# How do you factor examples b), e) and f) below?

## b) $y = 8 {x}^{2} - 10 x - 3$ e) $y = 5 {x}^{3} - 80 x$ f) $y = 36 {x}^{2} - 1$

##### 1 Answer
Jan 17, 2017

b) $y = 8 {x}^{2} - 10 x - 3 = \left(4 x + 1\right) \left(2 x - 3\right)$

e) $y = 5 {x}^{3} - 80 x = 5 x \left(x - 4\right) \left(x + 4\right)$

f) $y = 36 {x}^{2} - 1 = \left(6 x - 1\right) \left(6 x + 1\right)$

#### Explanation:

Example b)

$y = 8 {x}^{2} - 10 x - 3$

This quadratic is in the form $a {x}^{2} + b x + c$, with $a = 8$, $b = - 10$ and $c = - 3$.

In order to tell whether it factors 'nicely', let us first check the discriminant:

$\Delta = {b}^{2} - 4 a c = {\left(- 10\right)}^{2} - 4 \left(8\right) \left(- 3\right) = 100 + 96 = 196 = {14}^{2}$

Since this is a perfect square, the given quadratic will factor exactly.

Let's use an AC method to find the factors:

Look for a pair of factors of $A C = 8 \cdot 3 = 24$ which differ by $10$.

The pair $12 , 2$ works.

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

$y = 8 {x}^{2} - 10 x - 3$

$\textcolor{w h i t e}{y} = 8 {x}^{2} - 12 x + 2 x - 3$

$\textcolor{w h i t e}{y} = \left(8 {x}^{2} - 12 x\right) + \left(2 x - 3\right)$

$\textcolor{w h i t e}{y} = 4 x \left(2 x - 3\right) + 1 \left(2 x - 3\right)$

$\textcolor{w h i t e}{y} = \left(4 x + 1\right) \left(2 x - 3\right)$

$\textcolor{w h i t e}{}$
Example e)

$y = 5 {x}^{3} - 80 x$

For this example, we can separate out the common factor $5 x$, then use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = x$ and $b = 4$ as follows:

$y = 5 {x}^{3} - 80 x$

$\textcolor{w h i t e}{y} = 5 x \left({x}^{2} - 16\right)$

$\textcolor{w h i t e}{y} = 5 x \left({x}^{2} - {4}^{2}\right)$

$\textcolor{w h i t e}{y} = 5 x \left(x - 4\right) \left(x + 4\right)$

$\textcolor{w h i t e}{}$
Example f)

$y = 36 {x}^{2} - 1$

Notice that both the terms are perfect squares, so we can use the difference of squares identity again...

$y = 36 {x}^{2} - 1 = {\left(6 x\right)}^{2} - {1}^{2} = \left(6 x - 1\right) \left(6 x + 1\right)$