# How do you determine whether x+2 is a factor of the polynomial 5x^2+2x+6?

May 24, 2017

You see if $- 2$ is a solution to the polynomial set to 0.

#### Explanation:

Let's suppose you have this factorized polynomial:
$\left(x + 2\right) \left(2 x - 10\right) = 0$

A property of this factorized polynomial is that each of it's factors when set to 0 is the root to this polynomial.
For example, when $x = - 2$, $\left(x + 2\right) = 0$ so $- 2$ is the solution to this polynomial.
This property arises from the property in which when $0$ is multiplied with any number, the product is always going to be $0$.

Now, back to your polynomial. Let's set it to 0 first and solve for $x$.
$5 {x}^{2} + 2 x + 6 = 0$

Using the quadratic formula, we get $x = \frac{- 1 \pm 1 \sqrt{29} i}{5}$

So, $x + 2$ is not a factor of this equation. We can also evaluate whether $x = - 2$ will yield $0$, and it does not.

May 24, 2017

Use the remainder theorem to find $\left(x + 2\right)$ is not a factor.

#### Explanation:

RemainderTheorem

Given a polynomial $f \left(x\right)$ and constant $a$, $\left(x - a\right)$ is a factor of $f \left(x\right)$ if and only if $f \left(a\right) = 0$

$\textcolor{w h i t e}{}$
In our example:

$f \left(x\right) = 5 {x}^{2} + 2 x + 6$

and:

$f \left(- 2\right) = 5 {\left(\textcolor{b l u e}{- 2}\right)}^{2} + 2 \left(\textcolor{b l u e}{- 2}\right) + 6$

$\textcolor{w h i t e}{f \left(- 2\right)} = 20 - 4 + 6$

$\textcolor{w h i t e}{f \left(- 2\right)} = 22$

Since this is not zero, $\left(x + 2\right) = \left(x - \left(- 2\right)\right)$ is not a factor of $5 {x}^{2} + 2 x + 6$