## fraction decomposition - where is the mistake

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
jozwoz99
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Joined: Mon Jul 29, 2013 12:57 am
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### fraction decomposition - where is the mistake

Hi

When I try to decompose: (x + 1) / ((x + 2)^2)

I end up with: x + 1 = A(x + 2)^2 + B(x + 2)

this ends up with A = 1/2, B = - 1/2, so the answer seemingly should be:

1/2(x + 2) - 1/2(x + 2)^2

This doesn't make sense however as the result when these terms are added back is

(x + 1) / 2(x + 2)^2 rather than (x + 1) / (x + 2)^2

Can someone explain the result to me\

Thanks

jg.allinsymbols
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### Re: fraction decomposition - where is the mistake

Given $\frac{(x+1)}{(x+2)^2}$, decompose into denominations of this form:

$\frac{A}{(x+2)}\,+\,\frac{B}{(x+2)^2}$
Work the steps to perform the addition and then compare the numerator terms to solve for A and B.

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Toward you last steps, you should obtain the two simultaneous equations, Ax=x and 2A+B=1. If not, show your work and someone may identify your mistake.[url][/url]

nona.m.nona
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### Re: fraction decomposition - where is the mistake

jozwoz99 wrote:When I try to decompose: (x + 1) / ((x + 2)^2)

I end up with: x + 1 = A(x + 2)^2 + B(x + 2)

jg.allinsymbols
Posts: 72
Joined: Sat Dec 29, 2012 2:42 am
Contact:

### Re: fraction decomposition - where is the mistake

nona.m.nona wrote:
jozwoz99 wrote:When I try to decompose: (x + 1) / ((x + 2)^2)

I end up with: x + 1 = A(x + 2)^2 + B(x + 2)

$(x+1)/((x+2)^2)$
=$A/(x+2) + B/((x+2)^2)$
$(A/(x+2))*(x+2)/(x+2) + (B/((x+2)^2))$
$(Ax+2A)/((x+2)^2)+(B/((x+2)^2))$
$(Ax+2A+B)/((x+2)^2)$
And grouping to begin seeing correspondence of the numerators,
$((Ax)+(2A+B))/((x+2)^2)$

Now, carefully equate the corresponding parts of the numerator:
$Ax=x$ and $2A+B=1$ because you are comparing x+1 with Ax+2A+B
The coefficient on x is 1, so A=1.
Using this for A in the constants relationship, 2*1+B=1 means B=-1