Partial Fraction Decomposition Quadratic Factors Lesson

In the last lesson, we learned about the process of partial fraction decomposition, which allows us to write a rational expression as the sum of two or more rational expressions. We specifically focused on how to find the partial fraction decomposition of a rational expression when the denominator contained distinct linear factors or repeated linear factors. In this lesson, we will learn how to find the partial fraction decomposition of a rational expression when the denominator includes an irreducible quadratic factor.
Before we get into our procedure, let's list a few definitions here for reference.

• Distinct Factors: refers to each factor being different or unique
• Linear Factor: refers to a factor such as (ax + b), a ≠ 0
  • a and b represent real numbers
  • The factor contains a variable raised to the first power
  • The factor does not contain a variable raised to any higher power
  • Ex: (3x + 2) is a linear factor (highest exponent is 1)
  • Ex: (3x² + x + 2) is not a linear factor (highest exponent is 2)
• Quadratic Factor: refers to a factor such as ax² + bx + c, a ≠ 0
  • a, b, and c represent real numbers
  • The factor contains a variable raised to the second power
  • The factor does not contain a variable raised to any higher power
  • Ex: (7x² - 19x + 2) is a quadratic factor (highest exponent is 2)
  • Ex: (7x³ + 5x² - 19x + 2) is not a quadratic factor (highest exponent is 3)

In this section, you will see the term "irreducible" or "prime" quadratic factors. A polynomial is considered irreducible if it can't be factored into lower-degree polynomials within a given number system. For example, when we consider real numbers, a quadratic is irreducible if it has no real roots. Consider the following quadratic equation: $$x^2 + 1 = 0$$ Solve for x: $$x^2 = -1$$ $$x = \pm \sqrt{-1}$$ $$x = \pm i$$ The above example is irreducible over the real numbers since the roots are non-real complex numbers. The factorization would be given as: $$x^2 + 1 = (x + i)(x - i)$$ When we consider the rational numbers, a quadratic is irreducible if it can't be factored using rational numbers even if it has real roots that are irrational. Consider the following example: $$x^2 - 2 = 0$$ Solve for x: $$x^2 = 2$$ $$x = \pm \sqrt{2}$$ The square root of 2 is an irrational number. This means (x² - 2) is irreducible over the rational numbers but is reducible over the real numbers. The factorization would be given as: $$x^2 - 2 = \left(x + \sqrt{2}\right)\left(x - \sqrt{2}\right)$$ In the context of partial fraction decomposition, we will be working with rational expressions. Our goal is to decompose a rational expression into the sum of two or more simpler rational expressions. This means we will want to focus on being able to factor over the rational numbers.

• Always factor any quadratic that is reducible over the rational numbers
• Never factor any quadratic that is irreducible over the rational numbers

With the basic definitions out of the way, let's look at our procedure.

Partial Fraction Decomposition of f(x)/g(x):

1. If f(x)/g(x) is not a proper fraction, perform polynomial long division and apply the steps to the remainder
   • A proper fraction means the degree of the numerator is strictly less than the degree of the denominator
   • $$\text{Proper:} \, \frac{x}{x^2 + x + 5}$$
   • $$\text{Improper:} \, \frac{2x^2}{x + 3}$$
2. Factor the denominator g(x) completely into factors of the form:
   • (ax + b)ⁿ, where n is a positive integer (1, 2, 3,...)
   • (cx² + dx + e)^m, where m is a positive integer (1, 2, 3,...)
     • Where cx² + dx + e is irreducible over the rational numbers
3. For each distinct linear factor (ax + b), we produce a partial fraction:
   • $$\frac{A}{ax + b}$$
   • The capital letter "A" represents an unknown constant
4. For each repeated linear factor (ax + b)ⁿ, n > 1:
   • $$\frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_n}{(ax + b)^n}$$
5. For each distinct quadratic factor (cx² + dx + e), we produce a partial fraction:
   • $$\frac{Bx + C}{cx^2 + dx + e}$$
6. For each repeated quadratic factor (cx² + dx + e)^m, m > 1:
   • $$\frac{B_1x + C_1}{cx^2 + dx + e} + \frac{B_2x + C_2}{(cx^2 + dx + e)^2} + \cdots + \frac{B_mx + C_m}{(cx^2 + dx + e)^m}$$
7. Set the original rational expression equal to the sum of the partial fractions
8. Set up a system of linear equations and solve for A, B, C,...
   • Clear all denominators (multiply both sides by the LCD)
   • Write both sides in descending powers of x
   • Equate the coefficients of like powers of x
   • Equate constant terms
   • Solve the given system using algebraic methods

