# How do you factor 5x^2-17x+14?

Jan 25, 2017

$\left(5 x - 7\right) \left(x - 2\right)$

#### Explanation:

First you would find the zeros ${x}_{1} , {x}_{2}$ of the polynomial $5 {x}^{2} - 17 x + 14$

You would apply the quadratic formula:

${x}_{1 , 2} = \frac{17 \pm \sqrt{{17}^{2} - 4 \cdot 5 \cdot 14}}{2 \cdot 5}$

$= \frac{17 \pm \sqrt{289 - 280}}{10} = \frac{17 \pm 3}{10}$

${x}_{1} = \frac{14}{10} = \frac{7}{5}$ and ${x}_{2} = \frac{20}{10} = 2$

Then you would factor the given polynomial by applyng the relation:

$a {x}^{2} + b x + c = a \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

So you get:

$5 {x}^{2} - 17 x + 14 = 5 \left(x - \frac{7}{5}\right) \left(x - 2\right) = \left(5 x - 7\right) \left(x - 2\right)$

Jan 25, 2017

You can use an AC method to find:

$5 {x}^{2} - 17 x + 14 = \left(5 x - 7\right) \left(x - 2\right)$

#### Explanation:

This quadratic can be factored using an AC method:

Find a pair of factors of $A C = 5 \cdot 15 = 70$ with sum $B = 17$

The pair $10 , 7$ works.

Use this pair to split the middle term and factor by grouping:

$5 {x}^{2} - 17 x + 14 = 5 {x}^{2} - 10 x - 7 x + 14$

$\textcolor{w h i t e}{5 {x}^{2} - 17 x + 14} = \left(5 {x}^{2} - 10 x\right) - \left(7 x - 14\right)$

$\textcolor{w h i t e}{5 {x}^{2} - 17 x + 14} = 5 x \left(x - 2\right) - 7 \left(x - 2\right)$

$\textcolor{w h i t e}{5 {x}^{2} - 17 x + 14} = \left(5 x - 7\right) \left(x - 2\right)$