# How do you factor the expression 64x^2 + 81?

Feb 27, 2016

$\left(8 x + 9\right) \left(8 x - 9\right)$

Feb 27, 2016

$64 {x}^{2} + 81 = \left(8 x - 9 i\right) \left(8 x + 9 i\right)$

#### Explanation:

If $x$ is any Real number then ${x}^{2} \ge 0$. Hence $64 {x}^{2} + 81 \ge 81 > 0$.

As a result $64 {x}^{2} + 81$ has no linear factors with Real coefficients.

We can do something with Complex coefficients...

First notice that $64 {x}^{2} = {\left(8 x\right)}^{2}$ and $81 = {9}^{2}$

The difference of squares identity can be written:

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

The imaginary unit $i$ has the property that ${i}^{2} = - 1$.

Hence we find:

$64 {x}^{2} + 81$

$= {\left(8 x\right)}^{2} + {9}^{2}$

$= {\left(8 x\right)}^{2} - {\left(9 i\right)}^{2}$

$= \left(8 x - 9 i\right) \left(8 x + 9 i\right)$

If we want to, we can write a "sum of squares" identity:

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

Feb 27, 2016

$64 {x}^{2} + 81 = \left(8 x - 9 i\right) \left(8 x + 9 i\right)$

(there are no factors with only Real components)

#### Explanation:

Remember that
$\textcolor{w h i t e}{\text{XXX}} \left({\textcolor{red}{a}}^{2} - {\textcolor{b l u e}{b}}^{2}\right)$ can be factored as $\left(\textcolor{red}{a} - \textcolor{b l u e}{b}\right) \left(\textcolor{red}{a} + \textcolor{b l u e}{b}\right)$

$64 {x}^{2} + 81 = \left({\textcolor{red}{\left(8 x\right)}}^{2} - {\textcolor{b l u e}{\left(9 i\right)}}^{2}\right)$

$\textcolor{w h i t e}{\text{XXXXxX}} = \left(\textcolor{red}{8 x} - \textcolor{b l u e}{9 i}\right) \left(\textcolor{red}{8 x} + \textcolor{b l u e}{9 i}\right)$

If there were any factors with only Real components the equation
$\textcolor{w h i t e}{\text{XXX}} 64 {x}^{2} + 81 = 0$ would have Real solutions;
however we can tell from the fact that the discriminant
color(white)("XXX")(b^2-4ac) = (0^2-4(64)(81)) is less than zero that there are no Real roots.