How do you factor completely: 5x^2 + 6x − 85x2+6x−8?

2 Answers
Jim G.
May 25, 2018

(x+2)(5x-4)(x+2)(5x−4)

Explanation:

"using the a-c method to factor the quadratic"using the a-c method to factor the quadratic

"the factors of the product "5xx-8=-40the factors of the product 5×−8=−40

"which sum to + 6 are + 10 and - 4"which sum to + 6 are + 10 and - 4

"split the middle term using these factors"split the middle term using these factors

5x^2+10x-4x-8larrcolor(blue)"factor by grouping"5x2+10x−4x−8←factor by grouping

=color(red)(5x)(x+2)color(red)(-4)(x+2)=5x(x+2)−4(x+2)

"take out the "color(blue)"common factor "(x+2)take out the common factor (x+2)

=(x+2)(color(red)(5x-4))=(x+2)(5x−4)

5x^2+6x-8=(x+2)(5x-4)5x2+6x−8=(x+2)(5x−4)

KillerBunny
May 25, 2018

5(x-\frac{4}{5})(x+2)5(x−45)(x+2)

Explanation:

A quadratic equation ax^2+bx+cax2+bx+c can be factored once you know its roots x_1x1, x_2x2. So, there are three alternatives:

  • No (real) roots exist. In this case, the polynomial cannot be factorized any further
  • x_1=x_2=\hat{x}x1=x2=ˆx. In this case, the polynomial is the squared binomial a(x-\hat{x})^2a(x−ˆx)2
  • x_1 \ne x_2x1≠x2. In this case, the polynomial can be factored as a(x-x_1)(x-x_2)a(x−x1)(x−x2)

Let's look for the solutions of your equation:

x_{1,2} = \frac{-6\pm\sqrt{36+160}}{10} = \frac{-6\pm\sqrt{196}}{10} = \frac{-6\pm 14}{10}x1,2=−6±√36+16010=−6±√19610=−6±1410

So,
x_1 = \frac{-6 + 14}{10} = \frac{8}{10} = \frac{4}{5}x1=−6+1410=810=45

x_2 = \frac{-6 - 14}{10} = \frac{-20}{10} = -2x2=−6−1410=−2010=−2

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