5 (x^2+4) ^4(2x) (x-2) ^4 + (x^2+4) ^5(4) (x-2) ^3?
There is only one plus or minus sign that is not in parenthesis. I underlined it above. So there are two terms, also underlined.
The common bases between the two terms are (x^2+4) and (x-2). We factor these two terms with the smaller associated power.
That is we factor out (x^2+4)^4*(x-2)^3
Then inside the brackets we subtract 4 from each power of (x^2+4) and 3 from each power of (x-2). If a power becomes 0 then that factor is 1.
(x^2+4)^4*(x-2)^3 (5(x^2+4)^(4-4)*(2x)((x-2)^(4-3)+(x^2+4)^(5-4)*4*(x-2)^(3-3))
=(x^2+4)^4*(x-2)^3 (5(x^2+4)^(0)*(2x)(x-2)^(1)+(x^2+4)^(5-4)*4*(x-2)^(0))
=(x^2+4)^4*(x-2)^3 (5*(2x)(x-2)+(x^2+4)*4*)
=(x^2+4)^4*(x-2)^3 (10x(x-2)+4(x^2+4))
=(x^2+4)^4*(x-2)^3 (10x^2-20x)+(4x^2+16))
=(x^2+4)^4*(x-2)^3 (14x^2-20x+16)
=2(x^2+4)^4*(x-2)^3 (7x^2-10x+8)