Partial Fraction Decomposition with Distinct Quadratic Factors

For our first example, we will look at the scenario where our denominator can be factored into distinct quadratic factors only.
Example #1: Find the partial fraction decomposition. $$\frac{5x^2 - 4}{x^4 - 3x^2 - 10}$$ Step 1) We need to start with a proper fraction. This means the degree of the numerator is less than the degree of the denominator. In this case, the degree of the numerator is 2 and the degree of the denominator is 4, so we do have a proper fraction.
Step 2) Factor the denominator completely. $$x^4 - 3x^2 - 10$$ Since our polynomial is quadratic in form, we will factor using a substitution technique. $$x^4 - 3x^2 - 10 = (x^2)^2 - 3x^2 - 10$$ Let u = x² $$u^2 - 3u - 10 = (u - 5)(u + 2)$$ Replace u with x²: $$x^4 - 3x^2 - 10 = (x^2 - 5)(x^2 + 2)$$ Step 3) Set up each partial fraction based on the factored denominator. $$\frac{5x^2 - 4}{x^4 - 3x^2 - 10} = \frac{Ax + B}{x^2 - 5} + \frac{Cx + D}{x^2 + 2}$$ Step 4) Solve for all unknowns. We will clear all the denominators. We just need to multiply both sides by the LCD, which is the denominator of the original rational expression.
Scratch Work: $$\require{cancel}\frac{5x^2 - 4}{\cancel{(x^2 - 5)(x^2 + 2)}} \cdot \cancel{(x^2 - 5)(x^2 + 2)} = 5x^2 - 4$$ $$\frac{Ax + B}{\cancel{(x^2 - 5)}} \cdot \cancel{(x^2 - 5)}(x^2 + 2) = (Ax + B)(x^2 + 2)$$ $$\frac{Cx + D}{\cancel{(x^2 + 2)}} \cdot (x^2 - 5)\cancel{(x^2 + 2)} = (Cx + D)(x^2 - 5)$$ Update the Equation: $$5x^2 - 4 = (Ax + B)(x^2 + 2) + (Cx + D)(x^2 - 5)$$ We can create a linear system by changing the right side into the same form as the left side. $$5x^2 - 4 = Ax^3 + 2Ax + Bx^2 + 2B + Cx^3 - 5Cx + Dx^2 - 5D$$ Rearrange Terms: $$5x^2 - 4 = Ax^3 + Cx^3 + Bx^2 + Dx^2 + 2Ax - 5Cx + 2B - 5D$$ Factor Common Variables: $$5x^2 - 4 = (A + C)x^3 + (B + D)x^2 + (2A - 5C)x + (2B - 5D)$$ Equate coefficients of like powers of x and equate constant terms. Here we need to add in 0x³ and 0x on the left since those powers are missing. $$0x^3 + 5x^2 + 0x + (-4) = (A + C)x^3 + (B + D)x^2 + (2A - 5C)x + (2B - 5D)$$ Set up the system: $$1) \, A + C = 0$$ $$2) \, B + D = 5$$ $$3) \, 2A - 5C = 0$$ $$4) \, 2B - 5D = -4$$ Solve the first equation for A: $$A + C = 0$$ $$A = -C$$ Replace A with -C in equation #3: $$2A - 5C = 0$$ $$2(-C) - 5C = 0$$ $$-2C - 5C = 0$$ $$-7C = 0$$ $$C = 0$$ Since C = 0 and A = -C, we know that A = 0. Now we need to find B and D. We just need to work with equations #2 and #4. $$2) \, B + D = 5$$ $$4) \, 2B - 5D = -4$$ Multiply equation #2 by 5, then add the result to equation #4. $$5B + 5D = 25$$ $$\underline{2B - 5D = -4}$$ $$7B = 21$$ Solve for B: $$B = 3$$ Use the original equation #2 to find D. Replace B with 3. $$B + D = 5$$ $$3 + D = 5$$ $$D = 2$$ A = 0, B = 3, C = 0, and D = 2. We can replace these in our model and we are done. $$\frac{5x^2 - 4}{x^4 - 3x^2 - 10} = \frac{3}{x^2 - 5} + \frac{2}{x^2 + 2}$$

