# Ex.2.4 Q1 Polynomials Solution - NCERT Maths Class 9

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## Question

Determine which of the following polynomials has $$(x + 1)$$ a factor:

(i) \begin{align}{x}^{3}+{x}^{2}+x+1\end{align}

(ii) \begin{align}{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+x+1\end{align}

(iii) \begin{align}{{x}^{4}}+3{{x}^{3}}+3{{x}^{2}}+x+1\end{align}

(iv) \begin{align}{{x}^{3}}-{{x}^{2}}-(2+\sqrt{2})x+\sqrt{2}\end{align}

Video Solution
Polynomials
Ex 2.4 | Question 1

## Text Solution

Reasoning:

When a polynomial $$p(x)$$ is divided by $$x-a$$ and if $$p(a) = 0$$ then $$(x-a)$$ is a factor of $$p(x)$$. The root of $$x+1=0$$ is $$-1.$$

Steps:

(i) Let \begin{align} p(x)={{x}^{3}}+{{x}^{2}}+x+1\end{align}

\begin{align} \therefore p(-1)&={{(-1)}^{3}}+{{(-1)}^{2}}+(-1)+1 \\ & =-1+1-1+1=0 \\ \end{align}

Since the remainder of $$p\text{(-1) = 0}$$ , we conclude that $$(x+1)$$ is a factor of $${{x}^{3}}+{{x}^{2}}+x+1$$ .

(ii) Let $$p(x)={{x}^{4}}+{{x}^{3}}+{{x}^{2}}+x+1$$

\begin{align}\therefore p( - 1) &\!=\! {( - 1)^4}\!+\!{( - 1)^3}\!+\!{(\!-\!1)^2}\!+\!(\!-\!1)\!+\!1\\ &= \not 1 - \not 1 + \not 1 - \not 1+ 1\\ &= 1 \ne 0\end{align}

Since the remainder of  $$p( - 1) \ne 0$$, we conclude that $$(x+1)$$ in not a factor of  $$\,{x^4} + {x^3} + {x^2} + x + 1$$.

(iii) Let $$p(x) = {x^4} + 3{x^3} + 3{x^2} + x + 1$$

\begin{align} \therefore p( - 1) &\!=\!{( -\!1)^4}\!+\!3{(\!-\!1)^3}\!\!+\!3{(\!-\!1)^2}\!\!+\!( -\!1)\!\!+\!\!1 \\ &= 1 - 3 + 3 - 1 + 1 \\ &= 1 \ne 0\end{align}

Since the remainder of $$p( - 1) \ne 0$$ ,  $$(x+1)$$ is not a factor of $${x^4} + 3{x^3} + 3{x^2} + x + 1$$.

(iv) Let $$p(x) = {x^3} - {x^2} - (2 + \sqrt 2 )x + \sqrt 2$$

\begin{align} \therefore p( - 1) &\!=\!\!{(\!\!-\!1)^3}\!\!-\!{(\!-\!\!1)^2}\!\!-\!(2\!+\!\!\sqrt 2 )( -\!1)\!\!+\!\!\sqrt 2\!\\ &\!=\!-\!1\!-\!1\!+\!2 \!+ \!\sqrt 2\!+\!\sqrt 2\\ &= 2\sqrt 2 \end{align}

Since the remainder of $$p( - 1) = 0$$ , $$(x+1)$$ is  not a factor of $${x^3} - {x^2} - (2 + \sqrt 2 )x + \sqrt 2$$ .

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