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## #1 2013-11-19 08:20:58

ninjaman
Member
Registered: 2013-10-15
Posts: 55

### repeated linear factors, (i think)

hello im having a problem with the last part of this assignment that I have.

I have this so far

14x -4x^2        A(2-X)^2 + B(1+X)(2-X) + C(1+X)

A(4+4X+X^2) + B(2+1X-X^2) + C(1+X)

A4+4AX+AX^2 + B2+1BX-BX^2 +C(1+X)

I know that A is -2

so,            A-B = -4

somehow this is wrong, its most likely something that is staring me in the face but I cant think what it is

could it be that A+B, so it is the + that I left out. I thought that the -BX^2 meant -B so I put A-B

should it have been A+-B, which I think still gives the right answer?

my tutor explained it, but this part I didn't get and im struggling with the maths.

any help would really be appreciated!!!

thanks

all the best

simon:):):):):):)

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## #2 2013-11-19 09:26:31

bob bundy
Moderator
Registered: 2010-06-20
Posts: 7,930

### Re: repeated linear factors, (i think)

hi Simon,

It looks like you're trying to find A,B and C to make

So now compare the x^2 terms, the x terms and the non-x terms

x^2 :      A-B = -4 .......................1

x     :    -4A + B + C = 14 .............2

non-x:  4A + 2B + C = 0 ...............3

With three unknowns and three equations this should be possible.  Look for an easy way to find a letter.

Thinks    If I use 2 and 3 to eliminate C then I've got a pair of equations with just A and B.

So subtract 2 from 3

8A + B = -14 ..................4

You have to watch the signs here!

Now if I add equation 1 to 4 the Bs will go:

9A = -18      => A = -2

Now I can get B from equation 1

-2 - B = -4   =>  B = 2

Now I can get C from equation 3

4x-2 + 2 x 2 + C = 0 => C = 4

And there you have it.

Bob

ps.  Alternative method that works for this example but may not in general.

If you put x = 2 the (2-x) factors contribute zero so we get

14 x 2 - 4 x 2 x 2 = C x 3   => 28 - 16 = 3C  => 12 = 3C => C = 4

Put x = -1 and the (1+x) factors become zero

14 x -1 - 4 x -1 x -1 = A(2--1)^2 =>  -14 -4  = 9A  => -18 = 9A => A = -2

There's no easy value to get B so choose the easiest you can.  Put x = 0 and use the values we know for A and C

0 =   -2 x 2^2 + B x 1 x 2 + 4 x 1  =>  0 = -8 + 2B + 4 => 0 = -4 + 2B  => B = 2

Last edited by bob bundy (2013-11-19 09:40:30)

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #3 2013-11-19 09:51:04

ninjaman
Member
Registered: 2013-10-15
Posts: 55

### Re: repeated linear factors, (i think)

what I have to show is A+B or A-B, I thought it was A-B as I got a -B from the repeated factors bit.
should it be A-B?
so the answer is 2. but I cant get how im supposed to write the answer. that part I still don't understand. I get really confused when it comes to +/- signs and adding or subtracting them.

I have to write either A-B = -2 (-)2 meaning negative 2 take away positive 2. or A+B which is -2 + -2, which is negative 2 add negative 2.

I had to get the X^2 parts only to find this out.
I know the answers to A and C.
I didn't do the part for B.
I forgot to.
so I have to redo it.
im not sure if the A+B is specific to this sort of formula.

still confused!

thanks bob!!!

all the best simon

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## #4 2013-11-19 10:47:46

bob bundy
Moderator
Registered: 2010-06-20
Posts: 7,930

### Re: repeated linear factors, (i think)

hi Simon,

I thought I understood your problem but now I'm not so sure.  Here's a suggestion:

You said this is the last part of an assignment.  How about telling me (i) what the assignment is (ii) what you have successfully done so far.  Then I can see where we have to go next.

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #5 2013-11-19 11:47:33

ninjaman
Member
Registered: 2013-10-15
Posts: 55

### Re: repeated linear factors, (i think)

hello bob,

I have      14x-4x^2
(1+x)(2-x)^2

I got A(2-x)^2 + B(2-x)(1+x) + C(1+x)

when I found A = -2, C = 4(I think, I cant remember)

the problem was B, I forgot to put that in.
I couldn't remember how to figure out the repeated linear factors bit.
so I replace each X with the opposite value and worked through the problem.
I cant remember how to do B

cheers.
simon

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## #6 2013-11-19 21:36:38

bob bundy
Moderator
Registered: 2010-06-20
Posts: 7,930

### Re: repeated linear factors, (i think)

hi Simon,

Your values for A and C are correct.  You cannot choose a value for x that leads to B on its own.  Back in post 2 I showed how I would get B.

There's no easy value to get B so choose the easiest you can.  Put x = 0 and use the values we know for A and C

0 =   -2 x 2^2 + B x 1 x 2 + 4 x 1  =>  0 = -8 + 2B + 4 => 0 = -4 + 2B  => B = 2

Or you can use the x^2 terms:

x^2 :      A-B = -4 .......................1

Put in the value of A

-2 - B = -4

-2 - B + B = -4 + B            =>            -2 = -4 + B

-2 + 4 = -4 + 4 + B            =>             2 = B

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #7 2013-11-20 01:31:25

ninjaman
Member
Registered: 2013-10-15
Posts: 55

### Re: repeated linear factors, (i think)

hello bob

I HAVE TO use repeated linear factors, which is what I put in the first post.
I understand you saying use the easiest but repeated linear factors is what I have to use.
Thank you for your help. I have the answer, which I almost got in class. I ended it with negative 2.

its the method I don't understand.
That's what I have to show with the answer.

thanks for your help, all the best

simon

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## #8 2013-11-20 01:33:37

bob bundy
Moderator
Registered: 2010-06-20
Posts: 7,930

### Re: repeated linear factors, (i think)

Please write the exact wording of the question.

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #9 2013-11-20 07:10:22

ninjaman
Member
Registered: 2013-10-15
Posts: 55

### Re: repeated linear factors, (i think)

hello

it says,

"express the following algebraic fractions as partial fractions"

and that's it.
the lecturer specified that she wanted to see linear repeated factors in the working for this question.

cheers

simon

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## #10 2013-11-20 07:15:41

bob bundy
Moderator
Registered: 2010-06-20
Posts: 7,930

### Re: repeated linear factors, (i think)

So what was the algebraic fraction at the start of the question?

If this is correct, then stop worrying; you have correctly identified the three partial fractions:

and you have the repeated factor ok.

You can get A and C quickly by setting x = -1 and x = 2

There is no trick for getting B like this.  I know of two ways to get B and I've shown you both.  Choose the one you like best and finish the exercise.

Bob

Last edited by bob bundy (2013-11-20 07:51:38)

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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