x+ a. <\/p>\n\n\n\n

On comparing their coefficients,<\/p>\n\n\n\n

2k – 9 = 1
\u21d2 k = 10
\u21d2 k = 5 and,
-(8-k)k + 10 = a
\u21d2 a = -(8 – 5)5 + 10 =- 15 + 10 = -5
Hence, k = 5 and a = -5<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 2, class 10 Mathematics Chapter 2 solutions<\/p>\n\n\n\n

NCERT Solutions for Class 10th Mathematics: Chapter 2: Download PDF<\/strong><\/h2>\n\n\n\n

NCERT Solutions for Class 10th Mathematics: Chapter 2 Polynomials<\/p>\n\n\n\n

Download PDF<\/strong>: NCERT Solutions for Class 10th Mathematics: Chapter 2 Polynomials PDF<\/a><\/p>\n\n\n\n

Chapterwise NCERT Solutions for Class 10 Maths<\/strong>:<\/h4>\n\n\n\n