Question

divide.
$\frac{3x^{4}-22x^{3}+25x^{2}-9x+18}{x-6}$
$\frac{3x^{4}-22x^{3}+25x^{2}-9x+18}{x-6}=\square$

Explanation:

Step1: Divide leading terms

$\frac{3x^4}{x} = 3x^3$
Multiply divisor by $3x^3$: $3x^3(x-6)=3x^4-18x^3$
Subtract from dividend:
$(3x^4-22x^3+25x^2-9x+18)-(3x^4-18x^3)=-4x^3+25x^2-9x+18$

Step2: Divide new leading terms

$\frac{-4x^3}{x} = -4x^2$
Multiply divisor by $-4x^2$: $-4x^2(x-6)=-4x^3+24x^2$
Subtract from current polynomial:
$(-4x^3+25x^2-9x+18)-(-4x^3+24x^2)=x^2-9x+18$

Step3: Divide new leading terms

$\frac{x^2}{x} = x$
Multiply divisor by $x$: $x(x-6)=x^2-6x$
Subtract from current polynomial:
$(x^2-9x+18)-(x^2-6x)=-3x+18$

Step4: Divide new leading terms

$\frac{-3x}{x} = -3$
Multiply divisor by $-3$: $-3(x-6)=-3x+18$
Subtract from current polynomial:
$(-3x+18)-(-3x+18)=0$

Answer:

$3x^3 - 4x^2 + x - 3$