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

May 25, 2018

$\left(x + 2\right) \left(5 x - 4\right)$

Explanation:

$\text{using the a-c method to factor the quadratic}$

$\text{the factors of the product } 5 \times - 8 = - 40$

$\text{which sum to + 6 are + 10 and - 4}$

$\text{split the middle term using these factors}$

$5 {x}^{2} + 10 x - 4 x - 8 \leftarrow \textcolor{b l u e}{\text{factor by grouping}}$

$= \textcolor{red}{5 x} \left(x + 2\right) \textcolor{red}{- 4} \left(x + 2\right)$

$\text{take out the "color(blue)"common factor } \left(x + 2\right)$

$= \left(x + 2\right) \left(\textcolor{red}{5 x - 4}\right)$

$5 {x}^{2} + 6 x - 8 = \left(x + 2\right) \left(5 x - 4\right)$

May 25, 2018

$5 \left(x - \setminus \frac{4}{5}\right) \left(x + 2\right)$

Explanation:

A quadratic equation $a {x}^{2} + b x + c$ 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
• ${x}_{1} = {x}_{2} = \setminus \hat{x}$. In this case, the polynomial is the squared binomial $a {\left(x - \setminus \hat{x}\right)}^{2}$
• ${x}_{1} \setminus \ne {x}_{2}$. In this case, the polynomial can be factored as $a \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

Let's look for the solutions of your equation:

${x}_{1 , 2} = \setminus \frac{- 6 \setminus \pm \setminus \sqrt{36 + 160}}{10} = \setminus \frac{- 6 \setminus \pm \setminus \sqrt{196}}{10} = \setminus \frac{- 6 \setminus \pm 14}{10}$

So,
${x}_{1} = \setminus \frac{- 6 + 14}{10} = \setminus \frac{8}{10} = \setminus \frac{4}{5}$

${x}_{2} = \setminus \frac{- 6 - 14}{10} = \setminus \frac{- 20}{10} = - 2$