# How do you factor 3n^2 - 8n + 4?

May 19, 2015

$3 {n}^{2} - 8 n + 4 = \left(3 n - 2\right) \left(n - 2\right)$

How did I find this?

We are looking for factors in the form $\left(a n + b\right) \left(c n + d\right)$

$\left(a n + b\right) \left(c n + d\right) = a c {n}^{2} + \left(b c + a d\right) n + b d$

Comparing this with our original quadratic we find:
$a c = 3$
$b c + a d = - 8$
$b d = 4$

If $a$, $b$, $c$ and $d$ are integers then we are looking for integers that satisfy these equations.

$a c = 3$ has possible integer solutions:

$a = 3$, $c = 1$
$a = - 3$, $c = - 1$
$a = 1$, $c = 3$
$a = - 1$, $c = - 3$

We might as well pick the first of these, they will ultimately give very similar factorisations.

Let $a = 3$, $c = 1$

$b d$ > 0. So $b$ and $d$ are either both positive or both negative. If they were both positive, then $b c + a d = b + 3 d$ would be positive too, which it isn't. So both $b$ and $d$ are negative.

Notice that $b + 3 d = - 8$ is even. So either $b$ and $d$ are both odd or both even.

So we are looking for two negative whole numbers $b$ and $d$ such that $b d = 4$ and $b$ and $d$ are both even.

That only leaves the possibility $b = d = - 2$

May 19, 2015

$3 {n}^{2} - 8 n + 4$

We can Split the Middle Term of this expression to factorise it
In this technique, if we have to factorise an expression like $a {n}^{2} + b n + c$, we need to think of 2 numbers such that:

${N}_{1} \cdot {N}_{2} = a \cdot c = 3 \times 4 = 12$

and

${N}_{1} + {N}_{2} = b = - 8$
After trying out a few numbers we get ${N}_{1} = - 2$ and ${N}_{2} = - 6$
$- 2 \times - 6 = 12$ and $\left(- 2\right) + \left(- 6\right) = - 8$

$3 {n}^{2} - 8 n + 4 = 3 {n}^{2} - 6 n - 2 n + 4$

$= 3 n \left(n - 2\right) - 2 \left(n - 2\right)$

$= \textcolor{g r e e n}{\left(3 n - 2\right) \left(n - 2\right)}$