How do you factor #x^(2n) + 10xn + 16#?

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Answer 1 · 2015-06-28T08:45:19

Assuming the expression is #x^(2n)+10x^n+16#

#x^(2n)+10x^n+16 = (x^n + 2)(x^n + 8)#

Whether this can be factored further depends on the value of #n#

#x^(2n)+10x^n+16 = (x^n)^2+10(x^n)+16#

is quadratic in #x^n# and can be factored as:

#(x^n)^2+10(x^n)+16 = (x^n + 2)(x^n + 8)#

If you prefer, let #y = x^n#.

Then

#x^(2n)+10x^n+16#

#= y^2+10y+16#

#= (y+2)(y+8)#

#= (x^n+2)(x^n+8)#