{"id":623944,"date":"2023-09-01T01:30:56","date_gmt":"2023-09-01T01:30:56","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=623944"},"modified":"2023-09-04T09:44:52","modified_gmt":"2023-09-04T09:44:52","slug":"kc-sinha-exercise-2-3-mathematics-solution-class-10-chapter-2-polynomials","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-2-3-mathematics-solution-class-10-chapter-2-polynomials\/","title":{"rendered":"KC Sinha: Exercise 2.3 – Mathematics Solution Class 10 Chapter 2 Polynomials"},"content":{"rendered":"\n\n\n\n\n

Question 1 <\/h4>\n\n\n\n

Divide 2x3<\/sup> + 3x + 1 by x + 2 and find the quotient and the reminder. Is q(x) a factor of 2x3<\/sup> + 3x + 1 ?<\/strong><\/p>\n\n\n\n

Sol :
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process<\/p>\n\n\n\n

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

Quotient = 2x2<\/sup> \u2013 4x + 11
Remainder = \u2013 21
No, 2x2<\/sup> \u2013 4x + 11 is not a factor of 2x3<\/sup> + 3x + 1 because remainder \u2260 0<\/p>\n\n\n\n

\n\n\n\n

Question 2 <\/h4>\n\n\n\n

Divide 3x3<\/sup> + x2<\/sup> + 2x + 5 by 1 + 2x + x2<\/sup> and find the quotient and the remainder. Is 1 + 2x + x2<\/sup> a factor of 3x3<\/sup> + x2<\/sup> + 2x + 5<\/strong><\/p>\n\n\n\n

Sol :
Dividend = 3x3<\/sup> + x2<\/sup> + 2x + 5
Divisor = x2<\/sup> + 2x + 1
Now, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process<\/p>\n\n\n\n

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

Quotient = 3x \u2013 5
Remainder = 9x + 10
No, x2<\/sup> + 2x + 1 is not a factor of 3x3<\/sup> + x2<\/sup> + 2x + 5 because remainder \u2260 0<\/p>\n\n\n\n

\n\n\n\n

Question 3 A <\/h4>\n\n\n\n

Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :<\/strong><\/p>\n\n\n\n

p(x) = x3<\/sup> \u2013 3x2<\/sup> + 4x + 2 , g(x) = x \u2013 1<\/strong>
Sol :
p(x) = x3<\/sup> \u2013 3x2<\/sup> + 4x + 2,g(x) = x \u2013 1
Dividend = x3<\/sup> \u2013 3x2<\/sup> + 4x + 2
Divisor = x \u2013 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process<\/p>\n\n\n\n

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

Quotient [q(x)]= x2<\/sup> \u2013 2x + 2
Remainder [r(x)]= 4<\/p>\n\n\n\n

\n\n\n\n

Question 3 B <\/h4>\n\n\n\n

Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :<\/strong><\/p>\n\n\n\n

p(x) = 4x3<\/sup> \u2013 3x2<\/sup> + 2x + 3 , g(x) = x + 4<\/strong>
Sol :<\/p>\n\n\n\n

p(x) = 4x3<\/sup> \u2013 3x2<\/sup> + 2x + 3,g(x) = x + 4
Dividend = 4x3<\/sup> \u2013 3x2<\/sup> + 2x + 3
Divisor = x + 4
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process<\/p>\n\n\n\n

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

Quotient = 4x2<\/sup> \u2013 13x + 54
Remainder = \u2013 213<\/p>\n\n\n\n

\n\n\n\n

Question 3 C <\/h4>\n\n\n\n

Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :<\/strong><\/p>\n\n\n\n

p(x) = 2x4<\/sup> + 3x3<\/sup> + 4x2<\/sup> + 19x + 45, g(x) = x \u2013 2<\/strong>
Sol :
p(x) = 2x4<\/sup> + 3x3<\/sup> + 4x2<\/sup> + 19x + 45, g(x) = x \u2013 2
Dividend = 2x4<\/sup> + 3x3<\/sup> + 4x2<\/sup> + 19x + 45
Divisor = x \u2013 2
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,<\/p>\n\n\n\n

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

q(x) = 2x3<\/sup>+7x2<\/sup>+18x+55
r(x) = 155<\/p>\n\n\n\n

\n\n\n\n

Question 3 D <\/h4>\n\n\n\n

Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :<\/strong><\/p>\n\n\n\n

p(x) = x4<\/sup> + 2x3<\/sup> \u2013 3x2<\/sup> + x \u2013 1, g(x) = x \u2013 2<\/strong>
Sol :
p(x) = x4<\/sup> + 2x3<\/sup> \u2013 3x2<\/sup> + x \u2013 1, g(x) = x \u2013 2
Dividend = x4<\/sup> + 2x3<\/sup> \u2013 3x2<\/sup> + x \u2013 1
Divisor = x \u2013 2
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,<\/p>\n\n\n\n

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

q(x) = x3<\/sup>+4x2<\/sup>+5x+11
r(x) = 21<\/p>\n\n\n\n

\n\n\n\n

Question 3 E <\/h4>\n\n\n\n

Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :<\/strong><\/p>\n\n\n\n

p(x) = x3<\/sup> \u2013 3x2<\/sup>\u2013 x + 3, g(x) = x2<\/sup> \u2013 4x + 3<\/strong>
Sol :
p(x) = x3<\/sup> \u2013 3x2<\/sup>\u2013 x + 3, g(x) = x2<\/sup> \u2013 4x + 3
Dividend = x3<\/sup> \u2013 3x2<\/sup>\u2013 x + 3
Divisor = x2<\/sup> \u2013 4x + 3
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,<\/p>\n\n\n\n

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

q(x)= x + 1
r(x) = 0<\/p>\n\n\n\n

\n\n\n\n

Question 3 F <\/h4>\n\n\n\n

Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :<\/strong><\/p>\n\n\n\n

p(x) = x6<\/sup> + x4<\/sup> + x3<\/sup> + x2<\/sup> + 2x + 2, g(x) = x3<\/sup> + 1<\/strong>
Sol :
p(x) = x6<\/sup> + x4<\/sup> + x3<\/sup> + x2<\/sup> + 2x + 2, g(x) = x3<\/sup> + 1
Dividend = x6<\/sup> + x4<\/sup> + x3<\/sup> + x2<\/sup> + 2x + 2
Divisor = x3<\/sup> + 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,<\/p>\n\n\n\n

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

q(x) = x3<\/sup> + x
r(x) = x2<\/sup> + x + 2<\/p>\n\n\n\n

\n\n\n\n

Question 3 G <\/h4>\n\n\n\n

Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :<\/strong><\/p>\n\n\n\n

p(x) = x6<\/sup> + 3x2<\/sup> + 10 and g(x) = x3<\/sup> + 1<\/strong>
Sol :
p(x) = x6<\/sup> + 3x2<\/sup> + 10 , g(x) = x3<\/sup> + 1
Dividend = x6<\/sup> + 3x2<\/sup> + 10
Divisor = x3<\/sup> + 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process we get,<\/p>\n\n\n\n

