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

2 Answers
Jim G.
May 25, 2018

#(x+2)(5x-4)#

Explanation:

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

#"the factors of the product "5xx-8=-40#

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

#"split the middle term using these factors"#

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

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

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

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

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

KillerBunny
May 25, 2018

#5(x-\frac{4}{5})(x+2)#

Explanation:

A quadratic equation #ax^2+bx+c# can be factored once you know its roots #x_1#, #x_2#. 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}#. In this case, the polynomial is the squared binomial #a(x-\hat{x})^2#
  • #x_1 \ne x_2#. In this case, the polynomial can be factored as #a(x-x_1)(x-x_2)#

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}#

So,
#x_1 = \frac{-6 + 14}{10} = \frac{8}{10} = \frac{4}{5}#

#x_2 = \frac{-6 - 14}{10} = \frac{-20}{10} = -2#

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