# How do you factor completely 9x^2+16 ?

Dec 2, 2015

$9 {x}^{2} + 16$ has no simpler factors with Real coefficients, but it does have factors with Complex coefficients:

$9 {x}^{2} + 16 = \left(3 x - 4 i\right) \left(3 x + 4 i\right)$

#### Explanation:

If $x \in \mathbb{R}$ then ${x}^{2} \ge 0$, hence $9 {x}^{2} + 16 \ge 16 > 0$ for all $x \in \mathbb{R}$. Hence $9 {x}^{2} + 16$ has no linear factors with Real coefficients.

The difference of squares identity can be written:

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

If we let $a = 3 x$ and $b = 4 i$, then ${a}^{2} = 9 {x}^{2}$ and ${b}^{2} = 16 {i}^{2} = - 16$.

So:

$9 {x}^{2} + 16 = {\left(3 x\right)}^{2} - {\left(4 i\right)}^{2} = \left(3 x - 4 i\right) \left(3 x + 4 i\right)$