how do i do this.....?

Question

instructions: decompose each expressiono into partical fractions.
problem: (11x-7)/(2x^2-3x-2)

Answer 1

First factor the denominator: (2x + 1)(x - 2)

Your partial fractions will be A/(2x+1) + B/(x-2)

Multiply the first fraction by (x-2)/(x-2) and the
second by (2x+1)/(2x+1) to get the common denom.

So you have A(x-2) + B(2x+1) all over (x-2)(2x+1)

This is Ax - 2A + 2Bx + B all over 2x^2 - 3x - 2, the same denominator you started with, so the numerators must also be equal.

So, A + 2B = 11
and -2A + B = -7

Multiply the top by 2 and add it to the bottom to eliminate A

5B = 15 therefore B = 3

Plug this in to get A: A + 2(3) = 11; A = 5

Your partial fractions are now 5/(2x+1) + 3(x-2)

Hope that makes sense and I didn't do any typos

Answer 2

(11x-7)/(2x^2-3x-2)
(11x-7)/(2x+1)(x-2)
(11x-7)/(2x+1)(x-2) = A/(2x+1) + B/(x-2)

11x-7 = A(x-2) + B(2x+1)

For A, let x=-1/2
11(-1/2)-7 = A((-1/2)-2) + B(2(-1/2)+1)
(-11/2) - 7 = -5A/2 + 0
-25/2 = -5A/2
A = 5

For B, let x=2
11x-7 = A(x-2) + B(2x+1)
11(2)-7 = A(2-2) + B(2(2)+1)
22-7 = 0 + B(4+1)
15 = 5B
B = 3

So,
(11x-7)/(2x+1)(x-2) = 5/(2x+1) + 3/(x-2)

Answer 3

(11x-7)/(2x^2-3x-2)
=(11x-7)/(2x^2-4x+x-2)
=(11x-7)/(2x+1)(x-2)..eq1
let this be equal to
=[A/(2x+1)]+[B/(x-2)]
take lcm
=[A(x-2)+B(2x+1)]/(2x+1)(x-2)..eq2
compare numerators of
eq1 &eq2
(11x-7)=[A(x-2)+B(2x+1)]
compare coeff. of x
11=A+2B
compare constants
-7=-2A+B
solve this to find A&B
put in
[A/(2x+1)]+[B/(x-2)]