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

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To perform partial fraction decomposition of (x-3)/(x^2+4x+3), the denominator is factored into (x+3)(x+1). The next step involves expressing the fraction as A/(x+3) + B/(x+1) and eliminating the fractions by multiplying both sides by the denominator. By substituting specific values for x, such as -3 and -1, A and B can be determined easily. The resulting equations yield A = -2 and B = 3, confirming the decomposition. This method effectively simplifies the original expression into its partial fractions.
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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 alot
 
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.

 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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