# How do you solve 3x^2 + 9x = 12?

Apr 4, 2016

$\text{ } \left(3 x - 3\right) \left(x + 4\right)$

With practise these become easier to solve

#### Explanation:

Given: $\text{ } 3 {x}^{2} + 9 x = 12$

Write as:

$\text{ } 3 {x}^{2} + 9 x - 12 = 0$

We are looking for (?x+?)(?x+?)

The only whole number factors of 12 are
$\textcolor{b l u e}{\text{ } 1 , 12}$
$\textcolor{b l u e}{\text{ } 2 , 6}$
$\textcolor{b l u e}{\text{ } 3 , 4}$

This means we can only have one of:
" "(?x+1)(?x+12)
" "(?x+2)(?x+6)
" "(?x+3)(?x+4)

'................................................................

The only whole number factors of 3 are $\textcolor{m a \ge n t a}{1 , 3}$

So the ?x can only be these numbers giving:

" "(color(magenta)(1)x+?)(color(magenta)(3)x+?) -> (x+?)(3x+?)
'..............................................................

It is now a matter of trying them out and seeing what works.

I spot that from the factors of $\textcolor{m a \ge n t a}{3}$ if we have:

$\textcolor{m a \ge n t a}{\left(3 \textcolor{b l u e}{\times 4}\right) - \left(1 \textcolor{b l u e}{\times 3}\right)}$

we have:$\text{ } 12 - 3 = 9$ which will give us $+ 9 x$

So the factorisation must be

$\text{ } \left(\textcolor{m a \ge n t a}{3 x} \textcolor{b l u e}{- 3}\right) \left(\textcolor{m a \ge n t a}{x} \textcolor{b l u e}{+ 4}\right)$

'~~~~~~~~~~~~~~~~~~~~~
Check

$\text{ } \left(3 x - 3\right) \left(x + 4\right)$

$\text{ "=" } 3 {x}^{2} + 12 x - 3 x - 12$
$\text{ "=" } 3 {x}^{2} + 9 x - 12$