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17Calculus Precalculus - Quadratic Partial Fraction Expansion

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Partial Fraction Expansion - Quadratic Factors (Single and Repeating)

Okay, now we look at how to handle quadratic factors when they cannot be factored (in the real number system).

Terms with quadratic factors are of the form \(ax^2+bx+c\) where the highest power on the variable (\(x\) in this case) is two and \(a \neq 0\). We do not have the same requirement on \(b\) and \(c\), so \(ax^2\) and \(ax^2+c\) are also considered quadratic factors. However, the term \(ax^2+bx\) is NOT considered a quadratic factor, since it can be factored into two simple factors, i.e. \(ax^2 + bx = x(ax+b)\). In this case, we follow the techniques associated with simple factors.

When we have a quadratic factor, the numerator must be of the form \(Ax+B\). Notice that we now have an \(x\) in the numerator, not just constants. Also, notice that the highest power of \(x\) in the numerator is one less than the highest power in the denominator.

Quadratic Single Factors Example

\(\displaystyle{ \frac{1}{x(x^2+3)} = \frac{A}{x} + \frac{Bx+C}{x^2+3} }\)

Repeated Factors

If you understand how repeated factors work for linear terms discussed on the previous page, you should be able to anticipate how repeated factors work for quadratic factors.

factor in the denominator

partial fraction terms

\(ax^2+bx+c\)

\(\displaystyle{ \frac{A_1x+B_1}{ax^2+bx+c} }\)

\((ax^2+bx+c)^2\)

\(\displaystyle{ \frac{A_1x+B_1}{ax^2+bx+c} + \frac{A_2x+B_2}{(ax^2+bx+c)^2} }\)

\((ax^2+bx+c)^3\)

\(\displaystyle{ \frac{A_1x+B_1}{ax^2+bx+c} + \frac{A_2x+B_2}{(ax^2+bx+c)^2} + }\) \(\displaystyle{ + \frac{A_3x+B_3}{(ax^2+bx+c)^3} }\)

Do you see the pattern? For a denominator with the term \((ax^2+bx+c)^k\), we would have the factors

\(\displaystyle{ \frac{A_1x+B_1}{ax^2+bx+c} + \frac{A_2x+B_2}{(ax^2+bx+c)^2} + }\) \(\displaystyle{ \frac{A_3x+B_3}{(ax^2+bx+c)^3} + }\) \(\displaystyle{ . . . + \frac{A_kx+B_k}{(ax^2+bx+c)^k} }\)

Once these expansions are set up, the steps to find the constants are the same as with linear factors.

Okay, let's work the practice problems.

Practice

Unless otherwise instructed, expand the given fraction using partial fraction expansion. Give your answer in exact terms.

Basic

\(\displaystyle{\frac{5x^2+10+7}{(x^2+2)(x+3)}}\)

Problem Statement

Expand \(\displaystyle{\frac{5x^2+10+7}{(x^2+2)(x+3)}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

2926 video solution

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\(\displaystyle{\frac{x^4+3x^3+7x-1}{x(x^2+1)^2}}\)

Problem Statement

Expand \(\displaystyle{\frac{x^4+3x^3+7x-1}{x(x^2+1)^2}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

2927 video solution

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\(\displaystyle{\frac{6x^2+21x+11}{(x^2+3)(x+5)}}\)

Problem Statement

Expand \(\displaystyle{\frac{6x^2+21x+11}{(x^2+3)(x+5)}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

2928 video solution

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\(\displaystyle{\frac{3x^2+5x-4}{(x^2-7)(x+1)}}\)

Problem Statement

Expand \(\displaystyle{\frac{3x^2+5x-4}{(x^2-7)(x+1)}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

2929 video solution

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\(\displaystyle{\frac{3x^4-2x^3+6x^2-3x+3}{(x^2+2)^2(x+3)}}\)

Problem Statement

Expand \(\displaystyle{\frac{3x^4-2x^3+6x^2-3x+3}{(x^2+2)^2(x+3)}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

2930 video solution

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\(\displaystyle{\frac{x^2 - 1}{x(x^2+1)}}\)

Problem Statement

Expand \(\displaystyle{\frac{x^2 - 1}{x(x^2+1)}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

2931 video solution

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\(\displaystyle{\frac{2x^2-11x+17}{x^3-7x^2+15x-9}}\)

Problem Statement

Expand \(\displaystyle{\frac{2x^2-11x+17}{x^3-7x^2+15x-9}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

2932 video solution

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\(\displaystyle{\frac{7x^2-7x+18}{x^3+8}}\)

Problem Statement

Expand \(\displaystyle{\frac{7x^2-7x+18}{x^3+8}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

2933 video solution

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\(\displaystyle{\frac{-x^4+3x^3+6x^2+11x+20}{(x^2+2)^2(x+3)}}\)

