Exercise-2.2, Class 9th, Maths, Chapter 2, NCERT

1. Find the value of the polynomial $5x-4x^2+3$ at

(i) $x=0$
Answer. $5(0)-4(0{)}^2+3=3$

(ii) $x=-1$
Answer. $5(-1)-4(-1{)}^2+3=-5-4+3=-6$

(iii) $x=2$
Answer. $5(2)-4(2{)}^2+3=10-16+3=-3$

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

(i) $p(y)=y^2-y+1$
Answers. $p(0)=1,p(1)=1,p(2)=3$

(ii) $p(t)=2+t+2t^2-t^3$
Answers. $p(0)=2,p(1)=2+1+2-1=4,p(2)=2+2+8-8=4.$

(iii) $p(x)=x^3$
Answers. $p(0)=0,p(1)=1,p(2)=8$

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

(note: $(x-1)(x+1)=x^2-1$
Answers. $p(0)=-1,p(1)=0,p(2)=3$

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3. Verify whether the following are zeroes of the polynomial indicated against them.

(i) $p(x)=3x+1,x=-\frac{1}{3}$
Check / Answer. $p(-\frac{1}{3})=3(-\frac{1}{3})+1=-1+1=0$.

Yes, $-\frac{1}{3}$ is a zero.

(ii) $p(x)=5x-\pi ,x=\frac{4}{5}$
Check / Answer. $p(\frac{4}{5})=5\cdot \frac{4}{5}-\pi =4-\pi$.

Since $\pi \mathrm{\ne }4$, $4-\pi \mathrm{\ne }0$. No, $\frac{4}{5}$ is not a zero (numerically $4-\pi \approx 0.8584\mathrm{\ne }0$).

(iii) $p(x)=x^2-1,x=1,-1$
Check / Answer. $p(1)=1-1=0,p(-1)=1-1=0$. Yes, both $1$ and $-1$ are zeros.

(iv) $p(x)=(x+1)(x-2),x=-1,2$
Check / Answer. $p(-1)=0\cdot (-3)=0,p(2)=3\cdot 0=0$. Yes, both $-1$ and $2$ are zeros.

(v) $p(x)=x^2,x=0$
Check / Answer. $p(0)=0^2=0$. Yes, $0$ is a zero.

(vi) $p(x)=\mathrm{\ell }x+m,x=-{\displaystyle \frac{m}{\mathrm{\ell }}}$ (here $\mathrm{\ell }$ denotes the coefficient of $x$, assume $\mathrm{\ell }\mathrm{\ne }0$)
Check / Answer. $p\text{\hspace{0.17em}⁣}(-\frac{m}{\mathrm{\ell }})=\mathrm{\ell }(-\frac{m}{\mathrm{\ell }})+m=-m+m=0$. Yes, $x=-{\displaystyle \frac{m}{\mathrm{\ell }}}$ is a zero (provided $\mathrm{\ell }\mathrm{\ne }0$).