Partial Fraction Decomposition with Both Distinct Linear and Quadratic Factors

For our second example, we will look at the scenario where our denominator can be factored into both distinct linear and quadratic factors.
Example #2: Find the partial fraction decomposition. $$\frac{3x^3 + 8x^2 - 21x - 10}{x^4 - 5x^3 - 2x^2 + 10x}$$ Step 1) We need to start with a proper fraction. This means the degree of the numerator is less than the degree of the denominator. In this case, the degree of the numerator is 3 and the degree of the denominator is 4, so we do have a proper fraction.
Step 2) Factor the denominator completely. $$x^4 - 5x^3 - 2x^2 + 10x$$ $$= x(x^3 - 5x^2 - 2x + 10)$$ $$= x[x^2(x - 5) -2(x - 5)]$$ $$= x(x - 5)(x^2 - 2)$$ Step 3) Set up each partial fraction based on the factored denominator. $$\frac{3x^3 + 8x^2 - 21x - 10}{x(x - 5)(x^2 - 2)} = \frac{A}{x} + \frac{B}{x - 5} + \frac{Cx + D}{x^2 - 2}$$ Step 4) Solve for all unknowns. We will clear all the denominators. We just need to multiply both sides by the LCD, which is the denominator of the original rational expression.
Scratch Work: $$\frac{3x^3 + 8x^2 - 21x - 10}{\cancel{x(x - 5)(x^2 - 2)}} \cdot \cancel{x(x - 5)(x^2 - 2)} = 3x^3 + 8x^2 - 21x - 10$$ $$\frac{A}{\cancel{x}} \cdot \cancel{x}(x - 5)(x^2 - 2) = A(x - 5)(x^2 - 2)$$ $$\frac{B}{\cancel{(x - 5)}} \cdot x\cancel{(x - 5)}(x^2 - 2) = Bx(x^2 - 2)$$ $$\frac{Cx + D}{\cancel{(x^2 - 2)}} \cdot x(x - 5)\cancel{(x^2 - 2)} = (Cx^2 + Dx)(x - 5)$$ Update the Equation: $$3x^3 + 8x^2 - 21x - 10 = A(x - 5)(x^2 - 2) + Bx(x^2 - 2) + (Cx^2 + Dx)(x - 5)$$ In this case, since we have linear factors involved, we can choose suitable values for x to greatly speed up our work.
Let x = 5; this will remove A, C, and D from the equation. $$3(5)^3 + 8(5)^2 - 21(5) - 10 = B(5)((5)^2 - 2)$$ $$3(125) + 8(25) - 21(5) - 10 = B(5)(25 - 2)$$ $$375 + 200 - 105 - 10 = B(5)(23)$$ $$460 = 115B$$ $$B = 4$$ Let's replace B with 4 in our equation. $$3x^3 + 8x^2 - 21x - 10 = A(x - 5)(x^2 - 2) + 4x(x^2 - 2) + (Cx^2 + Dx)(x - 5)$$ Let x = 0; this will remove C and D from the equation. $$-10 = A(0 - 5)((0)^2 - 2)$$ $$-10 = A(-5)(-2)$$ $$-10 = 10A$$ $$A = -1$$ Let's replace A with -1 in our equation. $$3x^3 + 8x^2 - 21x - 10 = -(x - 5)(x^2 - 2) + 4x(x^2 - 2) + (Cx^2 + Dx)(x - 5)$$ At this point, we have C and D as the remaining unknowns. Let's simplify the right side. $$3x^3 + 8x^2 - 21x - 10 = -(x - 5)(x^2 - 2) + 4x(x^2 - 2) + (Cx^2 + Dx)(x - 5)$$ $$3x^3 + 8x^2 - 21x - 10 = -x^3 + 5x^2 + 2x - 10 + 4x^3 - 8x + Cx^3 - 5Cx^2 + Dx^2 - 5Dx$$ Rearrange Terms: $$3x^3 + 8x^2 - 21x - 10 = -x^3 + 4x^3 + Cx^3 + 5x^2 - 5Cx^2 + Dx^2 + 2x - 8x - 5Dx - 10$$ $$3x^3 + 8x^2 - 21x - 10 = Cx^3 + 3x^3 + 5x^2 - 5Cx^2 + Dx^2 - 6x - 5Dx - 10$$ Factor Common Variables: $$3x^3 + 8x^2 - 21x - 10 = (C + 3)x^3 + (5 - 5C + D)x^2 + (-6 - 5D)x - 10$$ Equate coefficients of like powers of x. We don't need to equate constant terms here, since we have -10 on each side. $$1) \, 3 = C + 3$$ $$2) \, 5 - 5C + D = 8$$ $$3) \, {-}6 - 5D = -21$$ Equation #2 isn't necessary, we can solve equation #1 for C, and equation #3 for D. $$C + 3 = 3$$ $$C = 0$$ $$-6 - 5D = -21$$ $$-5D = -15$$ $$D = 3$$ A = -1, B = 4, C = 0, and D = 3. We can replace these in our model and we are done. $$\frac{3x^3 + 8x^2 - 21x - 10}{x(x - 5)(x^2 - 2)} = -\frac{1}{x} + \frac{4}{x - 5} + \frac{3}{x^2 - 2}$$

