How do you divide $\frac{x^{3} - 2 x - 6}{x - 3}$ $(x^3-2x-6) / (x-3)$ ?

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Answer 1 · 2016-07-17T17:27:34

$x^{2} + 3 x + 7$ $x^2 +3x +7$ rem 15

Using synthetic division:
$(x^{3} + 0 x^{2} - 2 x - 6) \div (x - 3)$ $(x^3 + 0x^2 - 2x -6) div (x-3)$

Use the coefficients.

$| 1 0 -2 -6$ $" | 1 0 -2 -6"$
$3 | 3 9 21$ $" 3 | 3 9 21"$

$1 3 7 15$ $" 1 3 7 15"$

"Multiply the 3 by each number at the bottom"

The answers are the coefficients from $x^{2}, x^{1}, x^{0}, and R$ $x^2, x^1, x^0, and R$

$(x^{3} + 0 x^{2} - 2 x - 6) \div (x - 3)$ $(x^3 + 0x^2 - 2x -6) div (x-3)$

= $1 x^{2} + 3 x + 7$ $1x^2 +3x +7$ rem 15

How do you divide (x^3-2x-6) / (x-3)x3−2x−6x−3(x^3-2x-6) / (x-3)?