Answer:
[tex]\frac{5x+2}{3x^2+4x-4}= \frac{2}{3x-2}+\frac{1}{x+2}[/tex]
Step-by-step explanation:
Ooook, step 1: write the denominator as a product, or we're not going anywhere. Can't spot any two factor by staring at it, so quadratic formula it is.
[tex]x= \frac13(-2 \pm \sqrt{4+12})=\frac13(-2\pm4) = -2;\frac23[/tex]
We got our toots, let's split the denominator now: [tex]3x^2+4x-4 = (3x-2)(x+2)\\[/tex]
Now it's all a matter of writing down what we need:
[tex]\frac{5x-2}{3x^2+4x-4}=\frac{5x-2}{(3x-2)(x+2)} = \frac{A}{3x-2}+\frac{B}{x+2} = \frac{A(x+2)+B(3x-2)}{(3x-2)(x+2)}=\frac{Ax+2A+3Bx-2B}{(3x-2)(x+2)} =\frac{(A+3B)x+(2A-2B)}{(3x-2)(x+2)}[/tex]
Here comes the fun part. Two ways: the fast and loose, and the long one.
Fast and loose: let's look only at the numerators of the 3rd to last and first step: [tex]A(x+2)+B(3x-2) = 5x+2[/tex] If we plugged a convenient value, let's say [tex]x=-2[/tex] the coeffcent of A would disappear, and we would be able to find B. Let's do it. [tex]A(-2+2) + B(3(-2)-2) = 5(-2)+2 \rightarrow -8B =-8 B=1[/tex]
At this point we have B, let's pick a different value of x, let's say 1, to get A:
[tex]A(3) + 1(1) = 7 \rightarrow 3A=6 \rightarrow A=2[/tex]
More traditional: Pick the first and last numerators: [tex]5x+2 = (A+3B)x+(2A-2B)[/tex]. Stare at them long enough and pick your favourite excuse to justify the following system (two polynomials are equal if the coefficents are equal, polynomials create a vector space of base [tex](1,x,x^2...)[/tex]
[tex]\left \{ {A+3B=5} \atop {2A-2B=2}} \right.[/tex]
Again, [tex]A=2, B=1[/tex] is a solution.
Either way, the solution is:
[tex]\frac{5x+2}{3x^2+4x-4}= \frac{2}{3x-2}+\frac{1}{x+2}[/tex]