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

q(x) = x3<\/sup> \u2013 1
r(x) = 3x2<\/sup> + 11<\/p>\n\n\n\n

\n\n\n\n

Question 3 H <\/h4>\n\n\n\n

Divide the polynomial p(x) by the polynomial g(x) and find the quotient q(x) and remainder r(x) in each case :<\/strong><\/p>\n\n\n\n

p(x) = x4<\/sup> + 1, g(x) = x + 1<\/strong>
Sol :
p(x) = x4<\/sup> + 1, g(x) = x + 1
Dividend = x4<\/sup> + 1,
Divisor = x + 1<\/p>\n\n\n\n

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

q(x) = x3<\/sup> \u2013 x2<\/sup> + x \u2013 1
r(x) = 0<\/p>\n\n\n\n

\n\n\n\n

Question 4 <\/h4>\n\n\n\n

By division process, find the value of k for which x \u2013 1 is a factor of x3<\/sup> \u2013 6x2<\/sup> + 11x + k.<\/strong><\/p>\n\n\n\n

Sol :
On dividing x3<\/sup> \u2013 6x2<\/sup> + 11x + k by x \u2013 1 we get,<\/p>\n\n\n\n

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

Since x \u2013 1 is a factor of x3<\/sup> \u2013 6x2<\/sup> + 11x + k,
This means x \u2013 1 divides the given polynomial completely.
\u2192 6 x + k = 0
\u2192 k = \u2013 6x<\/p>\n\n\n\n

\n\n\n\n

Question 5 <\/h4>\n\n\n\n

By division process, find the value of c for which 2x + 1 is a factor of 4x4<\/sup> \u2013 3x2<\/sup> + 3x + c.<\/strong><\/p>\n\n\n\n

Sol :
On dividing 4x4<\/sup> \u2013 3x2<\/sup> + 3x + c by 2x + 1 we get,<\/p>\n\n\n\n

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

Since 2x + 1 is a factor of 4x4<\/sup> \u2013 3x2<\/sup> + 3x + c,
This means 2x + 1 divides the given polynomial completely,
\u2192 4x + c = o
\u2192 c = \u2013 4x<\/p>\n\n\n\n

\n\n\n\n

Question 6 A <\/h4>\n\n\n\n

Apply Division Algorithm to find the quotient q(x) and remainder r(x) on dividing p(x) by g(x) as given below:<\/strong><\/p>\n\n\n\n

p(x) = 2x2<\/sup> + 3x + 1, g(x) = x + 2<\/strong>
Sol :
p(x) = 2x2<\/sup> + 3x + 1, g(x) = x + 2
Dividend = 2x2<\/sup> + 3x + 1,
Divisor = x + 2
Apply the division algorithm we get,<\/p>\n\n\n\n

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

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

\n\n\n\n

Question 6 B <\/h4>\n\n\n\n

Apply Division Algorithm to find the quotient q(x) and remainder r(x) on dividing p(x) by g(x) as given below:<\/strong><\/p>\n\n\n\n

p(x) = x3<\/sup> \u2013 3x2<\/sup> + 5x \u2013 3, g(x) = x2<\/sup> \u2013 2<\/strong>
Sol :
p(x) = x3<\/sup> \u2013 3x2<\/sup> + 5x \u2013 3, g(x) = x2<\/sup> \u2013 2
Dividend = x3<\/sup> \u2013 3x2<\/sup> + 5x \u2013 3,
Divisor = x2<\/sup> \u2013 2
On applying division algorithm, we get,<\/p>\n\n\n\n

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

q(x) = x \u2013 3
r(x) = 7x \u2013 9<\/p>\n\n\n\n

\n\n\n\n

Question 6 C <\/h4>\n\n\n\n

Apply Division Algorithm to find the quotient q(x) and remainder r(x) on dividing p(x) by g(x) as given below:<\/strong><\/p>\n\n\n\n

p(x) = x4<\/sup> \u2013 1 , g(x) = x + 1<\/strong>
Sol :
p(x) = x4<\/sup> \u2013 1,g(x) = x + 1
Dividend = x4<\/sup> \u2013 1
Divisor = x + 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process<\/p>\n\n\n\n

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

Quotient = x3<\/sup> \u2013 x2<\/sup> + x \u2013 1
Remainder = 0<\/p>\n\n\n\n

\n\n\n\n

Question 6 D <\/h4>\n\n\n\n

Apply Division Algorithm to find the quotient q(x) and remainder r(x) on dividing p(x) by g(x) as given below:<\/strong><\/p>\n\n\n\n

p(x) = x3<\/sup> \u2013 3x2<\/sup> + 4x + 2 , g(x) = x \u2013 1<\/strong>
Sol :
p(x) = x3<\/sup> \u2013 3x2<\/sup> + 4x + 2,g(x) = x \u2013 1
Dividend = x3<\/sup> \u2013 3x2<\/sup> + 4x + 2
Divisor = x \u2013 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process<\/p>\n\n\n\n

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

Quotient = x \u2013 5
Remainder = 13x + 7<\/p>\n\n\n\n

\n\n\n\n

Question 6 E <\/h4>\n\n\n\n

Apply Division Algorithm to find the quotient q(x) and remainder r(x) on dividing p(x) by g(x) as given below:<\/strong><\/p>\n\n\n\n

p(x) = x3<\/sup> \u2013 6x2<\/sup> + 11x \u2013 6 , g(x) = x2<\/sup> \u2013 5x + 6<\/strong>
Sol :
p(x) = x3<\/sup> \u2013 6x2<\/sup> + 11x \u2013 6,g(x) = x2<\/sup> \u2013 5x + 6
Dividend = x3<\/sup> \u2013 6x2<\/sup> + 11x \u2013 6
Divisor = x2<\/sup> \u2013 5x + 6
Here, dividend and divisor both are in the standard form.<\/p>\n\n\n\n

Now, on dividing p(x) by g(x) we get the following division process<\/p>\n\n\n\n

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

Quotient = x \u2013 1<\/p>\n\n\n\n

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

\n\n\n\n

Question 6 F <\/h4>\n\n\n\n

Apply Division Algorithm to find the quotient q(x) and remainder r(x) on dividing p(x) by g(x) as given below:<\/strong><\/p>\n\n\n\n

p(x) = 6x3<\/sup> + 13x2<\/sup> + x \u2013 2 , g(x) = 2x + 1<\/strong>
Sol :
p(x) = 6x3<\/sup> + 13x2<\/sup> + x \u2013 2,g(x) = 2x + 1
Dividend = 6x3<\/sup> + 13x2<\/sup> + x \u2013 2
Divisor = 2x + 1
Here, dividend and divisor both are in the standard form.
Now, on dividing p(x) by g(x) we get the following division process<\/p>\n\n\n\n

