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

In summary: These are the two values we need to solve for A and B.In summary, the equation x- 3= (A+ B)x+ 3A+ B has two solutions, x=-1 and x=-3. If x= -1, 2A= -4; if x= -3, -2B= -6.
  • #1
Guzman10
2
0
(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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  • #2
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? :)
 
  • #3
Yes, thanks alot
 
  • #4
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.
 
  • #5
[tex]\frac{x- 3}{x^2+ 4x+ 3}= \frac{x- 3}{(x+ 1)(x+ 3)}= \frac{A}{x+1}+ \frac{B}{x+3}[/tex]

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

so we must have [tex](A+ B)x+ (3A+ B)= x- 3[/tex]. 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): [tex]x- 3= A(x+ 3)+ B(x+ 1)[/tex].

And now we can write [tex]x- 3= (A+ B)x+ 3A+ B[/tex] 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 [tex]-3= 3A+ B[/tex] again. If we take x= 1 we have [tex]-2= 4A+ 2B[/tex]. Dividing by 2 gives [tex]2A+ B= -1[/tex]. 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 [tex]-1- 3= -4= A(-1+ 3)+ B(-1+ 1)[/tex] so 2A= -4. Taking x= -3 we have [tex]-3- 3= -6= A(-3+ 3)+ B(-3+ 1)[/tex] so -2B= -6.
 

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

1. What is partial fraction decomposition?

Partial fraction decomposition is a method used to break down a rational function (a fraction where the numerator and denominator are polynomials) into smaller, simpler fractions.

2. Why is partial fraction decomposition useful?

Partial fraction decomposition is useful because it allows us to solve integrals and simplify complex rational functions, making them easier to work with and understand.

3. How do you perform partial fraction decomposition?

To perform partial fraction decomposition, you must first factor the denominator of the rational function. Then, you set up and solve a system of equations to determine the unknown coefficients of the smaller fractions.

4. What is the purpose of the constant in the denominator of the smaller fractions?

The constant in the denominator of the smaller fractions is used to ensure that the overall fraction is equivalent to the original rational function. This constant is determined through the process of solving the system of equations.

5. Can every rational function be decomposed using this method?

Yes, every rational function can be decomposed using partial fraction decomposition. However, the process may become more complex for functions with higher degrees in the numerator or denominator.

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