Question

Find the partial decomposition of
-3x^2 + 13x -12 all over x^3 - 4x^2 +4x

Answer 1

im thinking you have to factor out the denominator?

Answer 2

factor out an x in x^3 - 4x^2 +4x

Answer 3

\[x^3-4x^2+4x=x(x-2)^2\]

Answer 4

(-3x^2 + 13x -12)/(x(x-2)^2)

Answer 5

yea.. i forgot how to factor cube

Answer 6

A/x + B/(x-2) + C/(x-2)^2=(-3x^2 + 13x -12)/(x(x-2)^2)

Answer 7

whats the next step?

Answer 8

what do you think is the next step?

Answer 9

i wrote A(x-2)(x-2)^2 + Bx(x-2)^2 + Cx(x-2) = -3x^2 +13x -12

Answer 10

i think its wrong

Answer 11

Multiply both sides by (x (x-2)²)

Answer 12

looks good to me except for the
\[Bx(x-2)^2\] part. think that should be
\[Bx(x-2) \] yes?

Answer 13

\[A(x-2)^2 + Bx(x-2) + Cx = -3x^2 +13x -12
\]

Answer 14

right, now let x=0 and solve for A

Answer 15

\[A(-2)^2=-12\]
\[4A=-12\]
\[A=-3\] how'd i do?

Answer 16

that looks gooD!

Answer 17

now guess what you let x =?

Answer 18

2?

Answer 19

yup

Answer 20

got it

Answer 21

how about b? do i plug in for A and C?

Answer 22

did you find C?

Answer 23

C = 1

Answer 24

think b = 0

Answer 25

ok so now you have
\[-3(x-2)^2+Bx(x-2)+x=-3x^2+13x-12\]

Answer 26

yes B = 0

Answer 27

can you help me with another problem im going to make a new post

Answer 28

sure make a new one

Answer 29

btw answer is
\[\frac{1}{(x-2)^2}-\frac{3}{x}\]

Answer 30

ok