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

Quotient = 3x2<\/sup> + 5x \u2013 2
Remainder = 0<\/p>\n\n\n\n

\n\n\n\n

Question 7 A <\/h4>\n\n\n\n

Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial:<\/strong><\/p>\n\n\n\n

x \u2013 2, x3<\/sup> + 3x3<\/sup> \u2013 12x + 4<\/strong>
Sol :
Let us divide x3<\/sup> + 3x2<\/sup> \u2013 12x + 4 by x \u2013 2
The division process is<\/p>\n\n\n\n

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

Here, the remainder is 0, therefore x \u2013 2 is a factor of x3<\/sup> + 3x2<\/sup> \u2013 12x + 4<\/p>\n\n\n\n

\n\n\n\n

Question 7 B <\/h4>\n\n\n\n

Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial:<\/strong><\/p>\n\n\n\n

x2<\/sup> + 3x + 1,3x4<\/sup> + 5x3<\/sup> \u2013 7x2<\/sup> + 2x + 2<\/strong>
Sol :
Let us divide 3x4<\/sup> + 5x3<\/sup> \u2013 7x2<\/sup> + 2x + 2 by x2<\/sup> + 3x + 1
The division process is<\/p>\n\n\n\n

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

Here, the remainder is 0, therefore x2<\/sup> + 3x + 1 is a factor of 3x4<\/sup> + 5x3<\/sup> \u2013 7x2<\/sup> + 2x + 2<\/p>\n\n\n\n

\n\n\n\n

Question 7 C <\/h4>\n\n\n\n

Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial:<\/strong><\/p>\n\n\n\n

x2<\/sup> \u2013 3x + 4,2x4<\/sup> \u2013 11x3<\/sup> + 29x2<\/sup> \u2013 30x + 29<\/strong>
Sol :
Let us divide 2x4<\/sup> \u2013 11x3<\/sup> + 29x2<\/sup> \u2013 30x + 29 by x2<\/sup> \u2013 3x + 4
The division process is<\/p>\n\n\n\n

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

Here, the remainder is 58x \u2013 115 , therefore x2<\/sup> \u2013 3x + 4 is not a factor of 2x4<\/sup> \u2013 11x3<\/sup> + 29x2<\/sup> \u2013 30x + 29<\/p>\n\n\n\n

\n\n\n\n

Question 7 D <\/h4>\n\n\n\n

Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial:<\/strong><\/p>\n\n\n\n

x2<\/sup> \u2013 4x + 3,x3<\/sup> \u2013 x3<\/sup> \u2013 3x4<\/sup> \u2013 x + 3<\/strong>
Sol :
Let us divide x3<\/sup> \u2013 3x2<\/sup> \u2013 x + 3 by x2<\/sup> \u2013 4x + 3
The division process is<\/p>\n\n\n\n

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

Here, the remainder is 0, therefore x2<\/sup> \u2013 4x + 3 is a factor of x3<\/sup> \u2013 3x2<\/sup> \u2013 x + 3<\/p>\n\n\n\n

\n\n\n\n

Question 7 E <\/h4>\n\n\n\n

Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial:<\/strong><\/p>\n\n\n\n

t \u2013 1, t3<\/sup> + t2<\/sup> \u2013 2t + 1<\/strong>
Sol :
Let us divide t3<\/sup> + t2<\/sup> \u2013 2t + 1 by t \u2013 1
The division process is<\/p>\n\n\n\n

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

Here, the remainder is 1, therefore t \u2013 1 is not a factor of t3<\/sup> + t2<\/sup> \u2013 2t + 1<\/p>\n\n\n\n

\n\n\n\n

Question 7 F <\/h4>\n\n\n\n

Applying the Division Algorithm, check whether the first polynomial is a factor of the second polynomial:<\/strong><\/p>\n\n\n\n

t2<\/sup> \u2013 5t + 6,t2<\/sup> + 11t \u2013 6<\/strong>
Sol :
Let us divide t2<\/sup> + 11t \u2013 6 by t2<\/sup> \u2013 5t + 6
The division process is<\/p>\n\n\n\n

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

Here, the remainder is 16t \u2013 12,
Therefore, t2<\/sup> \u2013 5t + 6 is not a factor of t2<\/sup> + 11t \u2013 6<\/p>\n\n\n\n

\n\n\n\n

Question 8 <\/h4>\n\n\n\n

Give examples of polynomials p(x), g(x), q(x) and r(x) satisfying the Division Algorithm<\/strong><\/p>\n\n\n\n

p(x) = g(x).q(x) + r(x),deg r(x)<deg g(x)<\/strong>
And also satisfying<\/strong>
(i) deg p(x) = deg q(x) + 1<\/strong>
(ii)deg q(x) = 1<\/strong>
(iii) deg q(x) = deg r(x) + 1<\/strong>
Sol :
(i)Let p(x) = 12x2<\/sup> + 8x + 25, g(x) = 4,
q(x) = 3x2<\/sup> + 2x + 6 , r(x) = 0
Here, degree p(x) = degree q(x) = 2
Now, g(x).q(x) + r(x) = (3x2<\/sup> + 2x + 6)\u00d74 + 1
=<\/strong> 12x2<\/sup> + 8x + 24 + 1
= 12x2<\/sup> + 8x + 25
(ii) Let p(x) = t3<\/sup> + t2<\/sup> \u2013 2t, g(x) = t2<\/sup> + 2t,
q(x) = t \u2013 1 , r(x) = 0
Here, degree q(x) = 1
Now, g(x).q(x) + r(x) = (t2<\/sup> + 2t)\u00d7(t \u2013 1) + 0
=<\/strong> t3<\/sup> \u2013 t2<\/sup> + 2t2<\/sup> \u2013 2t
= t3<\/sup> + t2<\/sup> \u2013 2t
(iii) Let p(x) = x3<\/sup> + x2<\/sup> + x + 1 , g(x) = x2<\/sup> \u2013 1,
q(x) = x + 1 , r(x) = 2x + 2
Here, degree q(x) = degree r(x) + 1 = 1
Now, g(x).q(x) + r(x) = (x2<\/sup> \u2013 1)\u00d7(x + 1) + 2x + 2
=<\/strong> x3<\/sup> + x2<\/sup> \u2013 x \u2013 1 + 2x + 2
= x3<\/sup> + x2<\/sup> + x + 1<\/p>\n\n\n\n

