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

2 Answers
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:
#(x+2)(2x-10)=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#, #(x+2)=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#.
#5x^2+2x+6=0#

Using the quadratic formula, we get #x=(-1+-1sqrt(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 #(x+2)# is not a factor.

Explanation:

RemainderTheorem

Given a polynomial #f(x)# and constant #a#, #(x-a)# is a factor of #f(x)# if and only if #f(a) = 0#

#color(white)()#
In our example:

#f(x) = 5x^2+2x+6#

and:

#f(-2) = 5(color(blue)(-2))^2+2(color(blue)(-2))+6#

#color(white)(f(-2)) = 20-4+6#

#color(white)(f(-2)) = 22#

Since this is not zero, #(x+2) = (x-(-2))# is not a factor of #5x^2+2x+6#