How do you factor completely 64x^4-9 ?

Aug 2, 2016

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

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

$= \left(2 \sqrt{2} x - \sqrt{3}\right) \left(2 \sqrt{2} x + \sqrt{3}\right) \left(8 {x}^{2} + 3\right)$

$= \left(2 \sqrt{2} x - \sqrt{3}\right) \left(2 \sqrt{2} x + \sqrt{3}\right) \left(2 \sqrt{2} x - \sqrt{3} i\right) \left(2 \sqrt{2} x + \sqrt{3} i\right)$

Explanation:

The difference of squares identity can be written:

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

We will use this three times below.

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

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

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

If we allow irrational coefficients then we can factor this further:

$= \left({\left(2 \sqrt{2} x\right)}^{2} - {\left(\sqrt{3}\right)}^{2}\right) \left(8 {x}^{2} + 3\right)$

$= \left(2 \sqrt{2} x - \sqrt{3}\right) \left(2 \sqrt{2} x + \sqrt{3}\right) \left(8 {x}^{2} + 3\right)$

That is as far as we can go with Real coefficients, since $8 {x}^{2} + 3 > 0$ for all Real values of $x$. If we allow Complex coefficients then we can factor this further:

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

$= \left(2 \sqrt{2} x - \sqrt{3}\right) \left(2 \sqrt{2} x + \sqrt{3}\right) \left(2 \sqrt{2} x - \sqrt{3} i\right) \left(2 \sqrt{2} x + \sqrt{3} i\right)$