Question

$(6x^{3}-x^{2}+4x-9)div(2x-3)$

Explanation:

Step1: Divide leading terms

$\frac{6x^3}{2x} = 3x^2$
Multiply divisor by $3x^2$: $3x^2(2x-3)=6x^3-9x^2$
Subtract from dividend:
$(6x^3 - x^2 + 4x - 9) - (6x^3 - 9x^2) = 8x^2 + 4x - 9$

Step2: Divide new leading terms

$\frac{8x^2}{2x} = 4x$
Multiply divisor by $4x$: $4x(2x-3)=8x^2-12x$
Subtract from current polynomial:
$(8x^2 + 4x - 9) - (8x^2 - 12x) = 16x - 9$

Step3: Divide new leading terms

$\frac{16x}{2x} = 8$
Multiply divisor by $8$: $8(2x-3)=16x-24$
Subtract from current polynomial:
$(16x - 9) - (16x - 24) = 15$

Answer:

$3x^2 + 4x + 8 + \frac{15}{2x-3}$