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

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Discussion Overview

The discussion revolves around the process of partial fraction decomposition for the expression (x-3)/(x^2+4x+3). Participants explore methods to determine the coefficients A and B after factoring the denominator.

Discussion Character

  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant suggests starting by factoring the denominator and setting up the equation A/(x+3) + B/(x+1) to find A and B.
  • Another participant proposes eliminating the fractions by multiplying through by the denominator and suggests choosing specific values for x to simplify the equations.
  • A later reply provides a method to find A by substituting x = -3, leading to the conclusion that A = 3.
  • Further contributions detail the process of setting up equations based on coefficients, leading to a system of equations A + B = 1 and 3A + B = -3.
  • One participant mentions that using specific values for x, such as x = 0 or x = 1, can also yield equations to solve for A and B.
  • Another participant emphasizes that choosing values like x = -1 and x = -3 can simplify finding A and B by making one of the coefficients zero.

Areas of Agreement / Disagreement

Participants present various methods and approaches to find A and B, but there is no consensus on a single method or final values for A and B, as different participants arrive at different conclusions.

Contextual Notes

The discussion includes multiple approaches and equations, but the specific assumptions or steps taken by participants may not be fully detailed, leading to potential ambiguities in the solutions presented.

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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