Question

(6x^3-7x^2-16x+12): (2x+3)

Answer 1

\(\frac{6x^3-7x^2-16x+12}{2x+3}\)

\(=\frac{6x^3-12x^2+5x^2-10x-6x+12}{2x+3}\)

\(=\frac{6x^2\left(x-2\right)+5x\left(x-2\right)-6\left(x-2\right)}{2x+3}\)

\(=\frac{\left(x-2\right)\left(6x^2+5x-6\right)}{2x+3}\)

\(=\frac{\left(x-2\right)\left(6x^2+9x-4x-6\right)}{2x+3}\)

\(=\frac{\left(x-2\right)\left[3x\left(2x+3\right)-2\left(2x+3\right)\right]}{2x+3}\)

\(=\frac{\left(x-2\right)\left(2x+3\right)\left(3x-2\right)}{2x+3}\)

\(=\left(x-2\right)\left(3x-2\right)\)