Answer

Given (x+1)^4+(x+3)^4-16

Let

y = x+2
(1)

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)


Steven Zheng posted 4 hours ago

Scroll to Top