# Ex.2.4 Q2 Polynomials Solution - NCERT Maths Class 9

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## Question

Use the Factor Theorem to determine whether $$g(x)$$ is a factor of $$p(x)$$ in each of the following cases:

(i) $$p(x)\!=\!2{x^3}\!+\!{x^2}\!-\!2x\!-\!1,\,g(x)\!=\!x \!+\!1$$

(ii) $$p(x)\!=\!{x^3}\!+\!3{x^2}\!+\!3x\!+\!1,\,g(x)\!=\!x\!+\!2$$

(iii) $$p(x)\!=\!{x^3}\!-\!4{x^2}\!+\!x\!+\!6,\,g(x)\!=\!x\!-\!3$$

Video Solution
Polynomials
Ex 2.4 | Question 2

## Text Solution

Reasoning:

By factor theorem, $$(x-a)$$ is a factor of a polynomial $$p(x)$$ if $$p(a) = 0.$$

To find if $$g(x )= x+a$$ is a factor of $$p(x),$$ we need to find the root of $$g(x).$$

$$x + a = 0{\rm{ }} \to {\rm{ }}x{\rm{ }} = -a$$

Steps:

(i) Let $$p(x) = 2{x^3} + {x^2} - 2x - 1,\,g(x) = x + 1$$

$x + 1 = 0{\rm{ }} \to {\rm{ }}x{\rm{ }} = {\rm{ }}-1$

Now,

\begin{align}p( - 1) &= 2{( - 1)^3} + {( - 1)^2} - 2( - 1) - 1\\ & = - \not 2 + \not 1 + \not 2 - \not 1\\& = 0\end{align}

Since the remainder of $$p( - 1) = 0$$ , by factor theorem we can say $$g(x) = x+1$$ is a factor of $$p(x) = 2{x^3} + {x^2} - 2x - 1.$$

(ii) Let $$p(x) = {x^3} + 3{x^2} + 3x + 1,\,\,g(x) = x + 2$$

$x + 2 = 0{\rm{ }} \to {\rm{ }}x{\rm{ }} = {\rm{ }}-2$

Now,

\begin{align}p( - 2)&= {( - 2)^3} + 3{( - 2)^2} + 3( - 2) + 1\\ & = - 8 + 12 - 6 + 1\\ & = - 1 \ne 0\end{align}

Since the remainder of $$p( - 2) \ne 0$$ , by factor theorem we can say $$g(x) = x+2$$ is not a factor of $$p(x) = {x^3} + 3{x^2} + 3x + 1.$$

(iii) Let $$p(x) = {x^3} - 4{x^2} + x + 6,\,\,g(x) = x - 3$$

$x - 3 = 0{\rm{ }} \to {\rm{ }}x\,{\rm{ = }}\,{\rm{3}}$

Now,

\begin{align}p(3) &= {(3)^3} - 4{(3)^2} + 3 + 6\\ & = 27 - 36 + 3 + 6\\ & = 0\end{align}

Since the remainder of $$p(3) = 0$$ , by factor theorem we can say $$g(x) = x-3$$ is a factor of $$p(x) = {x^3} - 4{x^2} + x + 6.$$

Video Solution
Polynomials
Ex 2.4 | Question 2

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