# How do you factor 4x^2 + 9?

Jan 15, 2017

#### Answer:

There are no simpler factors with Real coefficients, but:

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

#### Explanation:

Note that ${x}^{2} \ge 0$ for any Real value of $x$.

Hence $4 {x}^{2} + 9 \ge 9$ for all Real values of $x$.

So $4 {x}^{2} + 9$ has no linear factors with Real coefficients.

It is possible to factor it with Complex coefficients.

The difference of squares identity can be written:

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

In conjunction with ${i}^{2} = - 1$, we find:

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