Partial Fraction Decomposition with Repeated Quadratic Factors

For our third and final example, we will look at a problem that requires polynomial long division and includes a repeated quadratic factor.
Example #3: Find the partial fraction decomposition. $$\frac{2x^5 - 11x^3 + 2x^2 + 23x + 10}{x^5 + 2x^4 - 6x^3 - 12x^2 + 9x + 18}$$ Step 1) We need to start with a proper fraction. This means the degree of the numerator is less than the degree of the denominator. In this case, the degree of the numerator is 5 and the degree of the denominator is also 5, so we do not have a proper fraction. Let's set up a long division. $$\frac{2x^5 - 11x^3 + 2x^2 + 23x + 10}{x^5 + 2x^4 - 6x^3 - 12x^2 + 9x + 18} = 2 + \frac{-4x^4 + x^3 + 26x^2 + 5x - 26}{x^5 + 2x^4 - 6x^3 - 12x^2 + 9x + 18}$$ For now, we will just work with the remainder. When we write our final answer, the quotient of 2 will need to be included. $$\frac{-4x^4 + x^3 + 26x^2 + 5x - 26}{x^5 + 2x^4 - 6x^3 - 12x^2 + 9x + 18}$$ Step 2) Factor the denominator completely. $$x^5 + 2x^4 - 6x^3 - 12x^2 + 9x + 18$$ $$= x^4(x + 2) - 6x^2(x + 2) + 9(x + 2)$$ $$= (x + 2)(x^4 - 6x^2 + 9)$$ The second factor is quadratic in form. $$x^4 - 6x^2 + 9 = (x^2)^2 - 6x^2 + 9$$ Let u = x² $$u^2 - 6u + 9 = (u - 3)^2$$ Replace u with x²: $$x^4 - 6x^2 + 9 = (x^2 - 3)^2$$ Return to the problem: $$(x + 2)(x^4 - 6x^2 + 9)= (x + 2)(x^2 - 3)^2$$ Step 3) Set up each partial fraction based on the factored denominator. $$\frac{-4x^4 + x^3 + 26x^2 + 5x - 26}{x^5 + 2x^4 - 6x^3 - 12x^2 + 9x + 18} = \frac{A}{x + 2} + \frac{Bx + C}{x^2 - 3} + \frac{Dx + E}{(x^2 - 3)^2}$$ Notice that we included one fraction for each power of (x² - 3): the first power (x² - 3) and the second power (x² - 3)².
Step 4) Solve for all unknowns. We will clear all the denominators. We just need to multiply both sides by the LCD, which is the denominator of the original rational expression. Here "original rational expression" refers to the rational expression formed from the remainder of the long division process.
