# How do you factor 15b^2 - 7b - 2?

Mar 27, 2018

$\left(3 b - 2\right) \left(5 b + 1\right)$

#### Explanation:

We have a quadratic expression in the form

$a {x}^{2} + b x + c$

$a = 15$
$b = - 7$
$c = - 2$

To factor this, first find two numbers that multiply to give $a c$ and add to give $b$.

For $a c = - 30$ and $b = - 7$

Maybe you can see by inspection that these two numbers are:

$- 10$ and $3$

How did I see that? Look at the factors of $30$

$1 , 2 , 3 , 5 , 6 , 10 , 15 , 30$

Now try adding or subtracting them to get $- 7$

Remember that one of them (and ONLY one of them) must be negative so that we get $- 30$ when we multiply them.

Now it's much easier to see that $- 10$ and $3$ are the numbers we want.

So why did we do this? Our original expression was

$15 {b}^{2} - 7 b - 2$

Let's now split the middle term using the numbers we just found.

$\Rightarrow 15 {b}^{2}$ $\textcolor{b l u e}{- 10 b}$ $\textcolor{red}{+ 3 b} - 2$

Now factor $5 b$ from the first two terms.

$5 b \left(3 b - 2\right) + \left(3 b - 2\right)$

Notice now that we have a common factor of $\left(3 b - 2\right)$.
Let's factor it out:

$\left(3 b - 2\right) \left(5 b + 1\right)$

And we're done!

NOTE:

It still would have worked if we had chosen to split the terms in the opposite order. Let's check:

$\Rightarrow 15 {b}^{2}$ $\textcolor{red}{+ 3 b}$ $\textcolor{b l u e}{- 10 b} - 2$

$\Rightarrow 3 b \left(5 b + 1\right) - 2 \left(5 b + 1\right)$

$\Rightarrow \left(5 b + 1\right) \left(3 b - 2\right)$