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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