Question

find the quotient and remainder using long division.
$\frac{x^{3}-6x^{2}+20}{x-5}$
the quotient is
the remainder is

Explanation:

Step1: Divide leading terms

$\frac{x^3}{x} = x^2$
Multiply divisor by $x^2$: $x^2(x-5) = x^3 - 5x^2$
Subtract from dividend:
$(x^3 - 6x^2 + 20) - (x^3 - 5x^2) = -x^2 + 20$

Step2: Divide new leading terms

$\frac{-x^2}{x} = -x$
Multiply divisor by $-x$: $-x(x-5) = -x^2 + 5x$
Subtract from current polynomial:
$(-x^2 + 20) - (-x^2 + 5x) = -5x + 20$

Step3: Divide new leading terms

$\frac{-5x}{x} = -5$
Multiply divisor by $-5$: $-5(x-5) = -5x + 25$
Subtract from current polynomial:
$(-5x + 20) - (-5x + 25) = -5$

Answer:

The quotient is $x^2 - x - 5$
The remainder is $-5$