Answer

Given (x+5)^4+(x+3)^4-82

Let

y = x+4
(1)

then

(x+5)^4+(x+3)^4-82

=(y-1)^4+(y+1)^4-82

=(y^2-2y+1)^2+(y^2+2y+1)^2-82

=(y^2+1)^2-4y(y^2+1)+4y^2+(y^2+1)^2+4y(y^2+1)+4y^2-82

=2(y^2+1)^2+8y^2-82

=2y^4+12y^2-80

=2(y^4+6y^2-40)

=2(y^2-4)(y^2+10)

=2(y-2)(y+2)(y^2+10)

Substitute (1) then

(x+5)^4+(x+3)^4-82= 2(x+6)(x+2)(x^2+8x+26)


Steven Zheng posted 1 hour ago

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