Question

If $p\left(x\right)=3x^2-4x+7$,find $p\left(\frac{1}{2}\right)$ and $p(-2)$

Answer 1

Given,

The polynomial $p\left(x\right)=3x^2-4x+7$

For $x=\frac{1}{2}$,

$p\left(\frac{1}{2}\right)=3{\left(\frac{1}{2}\right)}^2-4\left(\frac{1}{2}\right)+7$

$=\frac{3}{4}-2+7$

$=\frac{3-8+28}{4}$

$=\frac{23}{4}$

Again, for $x=-2$,

$p(-2)=3{(-2)}^2-4(-2)+7$

$=12+8+7$

$=27$.