Question

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

Answer 1

`(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)`

Answer 2

`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`