# How do you determine p(c) given p(x)=8x^3+12x^2+6x+1 and c=-1/2?

Aug 20, 2016

0

#### Explanation:

The first step is to obtain p(c). To do this substitute x = c into p(x).

$p \left(\textcolor{b l u e}{c}\right) = 8 {\left(\textcolor{b l u e}{c}\right)}^{3} + 12 {\left(\textcolor{b l u e}{c}\right)}^{2} + 6 \left(\textcolor{b l u e}{c}\right) + 1 = 8 {c}^{3} + 12 {c}^{2} + 6 c + 1$

We now have to evaluate p(c) when c $= - \frac{1}{2}$

$p \left(\textcolor{red}{- \frac{1}{2}}\right) = 8 {\left(\textcolor{red}{- \frac{1}{2}}\right)}^{3} + 12 {\left(\textcolor{red}{- \frac{1}{2}}\right)}^{2} + 6 \left(\textcolor{red}{- \frac{1}{2}}\right) + 1$

$\Rightarrow p \left(- \frac{1}{2}\right) = - 1 + 3 - 3 + 1 = 0$