Ex.2.3 Q1 Polynomials Solution - NCERT Maths Class 9

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Question

Find the remainder when \(\begin{align}x^{3}+3 x^{2}+3 x+1\end{align}\) is divided by

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

(ii) \(\begin{align} x-\frac{1}{2}\end{align}\)

(iii) \(\begin{align} x\end{align}\)

(iv) \(\begin{align} x+\pi\end{align}\)

(v)  \(\begin{align}5+2 x \end{align}\)

 Video Solution
Polynomials
Ex 2.3 | Question 1

Text Solution

Reasoning:

Let \(p(x)\) be any polynomial of degree greater than or equal to one and let a be any real number. If a polynomial \(p(x)\) is divided by \(\begin{align}x-a\end{align}\) then the remainder is \(p(a).\)

Steps:

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

(i) The root of \(x+1 = 0\) is \(-1\)

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

Hence by the remainder theorem, \(0\) is the remainder when \(\begin{align} x^{3}+3 x^{2}+3 x+1\end{align}\) is divided by \(x+1.\) We can also say that \(x+1\) is a factor of \(\begin{align}x^{3}+3 x^{2}+3 x+1\end{align}\) .

(ii) The root of \(\begin{align}x-\frac{1}{2}=0 \text { is } \frac{1}{2}\end{align}\)

\[\begin{align} p\left(\frac{1}{2}\right) &=\left(\frac{1}{2}\right)^{3}+3\left(\frac{1}{2}\right)^{2}+3\left(\frac{1}{2}\right)+1 \\ &=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1 \\ &=\frac{1+6+12+8}{8}=\frac{27}{8} \end{align}\]

Hence by the remainder theorem, \(\begin{align}\frac{27}{8}\end{align}\) is the remainder when \(\begin{align} x^{3}+3 x^{2}+3 x+1\end{align}\) is divided by \(\begin{align}x-\frac{1}{2}\end{align}\)

(iii) The root of \(\begin{align}x=0 \text { is } 0\end{align}\)

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

Hence by the remainder theorem, \(1\) is the remainder when \(\begin{align} x^{3}+3 x^{2}+3 x+1 \end{align}\) is divided by \(x .\)

(iv)The root of \(\begin{align}x+\pi=0 \text { is }-\pi \end{align}\)

\[\begin{align} p(-\pi) &=(-\pi)^{3}+3(-\pi)^{2}+3(-\pi)+1 \\ &=-\pi^{3}+3 \pi^{2}-3 \pi+1 \end{align}\]

Hence by the remainder theorem, \(\begin{align} -\pi^{3}+3 \pi^{2}-3 \pi+1\end{align}\)  is the remainder when \(\begin{align}x^{3}+3 x^{2}+3 x+1\end{align}\) is divided by \(\begin{align}x+\pi\end{align}\) .

(v) The root of \(\begin{align}5+2 x=0 \text { is } \frac{-5}{2}\end{align}\)

\[\begin{align}{p\left( {\frac{{ - 5}}{2}} \right)}&{ = \,\,\left[ \begin{array}{l}{\left( {\frac{{ - 5}}{2}} \right)^3} + 3{\left( {\frac{{ - 5}}{2}} \right)^2}+\\  3\left( {\frac{{ - 5}}{2}} \right) + 1\end{array} \right]}\\&{ = \frac{{ - 125}}{8} + \frac{{75}}{4} + \frac{{ - 15}}{2} + 1}\\&{ = \frac{{ - 125 + 150 - 60 + 8}}{8}}\\&{ = \frac{{ - 185 + 158}}{8}}\\&{ = \frac{{ - 27}}{8}}\end{align}\]

Hence by remainder theorem, \(\begin{align}\frac{-27}{8}\end{align}\) is the remainder when \(\begin{align}x^{3}+3 x^{2}+3 x+1\end{align}\) is divided by \(\begin{align}5+2 x\end{align}\) .

 Video Solution
Polynomials
Ex 2.3 | Question 1
  
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