Question

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

Answer 1

`( 12x^3+3x^2-108)/(3x-6) =(3(4x^3+x^2-36))/(3(x-2))=( 4x^3+x^2-9x^2+ 9x^2-36 )/( x-2 )=\\ \\=(4x^3-8x^2+9x^2 -36)/( x-2 )=( 4x^2(x-2) +9(x^2 - 4))/( x-2 )=\\ \\=( 4x^2(x-2) +9(x - 2)(x+2))/( x-2 )=( (x-2) [4x^2+ 9 (x+2)])/( x-2 )= \\ \\=4x^2+ 9( x+2)=4x^2+ 9 x+18`