Question

Let $p\left(x\right)=x^2-4x+3$ find the value of $p\left(0\right),p\left(1\right),P\left(2\right),p\left(3\right)$ and obtain zeroes of the polynomial $p\left(x\right)$.

Answer 1

Let $p\left(x\right)=x^2-4x+3$

$p\left(0\right)=0^2-4\left(0\right)+3=0+0+3=3$
$p\left(1\right)=1^2-4\left(1\right)+3=1-4+3=0$

$p\left(2\right)=2^2-4\left(2\right)+3=4-8+3=-1$

$p\left(3\right)=3^2-4\left(3\right)+3=9-12+3=0$

We observe that $P\left(1\right)$ and $p\left(3\right)$ are zeroes of the polynomial.