About Partial Fraction Decomposition Quadratic Factors:
Partial Fraction Decomposition allows us to break up a single rational expression into the sum of two or more rational expressions. Here, we will focus on the case where we have both linear factors and irreducible quadratic factors in the denominator.
Test Objectives
- Demonstrate the ability to perform long division with polynomials
- Demonstrate the ability to factor a polynomial
- Demonstrate the ability to find the partial fraction decomposition with quadratic factors
- Demonstrate the ability to find the partial fraction decomposition with repeated quadratic factors
#1:
Instructions: Find the partial fraction decomposition of each.
- Rearrange the denominator and create a group that is quadratic in form:
- $$(x^4 - x^2 - 6) + (2x^3 + 4x)$$
- Use substitution to factor the group on the left
- Factor out the GCF from the group on the right
- Factor out the common binomial factor
- Look for additional factoring
$$b)\hspace{.2em}\frac{13x^2 - 7x + 9}{3x^3 + 12x^2 + x + 4}$$
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#2:
Instructions: Find the partial fraction decomposition of each.
$$a)\hspace{.2em}\frac{-x^2 - 2x - 3}{x^3 - x^2 + 5x - 5}$$
$$b)\hspace{.2em}\frac{8x^2 + 3x + 23}{x^3 - 3x^2 + 4x - 12}$$
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#3:
Instructions: Find the partial fraction decomposition of each.
$$a)\hspace{.2em}\frac{-x^3 - 2x^2 + 3x - 1}{x^3 + 4x^2 + 3x + 12}$$
$$b)\hspace{.2em}\frac{x^3 + x^2 + 3x + 8}{x^3 - 4x^2 + 4x - 16}$$
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#4:
Instructions: Find the partial fraction decomposition of each.
$$a)\hspace{.2em}\frac{3x^4 + 2x^3 - 19x^2 - 2x + 28}{(2x - 1)(x^2 - 3)^2}$$
$$b)\hspace{.2em}\frac{-7x^4 + x^3 - 14x^2 + 4x - 7}{x(x^2 + 1)^2}$$
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#5:
Instructions: Find the partial fraction decomposition of each.
$$a)\hspace{.2em}\frac{3x^3 - 2x^2 - 3x - 2}{(x^2 - x - 1)^2}$$
Instructions: Find A, B, C, and D in terms of a and b.
$$b)\hspace{.2em}\frac{ax^3 + bx^2}{(x^2 + 3)^2} = \frac{Ax + B}{x^2 + 3} + \frac{Cx + D}{(x^2 + 3)^2}$$
Instructions: Find the partial fraction decomposition of each.
Note: Use the answers from part b to quickly find the answer for part c.
$$c)\hspace{.2em}\frac{5x^3 + 2x^2}{(x^2 + 3)^2}$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.2em}\frac{-6}{x - 1}+ \frac{3}{x + 3}+ \frac{7}{x^2 + 2}$$
$$b)\hspace{.2em}\frac{5}{x + 4} + \frac{-2x + 1}{3x^2 + 1} $$
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#2:
Solutions:
$$a)\hspace{.2em}\frac{-1}{x - 1}- \frac{2}{x^2 + 5}$$
$$b)\hspace{.2em}\frac{8}{x - 3}+ \frac{3}{x^2 + 4}$$
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#3:
Solutions:
$$a)\hspace{.2em}{-1} + \frac{1}{x + 4}+ \frac{x + 2}{x^2 + 3}$$
$$b)\hspace{.2em}1 + \frac{5}{x - 4}- \frac{1}{x^2 + 4}$$
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#4:
Solutions:
$$a)\hspace{.2em}\frac{3}{2x - 1}+ \frac{1}{x^2 - 3}+ \frac{2}{(x^2 - 3)^2}$$
$$b)\hspace{.2em}\frac{-7}{x}+ \frac{1}{x^2 + 1}+ \frac{3}{(x^2 + 1)^2}$$
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#5:
Solutions:
$$a)\hspace{.2em}\frac{3x + 1}{x^2 - x - 1} + \frac{x - 1}{(x^2 - x - 1)^2}$$
$$b)\hspace{.2em} A = a, B = b, C = -3a, \text{and} \, D = -3b$$ $$\frac{ax^3 + bx^2}{(x^2 + 3)^2} = \frac{ax + b}{x^2 + 3} + \frac{-3ax - 3b}{(x^2 + 3)^2}$$
$$c)\hspace{.2em}\frac{5x + 2}{x^2 + 3} + \frac{-15x - 6}{(x^2 + 3)^2}$$