Question

Resolve into partial fraction. (5x+2)/3x²+4x-4​

Answer 1

Answer:

$\frac{5x+2}{3x^2+4x-4}= \frac{2}{3x-2}+\frac{1}{x+2}$

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.

$x= \frac13(-2 \pm \sqrt{4+12})=\frac13(-2\pm4) = -2;\frac23$

We got our toots, let's split the denominator now: $3x^2+4x-4 = (3x-2)(x+2)$

Now it's all a matter of writing down what we need:

$\frac{5x-2}{3x^2+4x-4}=\frac{5x-2}{(3x-2)(x+2)} = \frac{A}{3x-2}+\frac{B}{x+2} = \frac{A(x+2)+B(3x-2)}{(3x-2)(x+2)}=\frac{Ax+2A+3Bx-2B}{(3x-2)(x+2)} =\frac{(A+3B)x+(2A-2B)}{(3x-2)(x+2)}$

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: $A(x+2)+B(3x-2) = 5x+2$ If we plugged a convenient value, let's say $x=-2$ the coeffcent of A would disappear, and we would be able to find B. Let's do it. $A(-2+2) + B(3(-2)-2) = 5(-2)+2 \rightarrow -8B =-8 B=1$

At this point we have B, let's pick a different value of x, let's say 1, to get A:

$A(3) + 1(1) = 7 \rightarrow 3A=6 \rightarrow A=2$

More traditional: Pick the first and last numerators: $5x+2 = (A+3B)x+(2A-2B)$. 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 $(1,x,x^2...)$

$\left \{ {A+3B=5} \atop {2A-2B=2}} \right.$

Again, $A=2, B=1$ is a solution.

Either way, the solution is:

$\frac{5x+2}{3x^2+4x-4}= \frac{2}{3x-2}+\frac{1}{x+2}$