Partial fractions help!

How do I express this as partial fractions : 10/(x+3)(2x+1)(x+1) ??

Im at :
10=A(2x+1)(x+1) +B(x+3)(x+1) + C(x+3)(2x+1)

But i just cant solve for any letter!

please help!

Reply 1

didgeridoo12uk

your aim is to make one of the brackets 0

so for example set x =-1, this eliminates A and B and allows you to work out C. by using different values of x you can get A and B.

(edited 13 years ago)

Reply 2

aura1947

x + 3 = 0

2x + 1 = 0

x + 1 = 0

Calculate those three values of x and then substitute them into the equation one at a time to find out the values of A, B and C.

Reply 3

ztibor

Original post by confuzzled92

How do I express this as partial fractions : 10/(x+3)(2x+1)(x+1) ??

Im at :
10=A(2x+1)(x+1) +B(x+3)(x+1) + C(x+3)(2x+1)

But i just cant solve for any letter!

please help!

Multiply factors by other factors
and get form of

Px^2+Qx+R=10

from this:
P=0
Q=0
R=10
Solve these equations simultaneously for A,B,C

Reply 4

confuzzled92

Original post by didgeridoo12uk

your aim is to make one of the brackets 0

so for example set x =-1, this eliminates A and B and allows you to work out C. by using different values of x you can get A and B.

But how would the left side still equal 10 if you were to insert just any x value? Shouldn't there only be one particular value of x which makes the whole thing 10?
It only makes sense for me to do this if the left side was also dependant on x...
Could you explain why its also alright to do it when the left is simply a constant?

Reply 5

inspiration109

10/(x+3)(2x+1)(x+1)= A/(x+3) + B/(2x+1) + C/(x+1)
10= A(x+1)(2x+1) +B(x+3)(x+1) +C(x+3)(2x+1)

when x=-1
10=C(-1+3)(2(-1)+1)
10=-2C
C=-5

when x=-3
10=A(-3+1)(2(-3)+1)
10=10A
A=1

to find B, sub in x=0

Reply 6

didgeridoo12uk

Original post by confuzzled92

right ok, you obviously haven't been taught the theory behind partial fractions very well at all.

you are given this function

\frac{10}{(x+3)(2x+1)(x+1)}

and want to change it into the form

\frac{A}{x+3} +\frac{B}{2x+1} + \frac{C}{x+1}

where you need to determine the values of A, B and C. The easiest way to do this is to set both forms of the functions equal to each other and work from there.

so we have

\frac{10}{(x+3)(2x+1)(x+1)}= \frac{A}{x+3} +\frac{B}{2x+1} + \frac{C}{x+1}

at the moment we have lots of x's on the denominator, this means we have the potential for some 'divide by zeros' to occur which is never good. So we're looking to have everything over a common denominator, this will allow us to cancel it out and make life a little easier. hopefully this next step is quite obvious/simple to follow.

\frac{10}{(x+3)(2x+1)(x+1)}= \frac{A(2x+1)(x+1)}{(x+3)(2x+1)(x+1)} +\frac{B(x+3)(x+1)}{(2x+1)(x+3)(x+1)} + \frac{C(x+3)(2x+1)}{(x+1)(x+3)(2x+1)}

everything is now over a common denominator so we can just get rid of it. yes i know you got to this stage but i wasn't sure if you really knew how/why you can jump straight to it

10=A(2x+1)(x+1) +B(x+3)(x+1) + C(x+3)(2x+1)

this equation like all of the ones before it is valid for all values of x. We still need to find out what A, B and C are though. there are a couple of ways of doing this. You can multiply all the brackets out and compare co-efficients

10=A(2x+1)(x+1) +B(x+3)(x+1) + C(x+3)(2x+1)

10=A(2x^2+3x +1) +B(x^2 + 4x + 3) + C(2x^2 +7x+3)

10=x^2(2A+B+2C) + x(3A+4B + 7C) + (A +3B + 3C)

0=x^2(2A+B+2C) + x(3A+4B + 7C) + (A +3B + 3C) -10

now we have effectively formed 3 unique equations and need to work out 3 variables so that will work out fine. and i'm sure you can handle that bit yourself

0=2A+B+2C

0= 3A+4B + 7C

0 = A +3B + 3C -10

the other (easier) way to do it, although depending on the original equations it may not always work is to substitute cleverly chosen values of x into the original equation.

10=A(2x+1)(x+1) +B(x+3)(x+1) + C(x+3)(2x+1)

now, if we substitute

x=-3

into the above equation, hopefully you can see that it will cause both B and C to be multiplied by 0, so we only have the A terms to deal with and so will very easily get a value for A. We can do the same thing to isolate the B term and C term. this will give us values for A, B and C that satisfy the original expression, allowing us to write the given function in the form that we want.

hope thats a bit clearer as to why you can sub whatever values of x into the equation you like. if at any point that stopped making sense just let me know and i'll try think of another way to explain it