Question

find the quotient and remainder using long division for $\frac{x^{3}-10x^{2}+26x-16}{x-4}$
the quotient is
the remainder is

Explanation:

Step1: Divide leading terms

$\frac{x^3}{x} = x^2$
Multiply divisor by $x^2$: $x^2(x-4) = x^3 - 4x^2$
Subtract from dividend:
$(x^3 - 10x^2 + 26x - 16) - (x^3 - 4x^2) = -6x^2 + 26x - 16$

Step2: Divide new leading terms

$\frac{-6x^2}{x} = -6x$
Multiply divisor by $-6x$: $-6x(x-4) = -6x^2 + 24x$
Subtract from current polynomial:
$(-6x^2 + 26x - 16) - (-6x^2 + 24x) = 2x - 16$

Step3: Divide new leading terms

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

Answer:

The quotient is $x^2 - 6x + 2$
The remainder is $-8$