How do you write the partial fraction decomposition of the rational expression # (8x^2 - 4x - 8)/(x^4 + 2x^3)#?

Question

Anna Allen · Accepted Answer

#(8x^2 - 4x - 8)/(x^3(x+ 2))=-4/(x+2)+4/x-4/x^3# We have: #(8x^2 - 4x - 8)/(x^4 + 2x^3)=(8x^2 - 4x - 8)/(x^3(x+ 2))# #(8x^2 - 4x - 8)/(x^3(x+ 2))=A/(x+2)+B/x+C/x^2+D/x^3# #color(white)((8x^2 - 4x - 8)/(x^3(x+ 2)))=(Ax+B(x+2))/(x(x+2))+C/x^2+D/x^3# #color(white)((8x^2 - 4x - 8)/(x^3(x+ 2)))=(Ax^2+Bx(x+2)+C(x+2))/(x^2(x+2))+D/x^3# #color(white)((8x^2 - 4x - 8)/(x^3(x+ 2)))=(Ax^3+Bx^2(x+2)+Cx(x+2)+D(x+2))/(x^3(x+2))# #8x^2 - 4x - 8=Ax^3+Bx^2(x+2)+Cx(x+2)+D(x+2)# Putting in #x=0# gives us: #-8=D(2)# #D=-4# #8x^2 - 4x - 8=Ax^3+Bx^2(x+2)+Cx(x+2)-4(x+2)# Now putting in #x=-2# gives us: #8(-2)^2 - 4(-2) - 8=A(-2)^3+B(-2)^2(-2+2)+C(-2)(-2+2)-4(-2+2)# #32=-8A# #A=-4# #8x^2 - 4x - 8=-4x^3+Bx^2(x+2)+Cx(x+2)-4(x+2)# There are no numbers that can give us just #B# or #C#, so we must find one in terms of the other, and so we shall put in #x=1# to get: #-4=-4+3B+3C-12# #12=3(B+C)# #B=4-C# (we'll shall put this back in the original equation to help find #C#. #8x^2 - 4x - 8=-4x^3+(4-C)x^2(x+2)+Cx(x+2)-4(x+2)# We need to put in a random value, i.e. #x=2# #8(2)^2 - 4(2)- 8=-4(2)^3+(4-C)(2)^2((2)+2)+C(2)((2)+2)-4((2)+2)# #16=8c+16(4-C)-48# #64=16(4-C)# #4=4-C# #C=0# Remember that #B=4-C# #B=4-0=4# #(8x^2 - 4x - 8)/(x^3(x+ 2))=-4/(x+2)+4/x+0/x^2-4/x^3# #color(white)((8x^2 - 4x - 8)/(x^3(x+ 2)))=-4/(x+2)+4/x-4/x^3#

Lily Anderson · Answer

undefined To write the partial fraction decomposition of the rational expression $ \frac{8x^2 - 4x - 8}{x^4 + 2x^3} $, follow these steps:

1. Factor the denominator $ x^4 + 2x^3 $ if possible.
2. Write the fraction in the form of partial fractions.
3. Find the values of the unknown constants.

Since $ x^4 + 2x^3 = x^3(x + 2) $, the partial fraction decomposition is of the form:

$$ \frac{8x^2 - 4x - 8}{x^4 + 2x^3} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x + 2} $$

To find the values of $ A $, $ B $, $ C $, and $ D $, perform partial fraction decomposition and solve for the unknowns.