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

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Answer 1 · 2016-09-08T15:31:18

$2$ $2$

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

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

$\Rightarrow p (c) = {(c)}^{3} + 2 {(c)}^{2} + 3 (c) + 4$ $=> p(c) = (c)^(3) + 2 (c)^(2) + 3 (c) + 4$

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

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

$\Rightarrow p (- 1) = {(- 1)}^{3} + 2 {(- 1)}^{2} + 3 (- 1) + 4$ $=> p(- 1) = (- 1)^(3) + 2 (- 1)^(2) + 3 (- 1) + 4$

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

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

$\Rightarrow p (- 1) = 2$ $=> p(- 1) = 2$

How do you determine p(c) given p(x)=x^3+2x^2+3x+4p(x)=x3+2x2+3x+4p(x)=x^3+2x^2+3x+4 and c=-1?