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

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Answer 1 · 2016-09-08T15:31:18

$2$

We have: $p(x) = x^(3) + 2 x^(2) + 3 x + 4$ ; $c = - 1$

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

$=> p(c) = (c)^(3) + 2 (c)^(2) + 3 (c) + 4$

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

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

$=> p(- 1) = (- 1)^(3) + 2 (- 1)^(2) + 3 (- 1) + 4$

$=> p(- 1) = - 1 + 2 - 3 + 4$

$=> p(- 1) = 1 + 1$

$=> p(- 1) = 2$

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