Plz help me with the partial fractions...?

Question

These are the rules:
Linear factors: f(x) / ((x+a)(x+b)) = ((A / (x+a)) + (B / (x+b))
Repeated linear factors: f(x) / (x+a)^2 = ((A / (x+a)) + (B / (x+a)^2)

i know that when we want to solve ex:
(2x+3) / (x-2)^2 we use the repeated linear factors method...

that's ok, but can't we say that:
(2x+3) / (x-2)^2 = (2x+3) / ((x-2)(x-2)) then solve it using the linear factors method ??? if no, plz say why...

Plz help me because i need a very clear answer

Thanx

Answer 1

So you want to know if

(2x + 3) / [(x - 2) (x - 2)] = A/(x - 2) + B/(x - 2)

For the sake of argument, let's see where this takes us.

Multiply both sides by (x - 2) (x - 2) to get

(2x + 3) = A(x - 2) + B(x - 2)
(2x + 3) = Ax - 2A + Bx - 2B
(2x + 3) = (A + B)x + (-2A - 2B)

Which means, equating the coefficients of both sides,

A + B = 2
-2A - 2B = 3

Right?

Solving this system of equations by elimination, let's multiply the first equation by 2.

2A + 2B = 4
-2A - 2B = 3

Adding the equations, we get

0 = 7

HUH? We just eliminated the variables we wanted to solve for, implying that there is no solution.

I don't know the reason WHY, for repeated linear factors, we have to do what we do, but look at what happened when we did; we ended up with NO solutions for A and B, meaning

(2x + 3) / [(x - 2) (x - 2)] CANNOT be expressed as

A/(x - 2) + B/(x - 2)

This is just a snippet of the reason why and not a general proof of why the linear factors method doesn't work. We tried it and look what happened.

The proper method, as you know, is by the rule you know:

(2x + 3) / [(x - 2)^2] = A/(x - 2) + B/(x - 2)^2

Multiply both sides by (x - 2)^2, to get

(2x + 3) = A(x - 2) + B
2x + 3 = Ax - 2A + B
2x + 3 = Ax + (-2A + B)

Therefore, equating the coefficients of both sides,

A = 2
-2A + B = 3

The system of equations yields A = 2, B = 7

(2x + 3) / [(x - 2)^2] = 2/(x - 2) + 7/(x - 2)^2

And this time we actually had a system of equations to *work* with.

Answer 2

a million/[x^3 - x] element the denominator. a million/[x(x^2 - a million)] a million/[x(x - a million)(x + a million)] this would nicely be of the kind a million/[x(x - a million)(x + a million)] = A/x + B/(x - a million) + C/(x + a million) that is a truth that's authentic for all x. to freshen up for A, B, and C, first i bypass to multiply both factors with the help of utilizing x(x - a million)(x + a million). a million = A(x - a million)(x + a million) + B(x)(x + a million) + C(x)(x - a million) by way of very truth this truth is authentic for all x, we are waiting to plug in any fee for x. pick x = 0: a million = A(0 - a million)(0 + a million) + 0 + 0 a million = A(-a million)(a million) a million = -A A = -a million. pick x = a million: a million = A(x - a million)(x + a million) + B(x)(x + a million) + C(x)(x - a million) a million = 0 + B(a million)(2) + 0 a million = 2B B = a million/2 pick x = -a million: a million = 0 + 0 + C(-a million)(-a million - a million) a million = C(-a million)(-2) a million = 2C C = a million/2 A = -a million, B = a million/2, C = a million/2, so a million/(X^3 - X) = -a million/x + (a million/2)/(x - a million) + (a million/2)/(x + a million)

Answer 3

If you solve it the way you stated, then you would get the same "A" and "B". By having one of the denominators be (x+a)^2 and the other being (x+a), then you can obtain a different "A" and "B" for the numerators. For your example,

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

2x+3 = A + Bx - 2 B (multiply both sides by (x-2) then simplify)

2x+3 = Bx +(A-2B) (collect x-terms and constants on right side)

2x = Bx (x-coefficients must be equal)
B=2

3 = (A-2B) (constants must be equal)
A = 7

If you solved it the other way, then you would end up with A and B being equal.

Answer 4

You can't, because then both fractions would have the same denominator (x - 2):

(2x + 3)/((x - 2)(x - 2)) = A/(x - 2) + B/(x - 2) = (A + B)/(x - 2)

So you could never find constants A and B that would be equivalent to the left hand side.

Answer 5

(2x+3)/(x-2)^2=A/(x-2)+B/(x-2)^2
2x+3=A(x-2)+B
2x+3=Ax+B-2A
comparing the coefficients
B-2A=3
A=2
B-4=3 or B=7
so thepartial fractions are
3/(x-2)+7/(x-2)^2