Question

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

Answer 1

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

$\therefore p\left(2\right)-p(-1)+p\left(\frac{1}{2}\right)$

$=\left[\right(2{)}^2-4\times \left(2\right)+3]-[(-1{)}^2-4\times (1)+3]+\left[\right(-\frac{1}{2}{)}^2-4\times \frac{1}{2}+3]$

$=[4-8+3]-[1+4+3]+[\frac{1}{4}-2+3]$

$=-1-8+\frac{5}{4}$

$=-\frac{31}{4}$

Hence, the value of $p\left(2\right)-p(-1)+p\left(\frac{1}{2}\right)is-\frac{31}{4}$.