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Polynomial division/ intergration

image.jpeg I'm trying to split this into partial fractions in order to intergrate. When I do so, I get -1 and a remainder of -x+5. But the answer of -X+5 isn't used? Why?
Reply 1
improper fraction
Reply 2
Original post by TeeEm
improper fraction

Can you expand on that please
Reply 3
Original post by Abby5001
Can you expand on that please


I am teaching now
@Zacken
Reply 4
Original post by Abby5001
image.jpeg I'm trying to split this into partial fractions in order to intergrate. When I do so, I get -1 and a remainder of -x+5. But the answer of -X+5 isn't used? Why?


What do you mean it isn't used? What are you trying to do? Can you post the full question?
Reply 5
Original post by Zacken
What do you mean it isn't used? What are you trying to do? Can you post the full question?

'Use partial fractions to intergrate the following'
image.jpeg
Reply 6
Original post by Abby5001
'Use partial fractions to intergrate the following'
image.jpeg


Okay, so you split it up into 3 terms? One is -1, the other two are fractions with denominators of x+3 and x-1. Do you get that?
Reply 7
Original post by Zacken
Okay, so you split it up into 3 terms? One is -1, the other two are fractions with denominators of x+3 and x-1. Do you get that?
I thought as the highest powers of the denominator and numerator were equal you would have to use polynomial division before you split it into partial fractions
Reply 8
Original post by Abby5001
I thought as the highest powers of the denominator and numerator were equal you would have to use polynomial division before you split it into partial fractions


Well, yes - that's where the -1 comes from.
Reply 9
Original post by Abby5001
I thought as the highest powers of the denominator and numerator were equal you would have to use polynomial division before you split it into partial fractions


May I suggest that you run through this video: https://www.youtube.com/watch?v=GRXAoppq-IA really quickly and see an example of how to deal with partial fraction decomposition with equal numerator and denominator?
You can only use partial fractions if the degree of the top of the fraction is strictly lower than the degree of the bottom.
Since both the top and the bottom are quadtratics, you will need to divide the fraction into a quotient and remainder, perhaps using algebraic long division. This is where the -1 came from.
Reply 11
Original post by Zacken
May I suggest that you run through this video: https://www.youtube.com/watch?v=GRXAoppq-IA really quickly and see an example of how to deal with partial fraction decomposition with equal numerator and denominator?
Ok but also when you divide the fraction you get a remainder of -x+5
Reply 12
Original post by Abby5001
Ok but also when you divide the fraction you get a remainder of -x+5


So now you need to partially decompose (-x+5)/(3+x)(1-x)

(assuming what you get correct)
Reply 13
Original post by Zacken
So now you need to partially decompose (-x+5)/(3+x)(1-x)

(assuming what you get correct)
which means what
Reply 14
Original post by Abby5001
which means what


After your division you get:

x2+(x+3)(1x)=1+x+5(x+3)(1x)\displaystyle \frac{x^2 + \cdots}{(x+3)(1-x)} = -1 + \frac{-x +5}{(x+3)(1-x)}

Now do partial fractions on the last term
Reply 15
I understand how to do all that I just didn't get why the solution bank ignored -X+5 it makes no sense to me

Original post by Zacken
After your division you get:

x2+(x+3)(1x)=1+x+5(x+3)(1x)\displaystyle \frac{x^2 + \cdots}{(x+3)(1-x)} = -1 + \frac{-x +5}{(x+3)(1-x)}

Now do partial fractions on the last term
Reply 16
Original post by Abby5001
I understand how to do all that I just didn't get why the solution bank ignored -X+5 it makes no sense to me


Forget the solution bank, if you know how to do it then that's good enough. I would do it this way as well.
Reply 17
Original post by Zacken
Forget the solution bank, if you know how to do it then that's good enough. I would do it this way as well.

Alright Thankyou

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