\n\n\n\n

Question 9 A <\/h4>\n\n\n\n

Find all the zeroes of the polynomial given below having given numbers as its zeroes.<\/strong><\/p>\n\n\n\n

x3<\/sup> \u2013 6x2<\/sup> + 11x \u2013 6;3<\/strong>
Sol :
Given zeroes is 3
So, (x \u2013 3) is the factor of x3<\/sup> \u2013 6x2<\/sup> + 11x \u2013 6
Let us divide x3<\/sup> \u2013 6x2<\/sup> + 11x \u2013 6 by x \u2013 3
The division process is<\/p>\n\n\n\n

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

Here, quotient = x2<\/sup> \u2013 3x + 2
= x2<\/sup> \u2013 2x \u2013 x + 2
= x(x \u2013 2) \u2013 1(x \u2013 2)
= (x \u2013 1)(x \u2013 2)
So, the zeroes are 1 and 2
Hence, all the zeroes of the given polynomial are 1, 2 and 3.<\/p>\n\n\n\n

\n\n\n\n

Question 9 B <\/h4>\n\n\n\n

Find all the zeroes of the polynomial given below having given numbers as its zeroes.<\/strong><\/p>\n\n\n\n

x4<\/sup> \u2013 8x3<\/sup> + 23x2<\/sup> \u2013 28x + 12;1,2<\/strong>
Sol :
Given zeroes are 1 and 2
So, (x \u2013 1)and (x \u2013 2) are the factors of x4<\/sup> \u2013 8x3<\/sup> + 23x2<\/sup> \u2013 28x + 12
\u27f9 (x \u2013 1)(x \u2013 2) = x2<\/sup> \u2013 3x + 2 is a factor of given polynomial.
Consequently, x2<\/sup> \u2013 3x + 2 is also a factor of the given polynomial.
Now, let us divide x4<\/sup> \u2013 8x3<\/sup> + 23x2<\/sup> \u2013 28x + 12 by x2<\/sup> \u2013 3x + 2
The division process is<\/p>\n\n\n\n

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

Here, quotient = x2<\/sup> \u2013 5x + 6
= x2<\/sup> \u2013 2x \u2013 3x + 6
= x(x \u2013 2) \u2013 3(x \u2013 2)
= (x \u2013 3)(x \u2013 2)
So, the zeroes are 3 and 2
Hence, all the zeroes of the given polynomial are 1, 2 ,2 and 3.<\/p>\n\n\n\n

\n\n\n\n

Question 9 C <\/h4>\n\n\n\n

Find all the zeroes of the polynomial given below having given numbers as its zeroes.<\/strong><\/p>\n\n\n\n

x3<\/sup> + 2x2<\/sup> \u2013 2; \u2013 2<\/strong>
Sol :
Given zeroes is \u2013 2
So, (x + 2) is the factor of x3<\/sup> + 2x2<\/sup> \u2013 x \u2013 2
Let us divide x3<\/sup> + 2x2<\/sup> \u2013 x \u2013 2 by x + 2
The division process is<\/p>\n\n\n\n

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

Here, quotient = x2<\/sup> \u2013 1
= (x \u2013 1)(x + 1)
So, the zeroes are \u2013 1 and 1
Hence, all the zeroes of the given polynomial are \u2013 1, \u2013 2 and 1.<\/p>\n\n\n\n

\n\n\n\n

Question 9 D <\/h4>\n\n\n\n

Find all the zeroes of the polynomial given below having given numbers as its zeroes.<\/strong><\/p>\n\n\n\n

x3<\/sup> + 5x2<\/sup> + 7x + 3; \u2013 3<\/strong>
Sol :
Given zeroes is \u2013 3
So, (x + 3) is the factor of x3<\/sup> + 5x2<\/sup> + 7x + 3
Let us divide x3<\/sup> + 5x2<\/sup> + 7x + 3 by x + 3
The division process is<\/p>\n\n\n\n

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

Here, quotient = x2<\/sup> + 2x + 1
= (x + 1)2<\/sup>
So, the zeroes are \u2013 1 and \u2013 1
Hence, all the zeroes of the given polynomial are \u2013 1, \u2013 1 and \u2013 3.<\/p>\n\n\n\n

\n\n\n\n

Question 9 E <\/h4>\n\n\n\n

Find all the zeroes of the polynomial given below having given numbers as its zeroes.<\/strong><\/p>\n\n\n\n

x4<\/sup> \u2013 6x3<\/sup> \u2013 26x2<\/sup> + 138x \u2013 35;2\u00b1\u221a3<\/strong>
Sol :
x4<\/sup> \u2013 6x3<\/sup> \u2013 26x2<\/sup> + 138x \u2013 35;2\u00b1\u221a3
Given zeroes are 2 + \u221a3 and 2 \u2013 \u221a3
So, (x \u2013 2 \u2013 \u221a3)and (x \u2013 2 + \u221a3) are the factors of x4<\/sup> \u2013 6x3<\/sup> \u2013 26x2<\/sup> + 138x \u2013 35
\u27f9 (x \u2013 2 \u2013 \u221a3)(x \u2013 2 + \u221a3)
= x2<\/sup> \u2013 2x + \u221a3 x \u2013 2x + 4 \u2013 2\u221a3 \u2013 \u221a3 x + 2\u221a3 \u2013 3
= x2<\/sup> \u2013 4x + 1 is a factor of given polynomial.
Consequently, x2<\/sup> \u2013 4x + 1 is also a factor of the given polynomial.
Now, let us divide x4<\/sup> \u2013 6x3<\/sup> \u2013 26x2<\/sup> + 138x \u2013 35 by x2<\/sup> \u2013 4x + 1
<\/p>\n\n\n\n

The division process is<\/p>\n\n\n\n

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

Here, quotient = x2<\/sup> \u2013 2x \u2013 35
= x2<\/sup> \u2013 7x + 5x \u2013 35
= x(x \u2013 7) + 5(x \u2013 7)
= (x + 5)(x \u2013 7)
So, the zeroes are \u2013 5 and 7
Hence, all the zeroes of the given polynomial are \u2013 5, 7, 2+\u221a3 and 2-\u221a3<\/p>\n\n\n\n

\n\n\n\n

Question 9 F <\/h4>\n\n\n\n

Find all the zeroes of the polynomial given below having given numbers as its zeroes.<\/strong><\/p>\n\n\n\n

x4<\/sup> + x3<\/sup> \u2013 34x2<\/sup> \u2013 4x + 120;2, \u2013 2.<\/strong>
Sol :
x4<\/sup> + x3<\/sup> \u2013 34x2<\/sup> \u2013 4x + 120;2, \u2013 2.
Given zeroes are \u2013 2 and 2
So, (x + 2)and (x \u2013 2) are the factors of x4<\/sup> + x3<\/sup> \u2013 34x2<\/sup> \u2013 4x + 120
\u27f9 (x + 2)(x \u2013 2) = x2<\/sup> \u2013 4 is a factor of given polynomial.
Consequently, x2<\/sup> \u2013 4 is also a factor of the given polynomial.
Now, let us divide x4<\/sup> + x3<\/sup> \u2013 34x2<\/sup> \u2013 4x + 120 by x2<\/sup> \u2013 4
The division process is<\/p>\n\n\n\n

