Write #(9x^2+26x+20)/((1+x)(2+x)^2)# as partial fraction, then expand as binomials up to and including #x^3#. Show the expansion is roughly #5-(7x)/2+Bx^2+Cx^3#?

For partial fractions I got:
#3/((1+x))+2/((2+x))+4/(2+x)^2#

For the expansion, I got: #5-(9x)/2+4x^2-(27x^3)/8#, I'm not sure why I'm getting #-(9x)/2# and not #-(7x)/2#

Answer 1

partial fraction decomposition is:

# (9x^2+26x+20)/((1+x)(2+x)^2) -= 3/(1+x) + 6/(2+x) - 4/(2+x)^2 #

The binomial expansion is:

# (9x^2+26x+20)/((1+x)(2+x)^2) = 5 -7/2x+3x^2 -23/8x^3 + ...#

Leading to #B=3# and #C=-23/8#

The partial fraction decomposition will be of the form:

# (9x^2+26x+20)/((1+x)(2+x)^2) -= A/(1+x) + B/(2+x) + C/(2+x)^2 # # " " = ( A(2+x)^2 + B(2+x)(2+x) + C(1+x) ) / ((1+x)(2+x)^2)#

Leading to the identity:

# 9x^2+26x+20 -= A(2+x)^2 + B(1+x)(2+x) + C(1+x) #
Where #A,B# are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:
Put # x = -1 => 9-26+20=A=> A = 3# Put # x = -2 => 36-52+20=-C=>C=-4# Coeff#(x^0) => 20=4A+2B+C=> B=6#

So we can now write:

# (9x^2+26x+20)/((1+x)(2+x)^2) -= 3/(1+x) + 6/(2+x) - 4/(2+x)^2 #

Now we can write the as

# S = 3/(1+x) + 6/(2+x) - 4/(2+x)^2 # # \ \ = 3/(1+x) + 6/(2(1+x/2)) - 4/(2(1+x/2))^2 # # \ \ = 3(1+x)^(-1) + 3(1+x/2)^(-1) - (1+x/2)^(-2) #

The Binomial Series tells us that:

# (1+x)^n = 1+nx + (n(n-1))/(2!)x^2 (n(n-1)(n-2))/(3!)x^3 + ...#
# S_1 = 3(1+x)^(-1) # # \ \ \ \ = 3{1+(-1)x+((-1)(-2))/(2)x^2 + ((-1)(-2)(-3))/6x^3 + ... } # # \ \ \ \ = 3{1-x+x^2 -x^3 + ... } # # \ \ \ \ = 3-3+3x^2 -3x^3 + ... #
# S_2 = 3(1+x/2)^(-1) # # \ \ \ \ = 3{1+(-1)(x/2)+((-1)(-2))/(2)(x/2)^2 + ((-1)(-2)(-3))/6(x/2)^3 + ... }# # \ \ \ \ = 3{1-1/2x+1/4x^2 -1/8x^3+ ... }# # \ \ \ \ = 3-3/2x+3/4x^2 -3/8x^3+ ... #
# S_3 = (1+x/2)^(-2) # # \ \ \ \ = 1+(-2)(x/2)+((-2)(-3))/2(x/2)^2+((-2)(-3)(-4))/6(x/2)^3+... # # \ \ \ \ = 1-x+3/4x^2-1/2x^3+... #

Combining these results:

S= {3-3x+3x^2 -3x^3 } + {3-3/2x+3/4x^2 -3/8x^3} - {

1-x+3/4x^2-1/2x^3}+O(x^4) #

# \ \ = (3+3-1) + (-3-3/2+1)x +(3+3/4-3/4)x^2 + (-3-3/8+1/2)x^3#
# \ \ = 5 -7/2x+3x^2 -23/8x^3 + ...#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To write (\frac{9x^2 + 26x + 20}{(1+x)(2+x)^2}) as partial fractions, first decompose it into the form (\frac{A}{1+x} + \frac{B}{2+x} + \frac{C}{(2+x)^2}). Then, find the values of A, B, and C. After that, expand the partial fractions and simplify to the form (5 - \frac{7x}{2} + Bx^2 + Cx^3).

First, write the partial fractions: [ \frac{9x^2 + 26x + 20}{(1+x)(2+x)^2} = \frac{A}{1+x} + \frac{B}{2+x} + \frac{C}{(2+x)^2} ]

Multiply both sides by ((1+x)(2+x)^2) to get: [ 9x^2 + 26x + 20 = A(2+x)^2 + B(1+x)(2+x) + C(1+x) ]

Expand both sides and simplify: [ 9x^2 + 26x + 20 = A(4 + 4x + x^2) + B(2 + 2x + x + x^2) + C(1 + x) ] [ 9x^2 + 26x + 20 = A(x^2 + 4x + 4) + B(x^2 + 3x + 2) + C(x + 1) ] [ 9x^2 + 26x + 20 = Ax^2 + 4Ax + 4A + Bx^2 + 3Bx + 2B + Cx + C ]

Equating coefficients, we get the following system of equations:

  1. (A + B = 9)
  2. (4A + 3B + C = 26)
  3. (4A + 2B + C = 20)

Solve the system to find: (A = -1), (B = 10), (C = -1)

Therefore, the partial fraction decomposition is: [ \frac{9x^2 + 26x + 20}{(1+x)(2+x)^2} = \frac{-1}{1+x} + \frac{10}{2+x} - \frac{1}{(2+x)^2} ]

Expanding the partial fractions: [ \frac{-1}{1+x} = -1 + x - x^2 + x^3 - ... ] [ \frac{10}{2+x} = 10 - 5x + 2.5x^2 - 1.25x^3 + ... ] [ \frac{-1}{(2+x)^2} = -1 + 2x - 3x^2 + 4x^3 - ... ]

Combining the expansions: [ -1 + x - x^2 + x^3 + 10 - 5x + 2.5x^2 - 1.25x^3 - 1 + 2x - 3x^2 + 4x^3 + ... ]

Simplify the terms up to and including (x^3): [ 5 - \frac{7x}{2} + 5.5x^2 + 1.75x^3 + ... ]

Therefore, the expansion is roughly (5 - \frac{7x}{2} + Bx^2 + Cx^3), where (B = 5.5) and (C = 1.75).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7