Question

simplify: 4x/(x^2-3x-18) + 2/(x-6)-2/(x+3)

Answer 1

Factorials work like this:

For a given binomial

Ax^2 + Bx + C

find A(x + a)(x + b) such that

a + b = B
ab = C

Answer 2

partial fractions?

Answer 3

yes

Answer 4

There are hidden gems in those problems. See that (x-6)? It's very relevant. (x+3)? Still relevant.

Answer 5

for \[\frac{4 x}{x^2-3 x-18} +\frac{2}{x-6}-\frac{2}{x+3}\]
factorise it
\[\frac{2x}{(x-6) (x+3)}(2x+(x+3)-(x-6))\]
\[\frac{2}{(x-6)(x+3)}(2x+9)\]

Answer 6

Partial fractions for that

Answer 7

\[\frac{2 (2x+9)}{(x-6) (3+x)}=\frac{A}{x-6}+\frac{B}{x+3}\]
Multiply by (x-6)(3+x)
\[2 (2 x+9)=A (x+3)+B (x-6)\]
\[\left( \begin{array}{c} A=\frac{14}{3} \\ B=-\frac{2}{3} \\ \end{array} \right)\]
So
\[ \frac{2 (2 x+9)}{(x-6) (x+3)}=\frac{14}{3 (x-6)}-\frac{2}{3 (x+3)}\]