Question

-15x/x^2+8x+16 + 12/x-4
adding expressions :)

Answer 1

\[\frac{-15x}{x^2}+8x+16+ \frac{12}{x-4}=\frac{(-15x)(x-4)+8x(x^2)(x-4)+16(x^2)(x-4)+12x^{2}}{(x-4)x^{2}}\]
\[\frac{-15x^{2}+60x+8x^{4}-32x^{3}+16x^{3}-64x^{2}+12x{2} }{{(x-4)x^{2}}}\]\[\frac{8x^{4}-16x^{3}-67x^{2}+60x}{(x-4)x^{2}}\]

\[\frac{\cancel{x}(8x^{3}-16x^{2}-67x^{1}+60)}{(x-4)x^{\cancel{2}}}\]

\[\frac{(8x^{3}-16x^{2}-67x^{1}+60)}{(x-4)x}\]


I can' t go more further .. regards

Answer 2

the second line should be :
\[\frac{-15x^{2}+60x+8x^{4}-32x^{3}+16x^{3}-64x^{2}+12x^{2} }{{(x-4)x^{2}}}\]

Answer 3

-15x/(x+4)^2 +12/(x-4)
-the denominator common is (x+4)^2 *(x-4) so than the first term need multiplie by
(x-4) and the second term by (x+4)^2

-15x(x-4) +12(x+4)^2
--------------------- =
(x+4)^2 *(x-4)