Answer:
[tex]\dfrac{2x(5x+1)}{4x-1}[/tex]
Step-by-step explanation:
Given expression:
[tex]\dfrac{\dfrac{4x^3-12x^2}{4x^2+7x-2}}{\dfrac{2x^2-6x}{5x^2+11x+2}}[/tex]
Consider the numerator and the denominator separately.
Numerator:
[tex]\dfrac{4x^3-12x^2}{4x^2+7x-2}\\ \\=\dfrac{4x^2(x-3)}{4x^2+8x-x-2}\\ \\=\dfrac{4x^2(x-3)}{4x(x+2)-(x+2)}\\ \\=\dfrac{4x^2(x-3)}{(x+2)(4x-1)}[/tex]
Denominator:
[tex]\dfrac{2x^2-6x}{5x^2+11x+2}\\ \\=\dfrac{2x(x-3)}{5x^2+10x+x+2}\\ \\=\dfrac{2x(x-3)}{5x(x+2)+(x+2)}\\ \\=\dfrac{2x(x-3)}{(x+2)(5x+1)}[/tex]
The whole expression:
[tex]\dfrac{\dfrac{4x^3-12x^2}{4x^2+7x-2}}{\dfrac{2x^2-6x}{5x^2+11x+2}}\\ \\ \\=\dfrac{\dfrac{4x^2(x-3)}{(x+2)(4x-1)}}{\dfrac{2x(x-3)}{(x+2)(5x+1)}}\\ \\ \\=\dfrac{4x^2(x-3)}{(x+2)(4x-1)}\cdot \dfrac{(x+2)(5x+1)}{2x(x-3)}\\ \\ \\=\dfrac{2x(5x+1)}{4x-1}[/tex]
