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

[Guzman10]

(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)

 

[Jameson]

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? :)

 

[Guzman10]

Yes, thanks alot

 

[BestMethod]

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.

 

[HallsofIvy]

$$\frac{x- 3}{x^2+ 4x+ 3}= \frac{x- 3}{(x+ 1)(x+ 3)}= \frac{A}{x+1}+ \frac{B}{x+3}$$

There are several different ways to do this. The most obvious is to add the fractions on the right side: [tex]

\frac{x- 3}{(x+ 1)(x+ 3)}​

=[/tex]$$\frac{A(x+3)}{(x+ 1)(x+ 3)}+ \frac{B(x+ 1)}{(x+1)(x+ 3)}=$$$$\frac{Ax+ 3A+ Bx+ B}{x^2+ 4x^2+ 3}$$

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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