Question

(1) Let p(x)=x^2- 4x +3. Find the value of p(0), p(1),p(2),p(3) and obtain zeroes of the
polynomial p(x).
​

Answer 1

MARK AS BRAINLIST

Hey there for the first part just put the values

0,1,2,3 in place of the x and solve

For second i am doing

$px = {x}^{2} - 4x + 3$

Here

A=1

B=-4

C=3

Now we factorise it

Into

This

Ac=3

Now middle term splitting

${x}^{2} - 3x - 1x + 3$

Now taking common

X(x-3)-1(x-3

Now

(x-1)(x-3)

So from here

Alpha = 1

And

Beta= 3

Answer 2

⠀⠀⠀$\huge\underline{ \mathrm{ \red{QueS{\pink{tiOn}}}}}$

If p(x)=x^2- 4x +3. Find the value of p(0), p(1) , p(2) , p(3) and obtain zeroes of the polynomial p(x).

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⠀⠀⠀$\huge{ \underline{ \purple{ \bold{ \underline{ \mathrm{ExPlanA{\green{TiOn }}}}}}}}$

$\large\underline{ \underline{ \red{ \bf {GivEen:-}}}}$$\bf\:{\red{p(x)={x}^{2}-4x+3}}$

⠀⠀⠀⠀⠀$\large\bf{ \pink{{To \:find}\begin{cases}\textsf{p(1)}\\\textsf{p(2)}\\\textsf{p(3)}\end{cases}}}$

⠀⠀⠀⠀⠀$\huge\underline{ \underline{ \orange{ \bold{sOluTiOn}}}}$

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀$\bf\:\underbrace{when \:x =1}$

➩⠀⠀⠀⠀⠀$\bf\:p(1)={1}^{2}-4\times1+3$

➩⠀⠀⠀⠀⠀$\bf\:p(1)=1-4+3$

➩⠀⠀⠀⠀⠀$\bf\:p(1)=-3+3=0$

⠀⠀⠀⠀⠀⠀⠀⠀$\bf\:\underbrace{when \:x =2}$

➩⠀⠀⠀⠀⠀$\bf\:p(2)={2}^{2}-4\times2+3$

➩⠀⠀⠀⠀⠀$\bf\:p(2)=4-8+3$

➩⠀⠀⠀⠀⠀$\bf\:p(2)= -1$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\bf\:\underbrace{when \:x =3}$

➩⠀⠀⠀⠀⠀$\bf\:p(3)={3}^{2}-4\times3+3$

➩⠀⠀⠀⠀⠀$\bf\:p(3)=9-12+3$

➩⠀⠀⠀⠀⠀$\bf\:p(3)=12-12$

➩⠀⠀⠀⠀⠀$\bf\:p(3)=0$

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⠀⠀⠀⠀⠀zeroes of polynomial

⠀⠀⠀⠀⠀$\bf\:p(x)={x}^{2}-4x+3$ :-

➩⠀⠀⠀⠀⠀$\bf\:{x}^{2}-x-3x+3$

➩⠀⠀⠀⠀⠀$\bf\:x(x-1)-3(x-1)$

➩⠀⠀⠀⠀⠀$\bf\:(x-1)(x-3)$

➩⠀⠀⠀⠀⠀$\underline{\boxed{\pink{\bf{\:x=1\: and\: x=3}}}}$