# How do you determine whether x+2 is a factor of the polynomial 2x^3+2x^2-x-6?

May 31, 2018

You divide using polynomial long division or synthetic division and if the remainder is 0 then it is a factor.

#### Explanation:

$2 {x}^{3} + 2 {x}^{2} - x - 6 \div i \mathrm{de} x + 2$

this division has a remainder of -12 so it is not a factor.

May 31, 2018

Just check if $x = - 2$ is a root of polynomial. See below

#### Explanation:

We know that if a polynomial $P \left(x\right)$ has a root for $x = a$, then $P \left(a\right) = 0$ and $x - a$ divides to $P \left(x\right)$ (this is the same that $x - a$ is a factor of $P \left(x\right)$

Lets see.

P(-2)=2·(-2)^3+2·(-2)^2-(-2)-6=-16+8+2-6=-12!=0

Then $x + 2$ is not a factor of $P \left(x\right)$

May 31, 2018

$x + 2$ not a factor of $f \left(x\right)$

#### Explanation:

Let $f \left(x\right) = 2 {x}^{3} + 2 {x}^{2} - x - 6$

We can use the Polynomial Remainder Theorem, which states when a polynomial, $f \left(x\right)$ is divided by $x - c$, the remainder is $f \left(c\right)$.

We are dividing $f \left(x\right)$ by $x + 2$, so $c = - 2$. Now, let's input this into $f \left(x\right)$. We get

$2 {\left(- 2\right)}^{3} + 2 {\left(- 2\right)}^{2} - \left(- 2\right) - 6$

$\implies 2 \left(- 8\right) + 2 \left(4\right) + 2 - 6$

$\implies - 16 + 8 + 2 - 6$

$\implies \textcolor{b l u e}{- 12}$

We have a remainder, which means $x + 2$ is not a factor of $f \left(x\right)$.

If $f \left(c\right)$ simplified to $0$, we would have no remainder, and $x + 2$ would be a factor, but since we have a remainder, $x + 2$ is not a factor.

Hope this helps!