Solution

$ \begin{aligned} & \text{ (i) } p(x)=5 x-4 x^{2}+3 \\ & p(0)=5(0)-4(0)^{2}+3 \\ & \quad=3 \\ & p(x)=5 x-4 x^{2}+3 \end{aligned} $

(ii)

$ \begin{aligned} & p(-1)=5(-1)-4(-1)^{2}+3 \\ & =-5-4(1)+3=-6 \\ & p(x)=5 x-4 x^{2}+3 \\ & \text{ (iii) } \\ & p(2)=5(2)-4(2)^{2}+3 \\ & =10-16+3=-3 \end{aligned} $

Solution

(i) $p(y)=y^{2}-y+1 p(0)=$

$(0)^{2}-(0)+1=1 p(1)=(1)^{2}-(1)+1=$

$1 p(2)=(2)^{2}-(2)+1=3$ (ii) $p(t)=$

$2+t+2 t^{2}-t^{3} p(0)=2+0+2(0)^{2}-$

$(0)^{3}=2 p(1)=2+(1)+2(1)^{2}-(1)^{3}$

$=2+1+2-1=4 p(2)=$

$2+2+2(2)^{2}-(2)^{3}$

$=2+2+8-8=4$

(iii) $p(x)=x^{3} p(0)=(0)^{3}=0 \quad p(1)=$

$(1)^{3}=1 p(2)=(2)^{3}=8$ (iv) $p(x)=(x-1)(x$

+1) $p(0)=(0-1)(0+1)=(-1)$

$(1)=-1 p(1)=(1-1)(1+1)=0(2)=0$

$p(2)=(2-1)(2+1)=1(3)=3$

Solution

(i) If $x=\frac{-1}{3}$ is a zero of given polynomial $p(x)=3 x+1$, then $\quad p(-\frac{1}{3})$ should be 0 .

Here, $p(\frac{-1}{3})=3(\frac{-1}{3})+1=-1+1=0$

Therefore, $x=\frac{-1}{3}$ is a zero of the given polynomial.

(ii) If $x=\frac{4}{5}$ is a zero of nolvnomial $p(x)=5 x-\pi$, then ${ }^{p(\frac{4}{5})}$ should be 0 .

Here, $p(\frac{4}{5})=5(\frac{4}{5})-\pi=4-\pi$

As $p(\frac{4}{5}) \neq 0$,

Therefore, $x=\frac{4}{5}$ is not a zero of the given polynomial.

(iii) If $x=1$ and $x=-1$ are zeroes of polynomial $p(x)=x^{2}-1$, then $p(1)$ and $p(-1)$ should be 0 .

Here, $p(1)=(1)^{2}-1=0$, and $p(-1)$

$=(-1)^{2}-1=0$

Hence, $x=1$ and -1 are zeroes of the given polynomial.

(iv) If $x=-1$ and $x=2$ are zeroes of polynomial $p(x)=(x+1)(x-2)$, then $p(-1)$ and $p(2)$ should be 0 .

Here, $p(-1)=(-1+1)(-1-2)=0(-3)=0$, and $p(2)$

$=(2+1)(2-2)=3(0)=0$

Therefore, $x=-1$ and $x=2$ are zeroes of the given polynomial.

(v) If $x=0$ is a zero of polynomial $p(x)=x^{2}$, then $p(0)$ should be zero.

Here, $p(0)=(0)^{2}=0$

Hence, $x=0$ is a zero of the given polynomial.

(vi) If $x=\frac{-m}{l}$ is a zero of polynomial $p(x)=I x+m$, then should be 0 .

Here,

$ p(\frac{-m}{l})=l(\frac{-m}{l})+m=-m+m=0 $

Therefore, $\quad x=-\frac{m}{l}$ is a zero of the given polynomial.

(vii) If $x=\frac{-1}{\sqrt{3}}$ and $\quad x=\frac{2}{\sqrt{3}}$ are zeroes of polynomial $p(x)=3 x^{2}-1$, then

$p(\frac{-m}{l})$

$p(\frac{-1}{\sqrt{3}})$ and $p(\frac{2}{\sqrt{3}})$ should be 0 .

Here, $p(\frac{-1}{\sqrt{3}})=3(\frac{-1}{\sqrt{3}})^{2}-1=3(\frac{1}{3})-1=1-1=0$, and

$p(\frac{2}{\sqrt{3}})=3(\frac{2}{\sqrt{3}})^{2}-1=3(\frac{4}{3})-1=4-1=3$

Hence, $\quad x=\frac{-1}{\sqrt{3}}$ is a zero of the given polynomial. However, $\quad x=\frac{2}{\sqrt{3}}$ is not a zero of the given polynomial.

(viii) If $\quad$ is a zero of polynomial $p(x)=2 x+1$, then $\quad$ should be 0 .

$ \begin{aligned} & x=\frac{1}{2} \\ & \text{ is a zero of polynor } \\ & (\frac{1}{2})=2(\frac{1}{2})+1=1+1=2 \end{aligned} $

As $p(\frac{1}{2}) \neq 0$,

Here, $p(\frac{1}{2})=2(\frac{1}{2})+1=1+1=2$

$ x=\frac{1}{2} \text{ is not a zero of the given polynomial. } $

Solution

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0 .

(i) $p(x)=x+5 p(x)$

$=0 x+5=0 x=-5$

Therefore, for $x$ $=-5$, the value of the polynomial is 0 and hence, $x=$ -5 is a zero of the given polynomial. (ii)

$ \begin{aligned} & p(x)=x-5 \\ & p(x)=0 x-5 \\ & =0 x=5 \end{aligned} $

Therefore, for $x=5$, the value of the polynomial is 0 and hence, $x=5$ is a zero of the given polynomial. (iii) $p(x)=2 x+5 p(x)=0$

$2 x+5=0$

$2 x=-5$

$x=-\frac{5}{2}$

Therefore, for $x=-\frac{5}{2}$, the value of the polynomial is 0 and hence, $x=\frac{-5}{2}$ is a zero of the given polynomial. (iv) $p(x)=3 x-2 p(x)=0$

$3 x-2=0$

$x=\frac{2}{3}$

Therefore, for $x=\frac{2}{3}$, the value of the polynomial is 0 and hence, $x=\frac{2}{3}$ is a zero of the given polynomial. (v) $p(x)=3 x p(x)=03 x=0 x=0$

Therefore, for $x=0$, the value of the polynomial is 0 and hence, $x=0$ is a zero of the given polynomial. (vi) $p(x)=a x p(x)=0 a x=0 x=0$

Therefore, for $x=0$, the value of the polynomial is 0 and hence, $x=0$ is a zero of the given polynomial. (vii) $p(x)=c x+d p(x)=0 c x+d=0$

$ x=\frac{-d}{c} $

Therefore, for $x=\frac{-d}{c}$, the value of the polynomial is 0 and hence, $x=\frac{-d}{c}$ is a zero of the given polynomial.