Partial fraction decomposition (x-3)/(x^2+4x+3)

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SUMMARY

The discussion focuses on the process of partial fraction decomposition for the expression (x-3)/(x^2+4x+3). The denominator is factored into (x+3)(x+1), leading to the equation A/(x+3) + B/(x+1). To solve for A and B, participants suggest substituting specific values for x that simplify the equations, ultimately determining A = -2 and B = 3. The method involves setting up a system of equations based on the coefficients of the resulting polynomial after clearing the fractions.

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Guzman10
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(x-3)/(x^2+4x+3)
After i factor the denominator what do i do next to find A and B?
=(x-3)/(x+3)(x+1)
=A/(x+3)+B/(x+1)
 
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First try to get rid of the fractions. Any ideas?

Then once you do that you can let $x$ equal any value you want. Choose $x$ such that $A$ or $B$ cancels out, then you can solve for the other one. Can you make any progress now? :)
 
Yes, thanks a lot
 
Now,to find a,multiplying both sides of the equality by (x+3);

(x-3)/(x+1) =A + B(x+3)/(x+1)

Now,setting x=-3 or x+3=0,makes the expression on the right containing (x+3) vanish. So,

(-3-3)/(-3+1)=A+0 .Then,

A=3

You could find B by a similar method.
 
\frac{x- 3}{x^2+ 4x+ 3}= \frac{x- 3}{(x+ 1)(x+ 3)}= \frac{A}{x+1}+ \frac{B}{x+3}

There are several different ways to do this. The most obvious is to add the fractions on the right side: <div style="text-align: left"><span style="font-family: 'Tahoma'">\frac{x- 3}{(x+ 1)(x+ 3)}</span>&#8203;</div><span style="font-family: 'Tahoma'"><br /> <span style="font-family: 'Tahoma'">=</span></span>\frac{A(x+3)}{(x+ 1)(x+ 3)}+ \frac{B(x+ 1)}{(x+1)(x+ 3)}=\frac{Ax+ 3A+ Bx+ B}{x^2+ 4x^2+ 3}

so we must have (A+ B)x+ (3A+ B)= x- 3. In order that this be true for all x we must have (A+ B)x= x and 3A+ B= -3. Solve the equations A+ B= 1 and 3A+ B= -3 for A and B.

Multiply both sides of the equation by (x+ 1)(x+ 3): x- 3= A(x+ 3)+ B(x+ 1).

And now we can write x- 3= (A+ B)x+ 3A+ B so we must have the A+ B= 1 and 3A+ B= -3 as before. 3A+ B=-3. A+ B= 1" 2A= -4. A= -2. B= 3

Or, since this is to be true for all x we can simply choose two values for x to get to equations. If we take x= 0 (just because it is an easy number) we have -3= 3A+ B again. If we take x= 1 we have -2= 4A+ 2B. Dividing by 2 gives 2A+ B= -1. That last is a new equation but satisfied by the same A and B.

The simplest method is to choose x= -1 and x= -3 because they make the coefficients of one of A and B 0. If x= -1 we have -1- 3= -4= A(-1+ 3)+ B(-1+ 1) so 2A= -4. Taking x= -3 we have -3- 3= -6= A(-3+ 3)+ B(-3+ 1) so -2B= -6.

 

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