# Partial Fraction Q (factorising denominator)Watch

#1
Hey Babes.

I am trying to factorise the denominator of this and am not sure how. Someone said find a factor and divide by this e.g. if 1 is a factor, divide the denominator by (x-1). However, I tried this and it failed to work really..

Q) By firstly factorising the denominator, express the following as a partial fraction:

(5x^2 - 15x - 8) / (x^3 - 4x^2 + x + 6)

I know what to do but simply can't factorise the bottom to get three linear factors or whatever you call them

Thanks
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9 years ago
#2
(x-1) isn't a factor, but there is a trivial factor there... just keep looking at other low integers.
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#3
(Original post by GHOSH-5)
(x-1) isn't a factor, but there is a trivial factor there... just keep looking at other low integers.
lol that was an example
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9 years ago
#4
If you need to factorise a cubic first you check, 1,2,0,-1,-2 and see if they are solutions

If you are factorizing it in an exam it will be one of these, or perhaps with a 3 if they are feeling nasty.

You should see 1 is to big. And 0 is too big but look what happens with -1
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9 years ago
#5
I think(x-2) is a factor, need to check.
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9 years ago
#6
(Original post by Danielisew)
Hey Babes.

I am trying to factorise the denominator of this and am not sure how. Someone said find a factor and divide by this e.g. if 1 is a factor, divide the denominator by (x-1). However, I tried this and it failed to work really..

Q) By firstly factorising the denominator, express the following as a partial fraction:

(5x^2 - 15x - 8) / (x^3 - 4x^2 + x + 6)

I know what to do but simply can't factorise the bottom to get three linear factors or whatever you call them

Thanks
Factor theorum and then divide
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#7
Ok f(3) = 0 ?

Therefore, is (x-3) a factor ?
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9 years ago
#8
As said above, there is an easy factor of the bottom (use remainder theorem with some small test values) then make sure the top doesn't factorise and cancel something out.
Then proceed with partial fractions method.
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9 years ago
#9
(Original post by Danielisew)
Ok f(3) = 0 ?

Therefore, is (x-3) a factor ?
it is, it appears.
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9 years ago
#10
(Original post by Danielisew)
Ok f(3) = 0 ?

Therefore, is (x-3) a factor ?
yes. Also try some of the other number I suggested and you should get all three with relative ease.
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#11
Babes, it worked with dividing by (x-3) as I get:

(x-3)(x+1)(x-2) And looking at the answer, these are in the denominator
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9 years ago
#12
It turns out that both f(2) and f(3) is 0 so both (x-2) and (x-3) are factors.

Hooray
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#13
(Original post by ~|Shock|~)
It turns out that both f(2) and f(3) is 0 so both (x-2) and (x-3) are factors.

Hooray
Yes

But if you find one, you can just divide by it and then find the other two when you factorise your answer from the division. Awfully sticky I know, but it is necessary to progress through the question
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9 years ago
#14
> solve([x^3-4*x^2+x+6], x);
{x = 2}, {x = 3}, {x = -1}
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