Question

How do you factor $4{\mathrm{x}}^4-64$?

Answer 1

Step-1: Take $4$out of the factor:

Factor ${\displaystyle 4}$ out of $4{\mathrm{x}}^4$ is written as,

$=4\left({\mathrm{x}}^4\right)-64$

Factor ${\displaystyle 4}$ out of ${\displaystyle -64}$ is written as,

$=4{\mathrm{x}}^4+4\cdot -16$

Factor ${\displaystyle 4}$ out of $4{\mathrm{x}}^4+4\cdot -16$ is written as,

$=4({\mathrm{x}}^4-16)$

${\mathrm{x}}^4$ is rewritten as ${\left({\mathrm{x}}^2\right)}^2$. Therefore,

$=4({\left({\mathrm{x}}^2\right)}^2-16)$

${\displaystyle 16}$ is rewritten as $4^2$.

Therefore,

$=4({\left({\mathrm{x}}^2\right)}^2-4^2)$

Step-2: Apply the difference of squares formula:

Since both terms are perfect squares, factor using the difference of squares formula, ${\mathrm{a}}^2-{\mathrm{b}}^2=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})$.

Here, consider $\mathrm{a}={\mathrm{x}}^2$ and $\mathrm{b}=4$.

$4\left(\right({\mathrm{x}}^2+4\left)\right({\mathrm{x}}^2-4\left)\right)$

$4\left(\right({\mathrm{x}}^2+4\left)\right(\mathrm{x}+2\left)\right(\mathrm{x}-2\left)\right)$ $\left[{\mathrm{a}}^2-{\mathrm{b}}^2=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})\right]$

Hence, the factor of $4{\mathrm{x}}^4-64$ is calculated as $4({\mathrm{x}}^2+4)(\mathrm{x}+2)(\mathrm{x}-2)$.