Question

Express 2x^2 +9x -6/(x^2 + x -6) as a partial fraction

Answer 1

divide first

Answer 2

Dont we split the bottom into the components? factoring the bottom to (x+3)(x-2) ?

Answer 3

not yet

Answer 4

the numerator and denominator are both polynomials of degree 2, so first you have to divide
you will get 2 + something and the "something" is what you will express as partial fractions

Answer 5

Ahhh i see, this is part 2 of my investigation. Algebraic long division. Okie doke i'll do that first.

Answer 6

So doing that we end up with 2 + 7x+6/x^2+x-6 ?

Answer 7

this problem has been cooked up to be easy
you will get nice integers for the partial fractions

Answer 8

right

Answer 9

if the numerator has degree greater than or equal to the denominator, you need to divide first, get a quotient and a remainder, then you can find partial fractions for the remainder

kind of like if you had \(\frac{24}{10}\) you might write is as \(2\tfrac{4}{10}\) and then mess around with the \(\frac{4}{10}\) part

Answer 10

Thankyou very much, i think i've got it now! I've got the answer as 2+ (3/x+3) + (4/x-2)

Answer 11

I'd never come across a question like that before, and i've got a test on it tomorrow :\ Hopefully i'll do ok :S

Answer 12

i think so
there is a very snappy way to do it, do you know how?

Answer 13

No i dont? I would like to see it if you have time?

Answer 14

forgetting about the 2, and looking only at
\[\frac{7x+6}{(x+3)(x-2)}\] you know this is going to be
\[\frac{A}{x+3}+\frac{B}{x-2}\] so to find B, put your finger over the \(x-2\) and the denominator
\[\frac{7x+6}{(x+3)\cancel{(x-2)}}\]and then replace \(x\) by 2 to get \(\frac{14+6}{5}=4\)

Answer 15

then to find A put your finger over the \(x+3\) in the denominator
\[\frac{7x+6}{\cancel{(x+3)}(x-2)}\] and replace \(x\) by \(-3\) to get
\[\frac{-21+6}{-5}=3\]

Answer 16

Oh wow.... does this work with every equation?

Answer 17

well it works for ones where the denominator has non repeated linear factors

Answer 18

so i wouldn't work in the case of say \(x^2+x+1\) in the denominator, or \((x-4)^2\)

Answer 19

it is really just a gimmick, it works exactly the same way as if you had written
\[A(x-2)+B(x+3)=7x+6\] and then replaced \(x\) by \(2\)
just solves in one step is all

Answer 20

Hmm, would you advise doing it the long way in a test? because i have a test on this tomorrow :\

Answer 21

whatever you are more comfortable with
it really only saves a second from writing \[A(x-2)+B(x+3)=7x+6\] and then putting \(x=2\) to get \(5B=7\times 2+6\)

Answer 22

the most common mistake is to forget which one you are solving for, so make sure on your paper to write \[\frac{A}{x+3}+\frac{B}{x-2}\] so you an remember which one is over each factor

Answer 23

Hmmm i've never learnt that before. I went with
ax-2a+bx+3b = 7x + 6
then x(a+b) -2a +3b
(a+b)=7 and 3b-2a=6
then solve those two simultaneously

Answer 24

ick

Answer 25

now you have to solve a system of equations

Answer 26

apparently you did not have much trouble doing it, but it seems to me the long way to do things, and it is almost always unnecessary

Answer 27

so we can just plug x=2 into the a term to cancel the a and then solve? Does this always work? assuming one term is cancelled?

Answer 28

multiply out
collect terms
equate like coefficients
solve the system

you might need this when you do laplace transforms, but for algebra it is too much work

Answer 29

No this isnt in our syllabus. Im in the highest highschool maths that is non-specialist. but this isnt in our course. They call it an "extended piece of work" or EPW which is essentially a mathematical investigation.

Answer 30

yeah you can almost always use this method

Answer 31

i think i recall that my math teacher was an "extended piece of work"

Answer 32

Hahahaha mine is great, we have a class of 6 (myself included)

Answer 33

here is an example, the answer will not be so nice but the method is easy
\[\frac{2x+3}{(x-4)(x+3)}\]

Answer 34

you know it is
\[\frac{A}{x-4}+\frac{B}{x+3}\] so you know that
\[A(x+3)+B(x-4)=2x+3\] and the point is that this has to be true NO MATTER WHAT X IS

Answer 35

therefore you are free to pick any \(x\) you like
and the obvious choice is \(x=4\) because that will get rid of the B term and leave you with
\[7A=2\times 4+3\] or
\[7A=11\] so \(A=\frac{11}{7}\)
much easier than multiplying out, collecting terms and then solving a system

Answer 36

but do whatever you are comfortable with

Answer 37

So just testing my skill. sub-ing x=-3 gives b=3/7

Answer 38

yes

Answer 39

easy right? just make sure to recall which one is A and which one is B, in other words, don't skip the step of writing \(\frac{A}{x-4}+\frac{B}{x+3}\) somewhere on your paper so you can refer back to it

Answer 40

good luck on your test

Answer 41

@satellite73 sorry mate, i lost connection then. But Thankyou very much for all of the help! I definitely owe you one. And so I just wanted to wish you all the best, and a happy and safe day/night, wherever you are. Its night here in aus. Thankyou again for all the help mate, you're a bloody legend!

Answer 42

you are quite welcome.

i am still laughing at "piece of work" maybe they don't have that idiom in aus

Answer 43

@satellite73 Hahaha no we do :P It was pretty funny. Just a quick question. How do you know which term goes first, so you dont solve the a and b wrong? so if you have (x-3)(x+4) how do you know which term to make a and which to make b? Im assuming a wrong choice would reverse the values?