How do you divide $(x^{3} - 3 x^{2} + 4 x - 12) \div (x^{2} + 4)$ $(x^3 - 3x^2 + 4x - 12) div (x^2 + 4)$ ?

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Answer 1 · 2016-06-01T02:52:03

$x - 3$ $x-3$

There is an easier way to do this if we realize $x^{3} - 3 x^{2} + 4 x - 12$ $x^3-3x^2+4x-12$ can be factored by grouping:

$x^{3} - 3 x^{2} + 4 x - 12 = x^{2} (x - 3) + 4 (x - 3) = (x^{2} + 4) (x - 3)$ $x^3-3x^2+4x-12=x^2(x-3)+4(x-3)=(x^2+4)(x-3)$

Therefore

$\frac{x^{3} - 3 x^{2} + 4 x - 12}{x^{2} + 4} = \frac{(x^{2} + 4) (x - 3)}{x^{2} + 4} = x - 3$ $(x^3-3x^2+4x-12)/(x^2+4)=((x^2+4)(x-3))/(x^2+4)=x-3$

How do you divide (x^3 - 3x^2 + 4x - 12) div (x^2 + 4)(x3−3x2+4x−12)÷(x2+4)(x^3 - 3x^2 + 4x - 12) div (x^2 + 4)?