Question

find the quotient and remainder using long division.
$\frac{x^{4}-x^{3}-50x+3}{x-4}$
the quotient is
the remainder is
question help: video 1 video 2

Explanation:

Step1: Divide leading terms

$\frac{x^4}{x} = x^3$
Multiply divisor by $x^3$: $x^3(x-4) = x^4 - 4x^3$
Subtract from dividend:
$(x^4 - x^3 - 50x + 3) - (x^4 - 4x^3) = 3x^3 - 50x + 3$

Step2: Divide new leading terms

$\frac{3x^3}{x} = 3x^2$
Multiply divisor by $3x^2$: $3x^2(x-4) = 3x^3 - 12x^2$
Subtract:
$(3x^3 - 50x + 3) - (3x^3 - 12x^2) = 12x^2 - 50x + 3$

Step3: Divide new leading terms

$\frac{12x^2}{x} = 12x$
Multiply divisor by $12x$: $12x(x-4) = 12x^2 - 48x$
Subtract:
$(12x^2 - 50x + 3) - (12x^2 - 48x) = -2x + 3$

Step4: Divide new leading terms

$\frac{-2x}{x} = -2$
Multiply divisor by $-2$: $-2(x-4) = -2x + 8$
Subtract:
$(-2x + 3) - (-2x + 8) = -5$

Answer:

The quotient is $x^3 + 3x^2 + 12x - 2$
The remainder is $-5$