Partial Fraction Q (factorising denominator)

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 :smile:

Thanks

Reply 1

Danielisew

OP

1

lol that was an example :frown:

Reply 2

The Muon

2

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

Reply 3

~|Shock|~

I think(x-2) is a factor, need to check.

Reply 4

modini

15

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 :smile:

Thanks

Factor theorum and then divide

Reply 5

Danielisew

OP

1

Ok f(3) = 0 ?

Therefore, is (x-3) a factor ? :smile:

Reply 6

benwellsday

12

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.

Reply 7

The Muon

2

Danielisew

Ok f(3) = 0 ?

Therefore, is (x-3) a factor ? :smile:

yes. Also try some of the other number I suggested and you should get all three with relative ease.

Reply 8

Danielisew

OP

1

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 :smile:

Reply 9

~|Shock|~

It turns out that both f(2) and f(3) is 0 so both (x-2) and (x-3) are factors.

Hooray :smile:

Reply 10

Danielisew

OP

1

~|Shock|~

It turns out that both f(2) and f(3) is 0 so both (x-2) and (x-3) are factors.

Hooray :smile:

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 :smile:

Reply 11

Student#1

> solve([x^3-4*x^2+x+6], x);
{x = 2}, {x = 3}, {x = -1}

Partial Fraction Q (factorising denominator)

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