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Ex.2.2 Q2 Polynomials Solution - NCERT Maths Class 9

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Question

Find \(p(0), p(1)\) and \(p(2)\) for each of the following polynomials:

(i) \(\begin{align}p(y)=y^{2}-y+1\end{align}\) 

(ii) \(\begin{align}p(t)=2+t+2 t^{2}-t^{3}\end{align}\) 

(iii) \(\begin{align}p(x)=x^{3}\end{align}\)

(iv) \(\begin{align}p(x)=(x-1)(x+1)\end{align}\)

 Video Solution
Polynomials
Ex 2.2 | Question 2

Text Solution

Steps:

(i) \({p(y)}={y^{2}-y+1}\)

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

(ii) \(p(t) =2+t+2\left(t^{2}\right)-t^{3}\)

\[\begin{align}p(0) &=2+0+2(0)^{2}-(0)^{3} \\ &=2+0+0-0=2 \\ p(1) &=2+1+2(1)^{2}-(1)^{3} \\ &=2+1+2-1=4 \\ p(2) &=2+2+2(2)^{2}-(2)^{3} \\ &=2+2+8-8=4 \end{align}\]

(iii)\(p(x) =x^{3} \)

\[\begin{align}p(0) &=(0)^{3}=0 \\ p(1) &=(1)^{3}=1 \\ p(2) &=(2)^{3}=8 \end{align}\]

(iv) \(p(x) =(x-1)(x+1)\)

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

 Video Solution
Polynomials
Ex 2.2 | Question 2
  
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