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

Here, quotient =x2<\/sup> + x \u2013 30
= x2<\/sup> + 6x \u2013 5x \u2013 30
= x(x + 6) \u2013 5(x + 6)
= (x + 6)(x \u2013 5)
So, the zeroes are \u2013 6 and 5
Hence, all the zeroes of the given polynomial are \u2013 2 , \u2013 6, 2 and 5.<\/p>\n\n\n\n

\n\n\n\n

Question 9 G <\/h4>\n\n\n\n

Find all the zeroes of the polynomial given below having given numbers as its zeroes.<\/strong><\/p>\n\n\n\n

2x4<\/sup> + 7x3<\/sup> \u2013 19x2<\/sup> \u2013 14x + 30;\u221a2, \u2013 \u221a2<\/strong>
Sol :
2x4<\/sup> + 7x3<\/sup> \u2013 19x2<\/sup> \u2013 14x + 30;\u221a(2, \u2013 \u221a2)
Given zeroes are \u221a2 and \u2013 \u221a2
So, (x \u2013 \u221a2)and (x + \u221a2) are the factors of 2x4<\/sup> + 7x3<\/sup> \u2013 19x2<\/sup> \u2013 14x + 30
\u27f9 (x \u2013 \u221a2)(x + \u221a2) = x2<\/sup> \u2013 2 is a factor of given polynomial.
Consequently, x2<\/sup> \u2013 2 is also a factor of the given polynomial.
Now, let us divide 2x4<\/sup> + 7x3<\/sup> \u2013 19x2<\/sup> \u2013 14x + 30 by x2<\/sup> \u2013 2
The division process is<\/p>\n\n\n\n

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

Here, quotient =2x2<\/sup> + 7x \u2013 15
= 2x2<\/sup> + 10x \u2013 3x \u2013 15
= 2x(x + 5) \u2013 3(x + 5)
= (2x \u2013 3)(x + 5)
So, the zeroes are \u2013 5 and $\\frac{3}{2}$
Hence, all the zeroes of the given polynomial are \u2013 5, \u2013 \u221a2, \u221a2 and $\\frac{3}{2}$<\/p>\n\n\n\n

\n\n\n\n

Question 9 H <\/h4>\n\n\n\n

Find all the zeroes of the polynomial given below having given numbers as its zeroes.<\/strong><\/p>\n\n\n\n

2x4<\/sup> \u2013 9x3<\/sup> + 5x2<\/sup> + 3x \u2013 1;2\u00b1\u221a3<\/strong>
Sol :
2x4<\/sup> \u2013 9x3<\/sup> + 5x2<\/sup> + 3x \u2013 1;2\u00b1\u221a3
Given zeroes are 2 + \u221a3 and 2 \u2013 \u221a3
So, (x \u2013 2 \u2013 \u221a3)and (x \u2013 2 + \u221a3) are the factors of 2x4<\/sup> \u2013 9x3<\/sup> + 5x2<\/sup> + 3x \u2013 1
\u27f9 (x \u2013 2 \u2013 \u221a3)(x \u2013 2 + \u221a3)
= x2<\/sup> \u2013 2x + \u221a3 x \u2013 2x + 4 \u2013 2\u221a3 \u2013 \u221a3 x + 2\u221a3 \u2013 3
= x2<\/sup> \u2013 4x + 1 is a factor of given polynomial.
Consequently, x2<\/sup> \u2013 4x + 1 is also a factor of the given polynomial.
Now, let us divide 2x4<\/sup> \u2013 9x3<\/sup> + 5x2<\/sup> + 3x \u2013 1 by x2<\/sup> \u2013 4x + 1
The division process is<\/p>\n\n\n\n

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

Here, quotient = 2x2<\/sup> \u2013 x \u2013 1
= 2x2<\/sup> \u2013 2x + x \u2013 1
= 2x(x \u2013 1) + 1(x \u2013 1)
= (2x + 1)(x \u2013 1)
So, the zeroes are $-\\frac{1}{2}$ and 1
Hence, all the zeroes of the given polynomial are $-\\frac{1}{2}$, 1, 2 + \u221a3 and 2 \u2013 \u221a3<\/p>\n\n\n\n

\n\n\n\n

Question 9 I <\/h4>\n\n\n\n

Find all the zeroes of the polynomial given below having given numbers as its zeroes.<\/strong><\/p>\n\n\n\n

2x3<\/sup> \u2013 4x \u2013 x2<\/sup> + 2;\u221a2, \u2013 \u221a2<\/strong><\/p>\n\n\n\n

Sol :
2x3<\/sup> \u2013 4x \u2013 x2<\/sup> + 2;\u221a2, \u2013 \u221a2
Given zeroes are \u221a2 and \u2013 \u221a2
So, (x \u2013 \u221a2) and (x + \u221a2) are the factors of 2x3<\/sup> \u2013 4x \u2013 x2<\/sup> + 2
\u27f9 (x \u2013 \u221a2)(x + \u221a2) = x2<\/sup> \u2013 2 is a factor of given polynomial.
Consequently, x2<\/sup> \u2013 2 is also a factor of the given polynomial.
Now, let us divide 2x3<\/sup> \u2013 4x \u2013 x2<\/sup> + 2by x2<\/sup> \u2013 2
The division process is<\/p>\n\n\n\n

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

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

So, the zeroes is $\\frac{1}{2}$
Hence, all the zeroes of the given polynomial are -\u221a2,\u221a2 and $\\frac{1}{2}$<\/p>\n\n\n\n

\n\n\n\n

Question 10 <\/h4>\n\n\n\n

Verify that 3,-1, <\/strong>$-\\frac{1}{3}$ <\/strong>are the zeroes of the cubic polynomial p(x) = 3x2<\/sup> \u2013 5x2<\/sup> \u2013 11x \u2013 3 and then verify the relationship between the zeroes and the coefficients.<\/strong><\/p>\n\n\n\n

Sol :
Let p(x) = 3 x3<\/sup> \u2013 5 x2<\/sup> \u2013 11x \u2013 3
Then, p( \u2013 1) = 3( \u2013 1)3<\/sup> \u2013 5( \u2013 1)2<\/sup> \u2013 11( \u2013 1) \u2013 3
= \u2013 3 \u2013 5 + 11 \u2013 3
= 0<\/p>\n\n\n\n

$p\\left(-\\frac{1}{3}\\right)=3\\left(-\\frac{1}{3}\\right)^{3}-5\\left(-\\frac{1}{3}\\right)^{2}-11\\left(-\\frac{1}{3}\\right)-3$<\/p>\n\n\n\n

