# How do you find all zeros of the function f(x)=8x^2+53x-21?

May 30, 2016

$x = \frac{3}{8}$ and $x = - 7$

#### Explanation:

We can use an AC method to factor the quadratic.

Find a pair of factors of $A C = 8 \cdot 21 = 168$ which differ by $53$.

The pair $56 , 3$ works in that $56 \cdot 3 = 168$ and $56 - 3 = 53$.

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

$8 {x}^{2} + 53 x - 21$

$= 8 {x}^{2} + 56 x - 3 x - 21$

$= \left(8 {x}^{2} + 56 x\right) - \left(3 x + 21\right)$

$= 8 x \left(x + 7\right) - 3 \left(x + 7\right)$

$= \left(8 x - 3\right) \left(x + 7\right)$

Hence the zeros of $f \left(x\right)$ are $x = \frac{3}{8}$ and $x = - 7$