# How do you multiply (8x^2+3)(8x^2-3)?

Aug 4, 2016

$64 {x}^{4} - 9$

#### Explanation:

$\textcolor{b l u e}{\text{Method 1}}$

$\textcolor{red}{\left(8 {x}^{2} + 3\right) \textcolor{p u r p \le}{\left(8 {x}^{2} - 3\right)}}$

Multiply everything inside one bracket by everything inside the other.

$\textcolor{p u r p \le}{\textcolor{red}{8 {x}^{2}} \left(8 {x}^{2} - 3\right) \text{ } \textcolor{red}{+ 3} \left(8 {x}^{2} - 3\right)}$

$64 {x}^{4} - 24 {x}^{2} \text{ } + 24 {x}^{2} - 9$

$64 {x}^{4} + 0 - 9$

$64 {x}^{4} - 9$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Method 2}}$
The following is a general equation that is worth committing to memory:

Consider the general case of: ${a}^{2} + {b}^{2} = \left(a + b\right) \left(a - b\right)$

Think of $\left(8 {x}^{2} + 3\right) \left(8 {x}^{2} - 3\right)$ as of form $\left(a + b\right) \left(a - b\right)$

This gives:$\text{ "[(8x^2)^2-3^3] = 64x^4-9 larr" a 1 line solution}$

'..................................................................................
$\textcolor{b r o w n}{\text{If this is still not clear then let "u=8x^2" giving:}}$

$\left(u + 3\right) \left(u - 3\right) = \left({u}^{2} - {3}^{2}\right)$

But $u = 8 {x}^{2}$ giving

$\left[{\left(8 {x}^{2}\right)}^{2} - {3}^{2}\right]$

$= 64 {x}^{4} - 9$