Partial Fraction Q (factorising denominator) Watch

Danielisew
Badges: 1
Rep:
?
#1
Report Thread starter 9 years ago
#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
0
quote
reply
Unbounded
Badges: 12
Rep:
?
#2
Report 9 years ago
#2
(x-1) isn't a factor, but there is a trivial factor there... just keep looking at other low integers.
0
quote
reply
Danielisew
Badges: 1
Rep:
?
#3
Report Thread starter 9 years ago
#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
0
quote
reply
The Muon
Badges: 2
Rep:
?
#4
Report 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
0
quote
reply
~|Shock|~
Badges: 0
Rep:
?
#5
Report 9 years ago
#5
I think(x-2) is a factor, need to check.
0
quote
reply
modini
Badges: 14
Rep:
?
#6
Report 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
0
quote
reply
Danielisew
Badges: 1
Rep:
?
#7
Report Thread starter 9 years ago
#7
Ok f(3) = 0 ?

Therefore, is (x-3) a factor ?
0
quote
reply
benwellsday
Badges: 12
Rep:
?
#8
Report 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.
0
quote
reply
Unbounded
Badges: 12
Rep:
?
#9
Report 9 years ago
#9
(Original post by Danielisew)
Ok f(3) = 0 ?

Therefore, is (x-3) a factor ?
it is, it appears.
0
quote
reply
The Muon
Badges: 2
Rep:
?
#10
Report 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.
0
quote
reply
Danielisew
Badges: 1
Rep:
?
#11
Report Thread starter 9 years ago
#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
0
quote
reply
~|Shock|~
Badges: 0
Rep:
?
#12
Report 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
0
quote
reply
Danielisew
Badges: 1
Rep:
?
#13
Report Thread starter 9 years ago
#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
0
quote
reply
Student#1
Badges: 0
Rep:
?
#14
Report 9 years ago
#14
> solve([x^3-4*x^2+x+6], x);
{x = 2}, {x = 3}, {x = -1}
0
quote
reply
X

Reply to thread

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of Lincoln
    Brayford Campus Undergraduate
    Wed, 12 Dec '18
  • Bournemouth University
    Midwifery Open Day at Portsmouth Campus Undergraduate
    Wed, 12 Dec '18
  • Buckinghamshire New University
    All undergraduate Undergraduate
    Wed, 12 Dec '18

Do you like exams?

Yes (146)
18.27%
No (486)
60.83%
Not really bothered about them (167)
20.9%

Watched Threads

View All