Problem Statement

Expand \(\displaystyle{\frac{-x^4+3x^3+6x^2+11x+20}{(x^2+2)^2(x+3)}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

2934 video solution

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\(\displaystyle{\frac{5x^2+7x+8}{(x+1)(x^2+2x+3)}}\)

Problem Statement

Expand \(\displaystyle{\frac{5x^2+7x+8}{(x+1)(x^2+2x+3)}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

2935 video solution

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\(\displaystyle{\frac{2x^2-x+4}{x^2+4x}}\)

Problem Statement

Expand \(\displaystyle{\frac{2x^2-x+4}{x^2+4x}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

2936 video solution

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\(\displaystyle{\frac{5x-1}{(3x^2-2)(x-1)}}\)

Problem Statement

Expand \(\displaystyle{\frac{5x-1}{(3x^2-2)(x-1)}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

2937 video solution

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\(\displaystyle{\frac{x^2+4x-2}{x^3+4x}}\)

Problem Statement

Expand \(\displaystyle{\frac{x^2+4x-2}{x^3+4x}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

2938 video solution

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\(\displaystyle{\frac{1}{x(x^2+1)}}\)

Problem Statement

Expand \(\displaystyle{\frac{1}{x(x^2+1)}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

2939 video solution

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\(\displaystyle{\frac{10}{(x-1)(x^2+9)}}\)

Problem Statement

Expand \(\displaystyle{\frac{10}{(x-1)(x^2+9)}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

2940 video solution

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\(\displaystyle{\frac{x^2+x+1}{(2x+1)(x^2+1)}}\)

Problem Statement

Expand \(\displaystyle{\frac{x^2+x+1}{(2x+1)(x^2+1)}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

There is more going on in this problem than partial fraction expansion, which you can ignore. You will get a chance to do this in calculus.

2941 video solution

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\(\displaystyle{\frac{x^4+1}{x(x^2+1)^2}}\)

Problem Statement

Expand \(\displaystyle{\frac{x^4+1}{x(x^2+1)^2}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

Krista King Math - 2942 video solution

video by Krista King Math

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\(\displaystyle{\frac{2x^2-x+9}{x(x^2+9)}}\)

Problem Statement

Expand \(\displaystyle{\frac{2x^2-x+9}{x(x^2+9)}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

2943 video solution

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\(\displaystyle{\frac{5x^2-8x+5}{(x-2)(x^2-x+1)}}\)

Problem Statement

Expand \(\displaystyle{\frac{5x^2-8x+5}{(x-2)(x^2-x+1)}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

MIP4U - 1495 video solution

video by MIP4U

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\(\displaystyle{\frac{3x^2-x+4}{x^3+2x^2+6x}}\)

Problem Statement

Expand \(\displaystyle{\frac{3x^2-x+4}{x^3+2x^2+6x}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

PatrickJMT - 1487 video solution

video by PatrickJMT

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\(\displaystyle{\frac{8x^3+13x}{(x^2+2)^2}}\)

Problem Statement

Expand \(\displaystyle{\frac{8x^3+13x}{(x^2+2)^2}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

MIP4U - 1496 video solution

video by MIP4U

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\(\displaystyle{\frac{10x^2+12x+20}{x^3-8}}\)

Problem Statement

Expand \(\displaystyle{\frac{10x^2+12x+20}{x^3-8}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

Khan Academy - 1489 video solution

video by Khan Academy

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Intermediate

\(\displaystyle{ \frac{4x^2}{4x^4+1} }\)

Problem Statement

Expand \(\displaystyle{ \frac{4x^2}{4x^4+1} }\) using partial fraction expansion. Give your answer in exact terms.

Hint

To factor \( 4x^4+1 \), start by adding and subtracting \( 4x^2 \).

Problem Statement

Expand \(\displaystyle{ \frac{4x^2}{4x^4+1} }\) using partial fraction expansion. Give your answer in exact terms.

Final Answer

\(\displaystyle{ \frac{4x^2}{4x^4+1} }\) \(\displaystyle{ = \frac{x}{2x^2-2x+1} - }\) \(\displaystyle{ \frac{x}{2x^2+2x+1} }\)

Problem Statement

Expand \(\displaystyle{ \frac{4x^2}{4x^4+1} }\) using partial fraction expansion. Give your answer in exact terms.

Hint

To factor \( 4x^4+1 \), start by adding and subtracting \( 4x^2 \).

Solution

This is part of a improper integral problem. You do not need to know how to work improper integrals to understand how to solve this problem.

Michael Penn - 3856 video solution

video by Michael Penn

Final Answer

\(\displaystyle{ \frac{4x^2}{4x^4+1} }\) \(\displaystyle{ = \frac{x}{2x^2-2x+1} - }\) \(\displaystyle{ \frac{x}{2x^2+2x+1} }\)

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Practice Instructions

Unless otherwise instructed, expand the given fraction using partial fraction expansion. Give your answer in exact terms.

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