How do you express #(x^2 - 8x + 44) / ((x + 2) (x - 2)^2)# in partial fractions?

Question

Benjamin Achtenberg · Accepted Answer

#(x^2-8x+44)/[(x+2)*(x-2)^2]=4/(x+2)-3/(x-2)+8/(x-2)^2#

#(x^2-8x+44)/[(x+2)*(x-2)^2]#

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

After expanding denominator,

#A*(x-2)^2+B*(x^2-4)+C*(x+2)=x^2-8x+44#

Set #x=-2#, #16A=64#, so #A=4#

Set #x=2#, #4C=32#, so #C=8#

Set #x=0#, #4A-4B+2C=44#, so #B=-3#

Thus,

#(x^2-8x+44)/[(x+2)*(x-2)^2]=4/(x+2)-3/(x-2)+8/(x-2)^2#

Chloe Alexander · Answer

undefined To express $\frac{x^2 - 8x + 44}{(x + 2)(x - 2)^2}$ in partial fractions, you first need to factor the denominator as $ (x + 2)(x - 2)^2 $. Then, you set up the partial fraction decomposition as follows:

$$
\frac{x^2 - 8x + 44}{(x + 2)(x - 2)^2} = \frac{A}{x + 2} + \frac{B}{x - 2} + \frac{C}{(x - 2)^2}
$$

Next, you clear the fractions by multiplying both sides by the denominator:

$$
x^2 - 8x + 44 = A(x - 2)^2 + B(x + 2)(x - 2) + C(x + 2)
$$

Now, you can solve for $A$, $B$, and $C$ by choosing appropriate values of $x$ that will make some terms disappear:

1. Substitute $x = -2$:
$$(-2)^2 - 8(-2) + 44 = A(-2 - 2)^2 + C(-2 + 2)$$

2. Substitute $x = 2$:
$$2^2 - 8(2) + 44 = B(2 + 2)(2 - 2) $$

3. Differentiate and substitute $x = 2$:
$$2 - 8 = 2A(0) + B(2 + 2) + C$$

Now, you can solve for $A$, $B$, and $C$. After finding their values, substitute them back into the partial fraction decomposition.