Verified answer

Factoring the denominator gives:

4/(x^3 - 2x^2) = 4/[x^2(x - 2)].

By partial fractions, we have that for some A, B, and C:

4/[x^2(x - 2)] = A/x + B/x^2 + C/(x - 2)

==> 4 = Ax(x - 2) + B(x - 2) + Cx^2.

i) Letting x = 0 ==> -2B = 4 ==> B = -2

ii) Letting x = 2 ==> 4C = 4 ==> C = 1

iii) Letting x = 1 ==> -A - B + C = 4 ==> -A + 2 + 1 = 4 ==> A = -1.

Therefore, 4/(x^3 - 2x^2) = -1/x - 2/x^2 + 1/(x - 2).

I hope this helps!

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4/(x^3 - 2x^2)

=> 4 / x^2(x - 2) = A/x + B/x^2 + C/(x - 2)

=> 4 = Ax(x-2) + B(x-2) + Cx^2

=> 4 = Ax^2 - 2Ax + Bx - 2B + Cx^2

=> 4 = x^2(A + C) + x (-2A + B) - 2B

-2B = 4 ==> B = -2

A + C = 0 ==> A = -C

-2A + B = 0 ==> -2A -2 = 0 ==> A = -1 and C = 1

A = -1, B = -2 and C = 1

4/(x^3 - 2x^2) = -( 1/x ) - ( 2/x^2 ) + ( 1/(x-2) )

4/x^2(x-2) = A/(x-2) + B/x + C/x^2

4 = Ax^2 + Bx(x-2) + C(x-2)

4 = Ax^2 + Bx^2 - 2Bx + Cx-2C

Compare the coefficients of x^2 on both sides

0 = A + B

A=-B

Compare the coefficients of x on both sides

0 = -2B+C

2B=C

Compare the constants on both sides

4 = -2C

C=-2

B=C/2 = -1

A=1

A=1; B=-1; C=-2

4/x^2(x-2) = A/(x-2) + B/x + C/x^2

4/x^2(x-2) = 1/(x-2) -1/x -2/x^2