Find the Remainder (6x^3-19x^2+20)÷(x^2-2x-3)

$(6 x^{3} - 19 x^{2} + 20) \div (x^{2} - 2 x - 3)$

Step 1

To calculate the remainder, first divide the polynomials.

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x^{2}$

$2 x$

$3$

$6 x^{3}$

$19 x^{2}$

$0 x$

$20$

Divide the highest order term in the dividend $6 x^{3}$ by the highest order term in divisor $x^{2}$ .

$6 x$

$x^{2}$

$2 x$

$3$

$6 x^{3}$

$19 x^{2}$

$0 x$

$20$

Multiply the new quotient term by the divisor.

$6 x$

$x^{2}$

$2 x$

$3$

$6 x^{3}$

$19 x^{2}$

$0 x$

$20$

$6 x^{3}$

$12 x^{2}$

$18 x$

The expression needs to be subtracted from the dividend, so change all the signs in $6 x^{3} - 12 x^{2} - 18 x$

$6 x$

$x^{2}$

$2 x$

$3$

$6 x^{3}$

$19 x^{2}$

$0 x$

$20$

$6 x^{3}$

$12 x^{2}$

$18 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$6 x$

$x^{2}$

$2 x$

$3$

$6 x^{3}$

$19 x^{2}$

$0 x$

$20$

$6 x^{3}$

$12 x^{2}$

$18 x$

$7 x^{2}$

$18 x$

Pull the next terms from the original dividend down into the current dividend.

$6 x$

$x^{2}$

$2 x$

$3$

$6 x^{3}$

$19 x^{2}$

$0 x$

$20$

$6 x^{3}$

$12 x^{2}$

$18 x$

$7 x^{2}$

$18 x$

$20$

Divide the highest order term in the dividend $- 7 x^{2}$ by the highest order term in divisor $x^{2}$ .

$6 x$

$7$

$x^{2}$

$2 x$

$3$

$6 x^{3}$

$19 x^{2}$

$0 x$

$20$

$6 x^{3}$

$12 x^{2}$

$18 x$

$7 x^{2}$

$18 x$

$20$

Multiply the new quotient term by the divisor.

$6 x$

$7$

$x^{2}$

$2 x$

$3$

$6 x^{3}$

$19 x^{2}$

$0 x$

$20$

$6 x^{3}$

$12 x^{2}$

$18 x$

$7 x^{2}$

$18 x$

$20$

$7 x^{2}$

$14 x$

$21$

The expression needs to be subtracted from the dividend, so change all the signs in $- 7 x^{2} + 14 x + 21$

$6 x$

$7$

$x^{2}$

$2 x$

$3$

$6 x^{3}$

$19 x^{2}$

$0 x$

$20$

$6 x^{3}$

$12 x^{2}$

$18 x$

$7 x^{2}$

$18 x$

$20$

$7 x^{2}$

$14 x$

$21$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

						$6 x$	-	$7$
$x^{2}$	-	$2 x$	-	$3$		$6 x^{3}$	-	$19 x^{2}$	+	$0 x$	+	$20$
					-	$6 x^{3}$	+	$12 x^{2}$	+	$18 x$
							-	$7 x^{2}$	+	$18 x$	+	$20$
							+	$7 x^{2}$	-	$14 x$	-	$21$
									+	$4 x$	-	$1$

The final answer is the quotient plus the remainder over the divisor.

$6 x - 7 + \frac{4 x - 1}{x^{2} - 2 x - 3}$

Step 2

Since the last term in the resulting expression is a fraction, the numerator of the fraction is the remainder.

$4 x - 1$

Find the Remainder (6x^3-19x^2+20)÷(x^2-2x-3)

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