# How do you determine p(c) given p(x)=x^3+2x^2+3x+4 and c=-1?

Sep 8, 2016

$2$

#### Explanation:

We have: $p \left(x\right) = {x}^{3} + 2 {x}^{2} + 3 x + 4$; $c = - 1$

First, let's substitute $x$ with $c$:

$\implies p \left(c\right) = {\left(c\right)}^{3} + 2 {\left(c\right)}^{2} + 3 \left(c\right) + 4$

Then, the value of $c$ is given as $- 1$.

Let's substitute $c$ with $- 1$:

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

$\implies p \left(- 1\right) = - 1 + 2 - 3 + 4$

$\implies p \left(- 1\right) = 1 + 1$

$\implies p \left(- 1\right) = 2$