Question

If $p\left(x\right)=x^2–4x+3$, then evaluate $p\left(2\right)-p(-1)+p\left(\frac{1}{2}\right)$.

Answer 1

Given $p\left(x\right)=x^2-4x+3$
Now, $p\left(2\right)=(2{)}^2-4\times 2+3=4-8+3=-1$
$p(-1)=(-1{)}^2-4(-1)+3=1+4+3=8$
And $p\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^2-4\times \frac{1}{2}+3$
$=\frac{1}{4}-2+3=\frac{1-8+12}{4}=\frac{5}{4}$
$\therefore p\left(2\right)-p(-1)+p\left(\frac{1}{2}\right)=-1-8+\frac{5}{4}$
$=-9+\frac{5}{4}=\frac{-36+5}{4}=\frac{-31}{4}$