Question:
If $p(x)=x^{2}-4 x+3$, then evaluate $p(2)-p(-1)+p(1 / 2)$.
Solution:
Given, $p(x)=x^{2}-4 x+3$
Now, $p(2)=(2)^{2}-4 \times 2+3=4-8+3=-1$
$p(-1)=(-1)^{2}-4(-1)+3=1+4+3=8$
and $p\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^{2}-4 \times \frac{1}{2}+3$
$=\frac{1}{4}-2+3=\frac{1-8+12}{4}=\frac{5}{4}$
$\therefore$ $p(2)-p(-1)+p\left(\frac{1}{2}\right)=-1-8+\frac{5}{4}$
$=-9+\frac{5}{4}=\frac{-36+5}{4}=\frac{-31}{4}$