How do you divide $\frac{- x^{3} - x^{2} + 9 x - 3}{x - 3}$ $(-x^3-x^2+9x-3)/(x-3)$ ?

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How do you divide $\frac{- x^{3} - x^{2} + 9 x - 3}{x - 3}$ $(-x^3-x^2+9x-3)/(x-3)$ ?

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Answer 1 · 2016-04-25T22:31:13

$- x^{2} + 2 x + 1$ $-x^2 + 2x +1$

Explanation:

You write the answer using everything you know. The division of a degree 3 by a degree one polynomial will be degree two. So you write
$- x^{3} - x^{2} + 9 x - 3 = (x - 3) (- x^{2} + 1 + a x)$ $-x^3 - x^2 + 9x -3 = (x-3)(-x^2 +1 + ax)$
This gives you
$a - 3 = - 1$ $a - 3= -1$
$a = 2$ $a=2$

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How do you divide $\frac{- x^{3} - x^{2} + 9 x - 3}{x - 3}$ $(-x^3-x^2+9x-3)/(x-3)$ ?

1 Answer

Explanation:

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How do you divide $\frac{- x^{3} - x^{2} + 9 x - 3}{x - 3}$ $(-x^3-x^2+9x-3)/(x-3)$ ?