$=\\left(-\\frac{1}{9}\\right)-\\left(\\frac{5}{9}\\right)+\\left(\\frac{11}{3}\\right)-3$<\/p>\n\n\n\n

$=\\left(\\frac{-1-5+33-27}{9}\\right)$<\/p>\n\n\n\n

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

p(3) = 3(3)3<\/sup> \u2013 5(3)2<\/sup> \u2013 11(3) \u2013 3
= 81 \u2013 45 \u2013 33 \u2013 3
= 0
Hence, we verified that 3, \u2013 1 and $-\\frac{1}{3}$ are the zeroes of the given polynomial.
So, we take \u03b1 = 3, \u03b2 = \u2013 1, $\\gamma=-\\frac{1}{3}$
Verification<\/u><\/strong>
\u03b1 + \u03b2 + \u03b3 $=3+(-1)+\\left(-\\frac{1}{3}\\right)=\\left(\\frac{5}{3}\\right)$<\/p>\n\n\n\n

$=-\\frac{\\text { Coefficient of } \\mathrm{x}^{2}}{\\text { Coefficient of } \\mathrm{x}^{3}}=-\\frac{5}{3}$<\/p>\n\n\n\n

\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 $=(3)(-1)+(-1)\\left(-\\frac{1}{3}\\right)+\\left(-\\frac{1}{3}\\right)(3)$
$=\\left(-\\frac{11}{3}\\right)$
$=\\frac{\\text { Coefficient of } \\mathrm{x}}{\\text { Coefficient of } \\mathrm{x}^{3}}=\\frac{-11}{3}=\\left(-\\frac{11}{3}\\right)$
and \u03b1\u03b2\u03b3 $=3 \\times-1 \\times\\left(-\\frac{1}{3}\\right)$
= 1
$=-\\frac{\\text { Constant term }}{\\text { Coefficient of } \\mathrm{x}^{3}}=-\\frac{-3}{3}=1$
Thus, the relationship between the zeroes and the coefficients is verified.<\/p>\n\n\n\n

\n\n\n\n

Question 11 A <\/h4>\n\n\n\n

Verify that the numbers given alongside of the cubic polynomial are their zeros. Also verify the relationship between the zeroes and the coefficients in each case :<\/strong><\/p>\n\n\n\n

x3<\/sup> \u2013 4x2<\/sup> + 5x \u2013 2;2, 1, 1<\/strong><\/p>\n\n\n\n

Sol :
Let p(x) = x3<\/sup> \u2013 4x2<\/sup> + 5x \u2013 2
Then, p(2) = (2)3<\/sup> \u2013 4(2)2<\/sup> + 5(2) \u2013 2
= 8 \u2013 16 + 10 \u2013 2
= 0
p(1) = (1)3<\/sup> \u2013 4(1)2<\/sup> + 5(1) \u2013 2
= 1 \u2013 4 + 5 \u2013 2
= 0
Hence, 2, 1 and 1 are the zeroes of the given polynomial x3<\/sup> \u2013 4x2<\/sup> + 5x \u2013 2.
Now, Let \u03b1 = 2 , \u03b2 = 1 and \u03b3 = 1
Then, \u03b1 + \u03b2 + \u03b3 = 2 + 1 + 1 = 4
$=-\\frac{\\text { Coefficient of } \\mathrm{x}^{2}}{\\text { Coefficient of } \\mathrm{x}^{3}}=-\\frac{-4}{1}=4$
\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = (2)(1) + (1)(1) + (1)(2)
= 2 + 1 + 2
= 5
$=\\frac{\\text { Coefficient of } \\mathrm{x}}{\\text { Coefficient of } \\mathrm{x}^{3}}=\\frac{5}{1}=5$
and \u03b1\u03b2\u03b3 = 2 \u00d7 1 \u00d7 1
= 2
$=-\\frac{\\text { Constant term }}{\\text { Coefficient of } \\mathrm{x}^{3}}=-\\frac{-2}{1}=2$
Thus, the relationship between the zeroes and the coefficients is verified.<\/p>\n\n\n\n

\n\n\n\n

Question 11 B <\/h4>\n\n\n\n

Verify that the numbers given alongside of the cubic polynomial are their zeros. Also verify the relationship between the zeroes and the coefficients in each case :<\/strong><\/p>\n\n\n\n

x3<\/sup> \u2013 6x2<\/sup> + 11x \u2013 6 ;1, 2, 3<\/strong>
Sol :
Let p(x) = x3<\/sup> \u2013 6x2<\/sup> + 11x \u2013 6
Then, p(1) = (1)3<\/sup> \u2013 6(1)2<\/sup> + 11(1) \u2013 6
= 1 \u2013 6 + 11 \u2013 6
= 0
p(2) = (2)3<\/sup> \u2013 6(2)2<\/sup> + 11(2) \u2013 6
= 8 \u2013 24 + 22 \u2013 6
= 0
p(3) = (3)3<\/sup> \u2013 6(3)2<\/sup> + 11(3) \u2013 6
= 27 \u2013 54 + 33 \u2013 6
= 0
Hence, 1, 2 and 3 are the zeroes of the given polynomial x3<\/sup> \u2013 6x2<\/sup> + 11x \u2013 6.
Now, Let \u03b1 = 1 , \u03b2 = 2 and \u03b3 = 3
Then, \u03b1 + \u03b2 + \u03b3 = 1 + 2 + 3 = 6
$=-\\frac{\\text { Coefficient of } \\mathrm{x}^{2}}{\\text { Coefficient of } \\mathrm{x}^{3}}=-\\frac{-6}{1}=6$
\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = (1)(2) + (2)(3) + (3)(1)
= 2 + 6 + 3
= 11
$=\\frac{\\text { Coefficient of } \\mathrm{x}}{\\text { Coefficient of } \\mathrm{x}^{3}}=\\frac{11}{1}=11$
and \u03b1\u03b2\u03b3 = 1 \u00d7 2 \u00d7 3
= 6
$=-\\frac{\\text { Constant term }}{\\text { Coefficient of } \\mathrm{x}^{3}}=-\\frac{-6}{1}=6$
Thus, the relationship between the zeroes and the coefficients is verified.<\/p>\n\n\n\n

