Answer
a. $P(x)=(x+2)(x-2)(x^2+2x+4)(x^2-2x+4)$,
b. $P(x)=(x-2)(x+2)(x+1+\sqrt 3i)(x+1-\sqrt 3i)(x-1-\sqrt 3i)(x-1+\sqrt 3i)$
Work Step by Step
$P(x)=x^6-64$,
a) Factor the polynomial into linear and irreducible quadratic factors:
Use the difference of squares formula:
$a^2-b^2=(a+b)(a-b)$
$P(x)=(x^3)^2-8^2=(x^3+8)(x^3-8)$
Use the difference of cubes formula:
$a^3-b^3=(a-b)(a^2+ab+b^2)$
$P(x)=(x^3+2^3)(x^3-2^3)=(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)$.
We got:
$P(x)=(x+2)(x-2)(x^2+2x+4)(x^2-2x+4)$
b. Factor $P$ completely:
$x^2+2x+4=0\Rightarrow x=-1\pm\sqrt 3i$
$x^2-2x+4=0\Rightarrow x=1\pm\sqrt 3i$
So $P$ can be written as:
$P(x)=(x-2)(x+2)(x+1+\sqrt 3i)(x+1-\sqrt 3i)(x-1-\sqrt 3i)(x-1+\sqrt 3i)$
