# How do you factor #5x^{2} + 11x + 7#?

##### 2 Answers

Use the discriminant to determine whether or not there are real roots, then use the quadratic formula to solve for those roots

#### Explanation:

The discriminant is the part of the quadratic formula under the square root

sub in the terms from the standard form of the equation

ex.

so in this case

If the discriminant is positive there are 2 real roots, if the discriminant equals 0 there is one real root, and if the discriminant is negative there are no real roots

Since the discriminant is negative, there are no real roots and you cannot factor this equation

#### Explanation:

This quadratic has negative discriminant, so can only be factored using Complex coefficients.

#Delta = b^2-4ac = 11^2-4(5)(7) = 121-140 = -19#

Multiply by

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We use this with

#20(5x^2+11x+7) = 100x^2+220x+140#

#color(white)(20(5x^2+11x+7)) = (10x)^2+2(11)(10x)+121+19#

#color(white)(20(5x^2+11x+7)) = (10x+11)^2+19#

#color(white)(20(5x^2+11x+7)) = (10x+11)^2-(sqrt(19)i)^2#

#color(white)(20(5x^2+11x+7)) = ((10x+11)-sqrt(19)i)((10x+11)+sqrt(19)i)#

#color(white)(20(5x^2+11x+7)) = (10x+11-sqrt(19)i)(10x+11+sqrt(19)i)#

So:

#5x^2+11x+7 = 1/20 (10x+11-sqrt(19)i)(10x+11+sqrt(19)i)#