\n\n\n\n

Question 11 C <\/h4>\n\n\n\n

Verify that the numbers given alongside of the cubic polynomial are their zeros. Also verify the relationship between the zeroes and the coefficients in each case :<\/strong><\/p>\n\n\n\n

x3<\/sup> + 2x2<\/sup> \u2013 x \u2013 2; \u2013 2 \u2013 2, 1<\/strong>
Sol :
Let p(x) = x3<\/sup> + 2x2<\/sup> \u2013 x \u2013 2
Then, p( \u2013 2) = ( \u2013 2)3<\/sup> + 2( \u2013 2)2<\/sup> \u2013 ( \u2013 2) \u2013 2
= \u2013 8 + 8 + 2 \u2013 2
= 0
p(1) = (1)3<\/sup> + 2(1)2<\/sup> \u2013 (1) \u2013 2
= 1 + 2 \u2013 1 \u2013 2
= 0
Hence, \u2013 2, \u2013 2 and 1 are the zeroes of the given polynomial x3<\/sup> + 2x2<\/sup> \u2013 x \u2013 2.
Now, Let \u03b1 = \u2013 2 , \u03b2 = \u2013 2 and \u03b3 = 1
Then, \u03b1 + \u03b2 + \u03b3 = \u2013 2 + ( \u2013 2) + 1 = \u2013 3
$=-\\frac{\\text { Coefficient of } \\mathrm{x}^{2}}{\\text { Coefficient of } \\mathrm{x}^{3}}=-\\frac{2}{1}=-2$
\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = ( \u2013 2)( \u2013 2) + ( \u2013 2)(1) + (1)( \u2013 2)
= 4 \u2013 2 \u2013 2
= 0
$=\\frac{\\text { Coefficient of } \\mathrm{x}}{\\text { Coefficient of } \\mathrm{x}^{3}}=\\frac{-1}{1}=-1$
and \u03b1\u03b2\u03b3 = ( \u2013 2) \u00d7 ( \u2013 2) \u00d7 1
= 4
$=-\\frac{\\text { Constant term }}{\\text { Coefficient of } \\mathrm{x}^{3}}=-\\frac{-2}{1}=2$
Thus, the relationship between the zeroes and the coefficients is not verified.<\/p>\n\n\n\n

\n\n\n\n

Question 11 D <\/h4>\n\n\n\n

Verify that the numbers given alongside of the cubic polynomial are their zeros. Also verify the relationship between the zeroes and the coefficients in each case :<\/strong><\/p>\n\n\n\n

x3<\/sup> + 5x2<\/sup> + 7x + 3 ; \u2013 3, 2 \u2013 1, \u2013 1<\/strong>
Sol :
Let p(x) = x3<\/sup> + 5x2<\/sup> + 7x + 3.
Then, p( \u2013 1) = ( \u2013 1)3<\/sup> + 5( \u2013 1)2<\/sup> + 7( \u2013 1) + 3
= \u2013 1 + 5 \u2013 7 + 3
= 0
p( \u2013 3) = ( \u2013 3)3<\/sup> + 5( \u2013 3)2<\/sup> + 7( \u2013 3) + 3
= \u2013 27 + 45 \u2013 21 + 3
= 0
Hence, \u2013 1, \u2013 1 and \u2013 3 are the zeroes of the given polynomial x3<\/sup> + 5x2<\/sup> + 7x + 3.
Now, Let \u03b1 = \u2013 1 , \u03b2 = \u2013 1 and \u03b3 = \u2013 3
Then, \u03b1 + \u03b2 + \u03b3 = \u2013 1 + ( \u2013 1) + ( \u2013 3) = \u2013 5
$=-\\frac{\\text { Coefficient of } \\mathrm{x}^{2}}{\\text { Coefficient of } \\mathrm{x}^{3}}=-\\frac{5}{1}=-5$
\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = ( \u2013 1)( \u2013 1) + ( \u2013 1)( \u2013 3) + ( \u2013 3)( \u2013 1)
= 1 + 3 + 3
= 7
$=\\frac{\\text { Coefficient of } \\mathrm{x}}{\\text { Coefficient of } \\mathrm{x}^{3}}=\\frac{7}{1}=7$
and \u03b1\u03b2\u03b3 = ( \u2013 1) \u00d7 ( \u2013 1) \u00d7 ( \u2013 3)
= \u2013 3
$=-\\frac{\\text { Constant term }}{\\text { Coefficient of } \\mathrm{x}^{3}}=-\\frac{3}{1}=-3$
Thus, the relationship between the zeroes and the coefficients is verified.<\/p>\n\n\n\n

\n\n\n\n

Question 12 <\/h4>\n\n\n\n

Find a cubic polynomial having 1, 2, 3 as its zeroes.<\/strong><\/p>\n\n\n\n

Sol :
Let the zeroes of the cubic polynomial be
\u03b1 = 1, \u03b2 = 2 and \u03b3 = 3
Then, \u03b1 + \u03b2 + \u03b3 = 1 + 2 + 3 = 6
\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = (1)(2) + (2)(3) + (3)(1)
= 2 + 6 + 3
= 11
and \u03b1\u03b2\u03b3 = 1 \u00d7 2 \u00d7 3
= 6
Now, required cubic polynomial
= x3<\/sup> \u2013 (\u03b1 + \u03b2 + \u03b3) x2<\/sup> + (\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1)x \u2013 \u03b1\u03b2\u03b3
= x3<\/sup> \u2013 (6) x2<\/sup> + (11)x \u2013 6
= x3<\/sup> \u2013 6 x2<\/sup> + 11x \u2013 6
So, x3<\/sup> \u2013 6 x2<\/sup> + 11x \u2013 6 is the required cubic polynomial which satisfy the given conditions.<\/p>\n\n\n\n

\n\n\n\n

Question 13 <\/h4>\n\n\n\n

Find a cubic polynomial having \u2013 3, \u2013 2, 2 as its zeroes.<\/strong><\/p>\n\n\n\n

Sol :
Let the zeroes of the cubic polynomial be
\u03b1 = \u2013 3, \u03b2 = \u2013 2 and \u03b3 = 2
Then, \u03b1 + \u03b2 + \u03b3 = \u2013 3 + ( \u2013 2) + 2
= \u2013 3 \u2013 2 + 2
= \u2013 3
\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = ( \u2013 3)( \u2013 2) + ( \u2013 2)(2) + (2)( \u2013 3)
= 6 \u2013 4 \u2013 6
= \u2013 4
and \u03b1\u03b2\u03b3 = ( \u2013 3) \u00d7 ( \u2013 2) \u00d7 2
= 6 \u00d7 2
= 12
Now, required cubic polynomial
= x3<\/sup> \u2013 (\u03b1 + \u03b2 + \u03b3) x2<\/sup> + (\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1)x \u2013 \u03b1\u03b2\u03b3
= x3<\/sup> \u2013 ( \u2013 3) x2<\/sup> + ( \u2013 4)x \u2013 12
= x3<\/sup> + 3 x2<\/sup> \u2013 4x \u2013 12
So, x3<\/sup> + 3x^2 \u2013 4x \u2013 12 is the required cubic polynomial which satisfy the given conditions.<\/p>\n\n\n\n

