Question

How would you completely factor 3x^4 -24x^2 +48 ?

Answer 1

If it factor it is 3(x+2)^2. (X-2)^2

Answer 2

$\bf 3x^4 -24x^2 +48 \impliedby \textit{common factor} \\\\ 3(x^4-8x^2+16)\quad \begin{cases} x^4\implies x^{2\cdot 2}\implies (x^2)^2 \\\\ 16\implies -4\cdot -4 \\\\ -8\implies -4+(-4) \end{cases} \\\\ thus \\\\ 3(x^4-8x^2+16)\implies 3[(x^2)^2-8x^2+16] \\\\ 3[(x^2-4)(x^2-4)]\impliedby 4=2^2 \\\\ 3[(x^2-2^2)(x^2-2^2)]$

$\bf -----------------------------\\\\ \textit{now, recall your }\textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\\\ -----------------------------\\\\ thus \\\\ 3[(x^2-2^2)(x^2-2^2)] \\\\ 3[(x-2)(x+2)(x-2)(x+2)]\implies 3[(x-2)^2(x+2)^2]$