How do you divide $(3x^2 + x – 15) /(x – 3)$ ?

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Answer 1 · 2018-04-09T10:33:13

Please look below.

The first step is to split the fraction into $2$ fractions where one is a multiple of the divisor and the second is the left over algebra. So

$(3x^2 + x -15)/(x-3)$

becomes

$(3x^2-9x)/(x-3) + (10x-15)/(x-3)$

$=(3x(x-3))/(x-3) + (10x - 15)/(x-3)$
$= 3x + (10x-15)/(x-3)$

Repeating this process will divide the polynomial completely

$= 3x + (10x - 30)/(x-3) + 15/(x-3)$

$= 3x + (10(x-3))/(x-3)+ 15/(x-3)$

$= 3x + 10 + 15/(x-3)$

The process end here since the remaining fraction cannot be simplified further. And so:

$(3x^2 + x -15)/(x-3) = 3x + 10 + 15/(x-3)$

How do you divide (3x^2 + x – 15) /(x – 3)(3x^2 + x – 15) /(x – 3)?