Improper fraction to proper fraction in algebra

Question

This question is in regards to the very first part of a question titled 'Express the following in partial fractions'

From my lecture notes I have got the following written down: $$\frac{x(x+3)}{x^2+x-12}=1+\frac{2x+12}{x^2+x-12}$$

I cant understand how the numerator goes from $x(x+3)$ , to, $1 + (2x+12)$.

I understand that the $x * x$ makes the $2x$, but how is $12$ derived from the $3$ part ?

I can do the rest of the question (factoring, cross multiply, compare numerators), just cant understand to $3$ to $12$ transition.

Any help is greatly appreciated.

-Connor

Answer 1

You should learn polynomial long division. In this case I think it is easiest done like this (add and subtract so that you get the polynomial in the denominator also in the numerator, and then collect...) $$ \frac{x(x+3)}{x^2+x-12}=\frac{(x^2+x-12)+(2x+12)}{x^2+x-12}=1+\frac{2x+12}{x^2+x-12}. $$

Answer 2

Working backwards:

$$1+\frac{2x+12}{x^2+x-12}= \frac{x^2+x-12}{x^2+x-12}+\frac{2x+12}{x^2+x-12}=\frac{x^2+x-12+2x+12}{x^2+x-12}=\frac{x^2+3x}{x^2+x-12}=\frac{x(x+3)}{x^2+x-12}$$