Question

If $p\left(x\right)=5x^3-3x^2+7x-9$, find (i) p(-1) (ii) p(2).

Answer 1

Given that $p\left(x\right)=5x^3-3x^2+7x-9$
(i) $p(-1)=5(-1{)}^3-3(-1{)}^2+7(-1)-9=-5-3-7-9$
$\therefore p(-1)=-24$
(ii) $p\left(2\right)=5(2{)}^3-3(2{)}^2+7\left(2\right)-9=40-12+14-9$
$\therefore p\left(2\right)=3$