Question

Find $p\left(0\right),p\left(1\right)$ and $p\left(2\right)$ for $p\left(x\right)=x^2-3x+2$ :

• A. $2,0,0$
• B. $-2,0,2$
• C. $3,2,0$
• D. $0,01$

Answer 1

The correct option is A $2,0,0$

Given polynomial is $p\left(x\right)=x^2-3x+2$

Replace $x$ by $0$, we get

$p\left(0\right)=(0{)}^2-3(0)+2$

$\Rightarrow p\left(0\right)=0+0+2$

$\Rightarrow p\left(0\right)=2$

Replace $x$ by $1$, we get

$p\left(1\right)=(1{)}^2-3(1)+2$

$\Rightarrow p\left(1\right)=1-3+2$

$\Rightarrow p\left(1\right)=0$

Replace $x$ by $2$, we get

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

$\Rightarrow p\left(2\right)=4-6+2$

$\Rightarrow p\left(2\right)=0$