Question

(12x³+3x²-108)÷ (3x-6)
 use long division to find the quotient below.

Answer 1

$\frac{ 12x^3+3x^2-108}{3x-6} =\frac{3(4x^3+x^2-36)}{3(x-2)}=\frac{ 4x^3+x^2-9x^2+ 9x^2-36 }{ x-2 }=\\ \\=\frac{4x^3-8x^2+9x^2 -36}{ x-2 }=\frac{ 4x^2(x-2) +9(x^2 - 4)}{ x-2 }=\\ \\=\frac{ 4x^2(x-2) +9(x - 2)(x+2)}{ x-2 }=\frac{ (x-2) [4x^2+ 9 (x+2)]}{ x-2 }= \\ \\=4x^2+ 9( x+2)=4x^2+ 9 x+18$