\n\n\n\n

Question 14 <\/h4>\n\n\n\n

Find a cubic polynomial with the sum of its zeroes are 0, \u2013 7 and \u2013 6 respectively.<\/strong><\/p>\n\n\n\n

Sol :
Let the zeroes be \u03b1, \u03b2 and \u03b3.
Then, we have
\u03b1 + \u03b2 + \u03b3 = 0
\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = \u2013 7
and \u03b1\u03b2\u03b3 = \u2013 6
Now, required cubic polynomial
= x3<\/sup> \u2013 (\u03b1 + \u03b2 + \u03b3) x2<\/sup> + (\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1)x \u2013 \u03b1\u03b2\u03b3
= x3<\/sup> \u2013 (0) x2<\/sup> + ( \u2013 7)x \u2013 ( \u2013 6)
= x3<\/sup> \u2013 7x + 6
So, x3<\/sup> \u2013 7x + 6 is the required cubic polynomial.<\/p>\n\n\n\n

\n\n\n\n

Question 15 A <\/h4>\n\n\n\n

Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and product of its zeroes as the numbers given below:<\/strong><\/p>\n\n\n\n

2, \u2013 7, \u2013 14<\/strong>
Sol :
Let the zeroes be \u03b1, \u03b2 and \u03b3.
Then, we have
\u03b1 + \u03b2 + \u03b3 = 2
\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = \u2013 7
and \u03b1\u03b2\u03b3 = \u2013 14
<\/p>\n\n\n\n

Now, required cubic polynomial<\/p>\n\n\n\n

=x3<\/sup>-(\u03b1+\u03b2+\u03b3)x2<\/sup>+(\u03b1\u03b2+\u03b2\u03b3+\u03b3\u03b1)-\u03b1\u03b2\u03b3<\/p>\n\n\n\n

=x3<\/sup>-(2)x2<\/sup>+(-7x)-(-14)<\/p>\n\n\n\n

Sox3<\/sup> \u2013 2x2<\/sup> \u2013 7x + 14 is the required cubic polynomial.<\/p>\n\n\n\n

\n\n\n\n

Question 15 B <\/h4>\n\n\n\n

Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and product of its zeroes as the numbers given below:<\/strong><\/p>\n\n\n\n

$-4 \\frac{1}{2}, \\frac{1}{3}$<\/strong>
Sol :
Let the zeroes be \u03b1, \u03b2 and \u03b3.
Then, we have
$\\alpha+\\beta+\\gamma=-4$
$\\alpha \\beta+\\beta \\gamma+\\gamma \\alpha=\\frac{1}{2}$
$\\alpha \\beta \\gamma=\\frac{1}{3}$
Now, required cubic polynomial
$=\\mathrm{x}^{3}-(\\alpha+\\beta+\\gamma) \\mathrm{x}^{2}+(\\alpha \\beta+\\beta \\gamma+\\gamma \\alpha) \\mathrm{x}-\\alpha \\beta \\gamma$
$=x^{3}-(-4) x^{2}+\\left(\\frac{1}{2}\\right) x-\\left(\\frac{1}{3}\\right)$
$=\\frac{6 x^{3}+24 x^{2}+3 x-2}{6}$
So, 6x3<\/sup> + 24x2<\/sup> + 3x \u2013 2 is the required cubic polynomial which satisfy the given conditions.<\/p>\n\n\n\n

\n\n\n\n

Question 15 C <\/h4>\n\n\n\n

Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and product of its zeroes as the numbers given below:<\/strong><\/p>\n\n\n\n

$\\frac{5}{7}, \\frac{1}{7}, \\frac{1}{7}$<\/strong><\/p>\n\n\n\n

Sol :
Let the zeroes be \u03b1, \u03b2 and \u03b3.
Then, we have
$\\alpha+\\beta+\\gamma=\\frac{5}{7}$
$\\alpha \\beta+\\beta \\gamma+\\gamma \\alpha=\\frac{1}{7}$
$\\alpha \\beta \\gamma=\\frac{1}{7}$
<\/p>\n\n\n\n

Now, required cubic polynomial
$=\\mathrm{x}^{3}-(\\alpha+\\beta+\\gamma) \\mathrm{x}^{2}+(\\alpha \\beta+\\beta \\gamma+\\gamma \\alpha) \\mathrm{x}-\\alpha \\beta \\gamma$
$=x^{3}-\\left(\\frac{5}{7}\\right) x^{2}+\\left(\\frac{1}{7}\\right) x-\\left(\\frac{1}{7}\\right)$
$=\\frac{7 x^{3}-5 x^{2}+x-1}{7}$
So, 7x3<\/sup> \u2013 5x2<\/sup> + x \u2013 1 is the required cubic polynomial which satisfy the given conditions.<\/p>\n\n\n\n

\n\n\n\n

Question 15 D <\/h4>\n\n\n\n

Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and product of its zeroes as the numbers given below:<\/strong><\/p>\n\n\n\n

$\\frac{2}{5}, \\frac{1}{10}, \\frac{1}{2}$<\/strong><\/p>\n\n\n\n

Sol :
Let the zeroes be \u03b1, \u03b2 and \u03b3.
Then, we have
$\\alpha+\\beta+\\gamma=\\frac{2}{5}$
$\\alpha \\beta+\\beta \\gamma+\\gamma \\alpha=\\frac{1}{10}$
$\\alpha \\beta \\gamma=\\frac{1}{2}$
Now, required cubic polynomial
$=\\mathrm{x}^{3}-(\\alpha+\\beta+\\gamma) \\mathrm{x}^{2}+(\\alpha \\beta+\\beta \\gamma+\\gamma \\alpha) \\mathrm{x}-\\alpha \\beta \\gamma$
$=x^{3}-\\left(\\frac{2}{5}\\right) x^{2}+\\left(\\frac{1}{10}\\right) x-\\left(\\frac{1}{2}\\right)$
$=\\frac{10 x^{3}-4 x^{2}+x-5}{10}$
So, 10x3<\/sup> \u2013 4x2<\/sup> + x \u2013 5 is the required cubic polynomial which satisfy the given conditions.<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 10 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Question 1  Divide 2×3 + 3x + 1 by x + 2 and find the quotient and the reminder. Is q(x) a factor of 2×3 + 3x + 1 ? Sol :Here, dividend and divisor both are in the standard form.Now, on dividing p(x) by g(x) we get the following division process Quotient = 2×2 \u2013 4x + […]<\/p>\n","protected":false},"author":302,"featured_media":623922,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[24],"tags":[],"boards":[],"class_list":["post-623944","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-10","entry"],"yoast_head":"\nKC Sinha: Exercise 2.3 - Mathematics Solution Class 10 Chapter 2 Polynomials - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Question 1  Divide 2x3 + 3x + 1 by x + 2 and find the quotient and the reminder. 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