Factor
$$(x+3)^4+(x+1)^4-16$$

Question

Uniteasy Admin · Accepted Answer

Given $$(x+1)^4+(x+3)^4-16$$
Let
$$y = x+2$$n
then
$$(x+1)^4+(x+3)^4-16$$
$$=(y-1)^4+(y+1)^4-16$$
$$=(y^2-2y+1)^2+(y^2+2y+1)^2-16$$
$$=(y^2+1)^2-4y(y^2+1)+4y^2+(y^2+1)^2+4y(y^2+1)+4y^2-16$$
$$=2(y^2+1)^2+8y^2-16$$
$$=2y^4+12y^2-14$$
$$=2(y^4+6y^2-7)$$
$$=2(y^2-1)(y^2+7)$$
$$ =2(y-1)(y+1)(y^2+7) $$
Substitute (1) then
$$(x+1)^4+(x+3)^4-16 = 2(x+3)(x+1)(x^2+4x+11)$$

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