# 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#?

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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#

For partial fractions I got:

For the expansion, I got:

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

The partial fraction decomposition will be of the form:

Leading to the identity:

So we can now write:

Now we can write the as

The Binomial Series tells us that:

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) #

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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:

- (A + B = 9)
- (4A + 3B + C = 26)
- (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).

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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.

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