Answer:
The partial fraction decomposition is
.
Step-by-step explanation:
Partial-fraction decomposition is the process of starting with the simplified answer and taking it back apart, of "decomposing" the final expression into its initial polynomial fractions.
To find the partial fraction decomposition of
:
First, the form of the partial fraction decomposition is
![\frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}=\frac{A}{x + 1}+\frac{B}{\left(x + 1\right)^{2}}+\frac{C}{x + 2}](https://tex.z-dn.net/?f=%5Cfrac%7B-%204%20x%5E%7B2%7D%20%2B%2013%20x%20-%2012%7D%7B%5Cleft%28x%20%2B%201%5Cright%29%5E%7B2%7D%20%5Cleft%28x%20%2B%202%5Cright%29%7D%3D%5Cfrac%7BA%7D%7Bx%20%2B%201%7D%2B%5Cfrac%7BB%7D%7B%5Cleft%28x%20%2B%201%5Cright%29%5E%7B2%7D%7D%2B%5Cfrac%7BC%7D%7Bx%20%2B%202%7D)
Write the right-hand side as a single fraction:
![\frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}=\frac{\left(x + 1\right)^{2} C + \left(x + 1\right) \left(x + 2\right) A + \left(x + 2\right) B}{\left(x + 1\right)^{2} \left(x + 2\right)}](https://tex.z-dn.net/?f=%5Cfrac%7B-%204%20x%5E%7B2%7D%20%2B%2013%20x%20-%2012%7D%7B%5Cleft%28x%20%2B%201%5Cright%29%5E%7B2%7D%20%5Cleft%28x%20%2B%202%5Cright%29%7D%3D%5Cfrac%7B%5Cleft%28x%20%2B%201%5Cright%29%5E%7B2%7D%20C%20%2B%20%5Cleft%28x%20%2B%201%5Cright%29%20%5Cleft%28x%20%2B%202%5Cright%29%20A%20%2B%20%5Cleft%28x%20%2B%202%5Cright%29%20B%7D%7B%5Cleft%28x%20%2B%201%5Cright%29%5E%7B2%7D%20%5Cleft%28x%20%2B%202%5Cright%29%7D)
The denominators are equal, so we require the equality of the numerators:
![- 4 x^{2} + 13 x - 12=\left(x + 1\right)^{2} C + \left(x + 1\right) \left(x + 2\right) A + \left(x + 2\right) B](https://tex.z-dn.net/?f=-%204%20x%5E%7B2%7D%20%2B%2013%20x%20-%2012%3D%5Cleft%28x%20%2B%201%5Cright%29%5E%7B2%7D%20C%20%2B%20%5Cleft%28x%20%2B%201%5Cright%29%20%5Cleft%28x%20%2B%202%5Cright%29%20A%20%2B%20%5Cleft%28x%20%2B%202%5Cright%29%20B)
Expand the right-hand side:
![- 4 x^{2} + 13 x - 12=x^{2} A + x^{2} C + 3 x A + x B + 2 x C + 2 A + 2 B + C](https://tex.z-dn.net/?f=-%204%20x%5E%7B2%7D%20%2B%2013%20x%20-%2012%3Dx%5E%7B2%7D%20A%20%2B%20x%5E%7B2%7D%20C%20%2B%203%20x%20A%20%2B%20x%20B%20%2B%202%20x%20C%20%2B%202%20A%20%2B%202%20B%20%2B%20C)
The coefficients near the like terms should be equal, so the following system is obtained:
![\begin{cases} A + C = -4\\3 A + B + 2 C = 13\\2 A + 2 B + C = -12 \end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%20A%20%2B%20C%20%3D%20-4%5C%5C3%20A%20%2B%20B%20%2B%202%20C%20%3D%2013%5C%5C2%20A%20%2B%202%20B%20%2B%20C%20%3D%20-12%20%5Cend%7Bcases%7D)
Solving this system, we get that
.
Therefore,
![\frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}=\frac{50}{x + 1}+\frac{-29}{\left(x + 1\right)^{2}}+\frac{-54}{x + 2}](https://tex.z-dn.net/?f=%5Cfrac%7B-%204%20x%5E%7B2%7D%20%2B%2013%20x%20-%2012%7D%7B%5Cleft%28x%20%2B%201%5Cright%29%5E%7B2%7D%20%5Cleft%28x%20%2B%202%5Cright%29%7D%3D%5Cfrac%7B50%7D%7Bx%20%2B%201%7D%2B%5Cfrac%7B-29%7D%7B%5Cleft%28x%20%2B%201%5Cright%29%5E%7B2%7D%7D%2B%5Cfrac%7B-54%7D%7Bx%20%2B%202%7D)
x^3 = 5x
Subtract 5x from both sides:
x^3 - 5x = 0
Factor X out of x^3:
x(x^2-5x) = 0
Set X to 0 and solve for x:
x = 0
Set (X^2-5) to equal 0 and solve for x:
x^2 - 5 = 0
Add 5 to both sides:
x^2 = 5
Take the square root of both sides:
x = √5, -√5
The three solutions of x are 0, √5, -√5
In order from smallest to largest = -√5, 0, √5
The largest possible value is √5
12X-6=180
12X=186
X=186/12=15.5