QuestionFind the partial fraction decomposition. $\frac{-2 x-12}{x^{3}-2 x^{2}}$

Studdy Solution

STEP 1

What is this asking? Rewrite this big fraction as the sum of smaller, simpler fractions! Watch out! Don't forget to factor the denominator completely before starting the decomposition.
Also, remember to account for repeated factors!

STEP 2

1. Factor the Denominator
2. Set up the Decomposition
3. Solve for the Constants
4. Write the Final Decomposition

STEP 3

Alright, let's factor that denominator!
We've got $x^3 - 2x^2$ .
We can see that $x^2$ is a common factor, so we can pull it out: $x^2(x - 2)$ .
Boom!

STEP 4

Now, we set up our partial fraction decomposition.
Since we have a repeated factor $x^2$ , we need two fractions for it: one with $x$ and one with $x^2$ .
We also have a factor of $(x-2)$ , so we need a fraction for that too.
This gives us: $\frac{-2x - 12}{x^3 - 2x^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-2}$ Where $A$ , $B$ , and $C$ are constants we need to find.
Exciting!

STEP 5

Let's multiply both sides of our equation by the original denominator, $x^2(x-2)$ , to get rid of those pesky fractions: $-2x - 12 = A \cdot x(x-2) + B \cdot (x-2) + C \cdot x^2$

STEP 6

Now, let's be clever and choose some strategic values for $x$ to make our lives easier.
If we let $x = 0$ , we get: $-12 = -2B$ So, $B = \textbf{6}$ !

STEP 7

Next, let $x = 2$ .
This gives us: $-16 = 4C$ So, $C = \textbf{-4}$ !

STEP 8

We've got $B$ and $C$ , now we just need $A$ .
Let's expand our equation and group like terms: $-2x - 12 = Ax^2 - 2Ax + Bx - 2B + Cx^2$ $-2x - 12 = (A+C)x^2 + (-2A+B)x - 2B$ Since we already know $B = 6$ and $C = -4$ , we can substitute those values: $-2x - 12 = (A-4)x^2 + (-2A+6)x - 12$ Now, look at the coefficients of $x$ : $-2 = -2A + 6$ .
Solving for $A$ , we get $A = \textbf{4}$ !

STEP 9

We found $A = 4$ , $B = 6$ , and $C = -4$ .
Let's plug these back into our decomposition: $\frac{-2x - 12}{x^3 - 2x^2} = \frac{4}{x} + \frac{6}{x^2} + \frac{-4}{x-2}$

STEP 10

Our final answer is: $\frac{4}{x} + \frac{6}{x^2} - \frac{4}{x-2}$

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