Question

partial fraction decomposition:

Answer 1

oh little joy!

Answer 2

my work thus far:

Answer 3

wow you have choices
you could check instead but you can to it step by step

Answer 4

i know haha. But my final won't be multiple choice so might as well learn it right? haha

Answer 5

might be a snappier way to do it

Answer 6

if i were to that would i just get the denominators the same and then add them to see which equals the original? (I still want to do it by had tho lol..need to get it down)

Answer 7

\[\frac{A}{x-2}+\frac{Bx+C}{x^2+2x+2}\]for sure
then \[A(x^2+2x+2)+(Bx+C)(x-2)=10\]

Answer 8

now instead of equating like coefficients, let \(x=2\)

Answer 9

you get
\[A(2^2+2\times 2+2)=10\] or
\[10A=10\] making \(A=1\) that gives you a running head start

Answer 10

ok so A is one...and now we need B and C..

Answer 11

right and now it is easier than solving a 3 by 3 system because you get
\[x^2+2x+2+(Bx+D)(x-2)=10\]

Answer 12

typo there but you get the idea right?

Answer 13

ya C instead of D i get u tho

Answer 14

what could i plug in to get b or c ?

Answer 15

nothing now becuase \(x^2+2x+2\) has no real roots so you cannot use that trick

Answer 16

unless you want to use complex numbers (it works but is messy)

Answer 17

now you can probably get B by thinking
on the left you have \[x^2+Bx^2\] on the right you have no \(x^2\) so \(B=-1\)

Answer 18

that is the "equate like coefficients' method, but in your head

Answer 19

ya i get that part!

Answer 20

so now plug a and b to get C?

Answer 21

yup

Answer 22

easier to get some quickly and avoid solving a system of equations

Answer 23

ok stuck. i have C and X and idk what to do..

Answer 24

we know A = 1, B = -1 right?

Answer 25

yes

Answer 26

so \[x^2+2x+2+(Bx+C)(x-2)=10\] becomes
\[x^2+2x+2+(-x+C)(x-2)=10\]

Answer 27

yes i got that part. But i have to variables

Answer 28

you can still multiply out if you like, or just look at the constant on the left
it will be \(2-2C\) that is the only term without an \(x\) or an \(x^2\) in it

Answer 29

that means
\[2-2C=10\]

Answer 30

so C is -4?

Answer 31

if this is too confusing you can go ahead and multiply all that mess out, hope you don't make aqn algebra mistake, then equate like coefficients and solve a 3 by 3 system of equations
i find that almost impossible to do without the computer

Answer 32

yes of course
\[2-2C=10\\
-2C=8\\
C=-4\]

Answer 33

so the answer is the last equation?

Answer 34

yes

Answer 35

if you are more comfortable with your method, you got to the step where you have to solve
\[A+B=0\\
2A-2B+C=0\\
A-C=5\] if you care to solve that it will work assuming you did not make an algebra mistake

Answer 36

sounds good thank you so much :)

Answer 37

yw