Scratch Work: $$\frac{-4x^4 + x^3 + 26x^2 + 5x - 26}{\cancel{(x + 2)(x^2 - 3)^2}} \cdot \cancel{(x + 2)(x^2 - 3)^2} = -4x^4 + x^3 + 26x^2 + 5x - 26$$ $$\frac{A}{\cancel{(x + 2)}} \cdot \cancel{(x + 2)}(x^2 - 3)^2 = A(x^2 - 3)^2$$ $$\frac{Bx + C}{\cancel{(x^2 - 3)}} \cdot (x + 2)(x^2 - 3)^{1 \cancel{2}} = (Bx + C)(x + 2)(x^2 - 3)$$ $$\frac{Dx + E}{\cancel{(x^2 - 3)^2}} \cdot (x + 2)\cancel{(x^2 - 3)^2} = (Dx + E)(x + 2)$$ Update the Equation: $$-4x^4 + x^3 + 26x^2 + 5x - 26 = A(x^2 - 3)^2 + (Bx + C)(x + 2)(x^2 - 3) + (Dx + E)(x + 2)$$ Let x = -2; this will remove B, C, D, and E from the equation. $$-4(-2)^4 + (-2)^3 + 26(-2)^2 + 5(-2) - 26 = A((-2)^2 - 3)^2$$ $$-4(16) + (-8) + 26(4) + 5(-2) - 26 = A(4 - 3)^2$$ $$-64 - 8 + 104 - 10 - 26 = A$$ $$A = -4$$ Let's replace A in our equation. $$-4x^4 + x^3 + 26x^2 + 5x - 26 = -4(x^2 - 3)^2 + (Bx + C)(x + 2)(x^2 - 3) + (Dx + E)(x + 2)$$ Since simplifying the right side is extremely tedious, we will do it separately and then reassemble the equation when we are done.
Scratch Work (Right Side Only):
Part 1: $$-4(x^2 - 3)^2 = -4x^4 + 24x^2 - 36$$ Part 2: $$(Bx + C)(x + 2)(x^2 - 3) = (Bx^2 + 2Bx + Cx + 2C)(x^2 - 3)$$ $$= Bx^4 + 2Bx^3 + Cx^3 + 2Cx^2 - 3Bx^2 - 6Bx - 3Cx - 6C$$ Part 3: $$(Dx + E)(x + 2) = Dx^2 + 2Dx + Ex + 2E$$ Update the Right Side: $$-4x^4 + 24x^2 - 36 + Bx^4 + 2Bx^3 + Cx^3 + 2Cx^2 - 3Bx^2 - 6Bx - 3Cx - 6C + Dx^2 + 2Dx + Ex + 2E$$ Rearrange Terms: $$Bx^4 - 4x^4 + 2Bx^3 + Cx^3 + 24x^2 + 2Cx^2 - 3Bx^2 + Dx^2 - 6Bx - 3Cx + 2Dx + Ex - 36 - 6C + 2E$$ Factor Common Variables: $$(B - 4)x^4 + (2B + C)x^3 + (24 + 2C - 3B + D)x^2 + (-6B - 3C + 2D + E)x + (- 36 - 6C + 2E)$$ Now that we are done working on the right side, we can set the left and right sides equal. $$-4x^4 + x^3 + 26x^2 + 5x + (-26)$$ $$= (B - 4)x^4 + (2B + C)x^3 + (24 + 2C - 3B + D)x^2 + (-6B - 3C + 2D + E)x + (- 36 - 6C + 2E)$$ Equate coefficients of like powers of x and equate constant terms. $$1) \, {-}4 = B - 4$$ $$2) \, 2B + C = 1$$ $$3) \, 24 + 2C - 3B + D = 26$$ $$4) \, {-}6B - 3C + 2D + E = 5$$ $$5) \, {-}36 - 6C + 2E = -26$$ Solve the first equation for B: $$-4 = B - 4$$ $$B = 0$$ Plug in for B in equation #2 to find C: $$2(0) + C = 1$$ $$C = 1$$ Plug in for B and C in equation #3 to find D: $$24 + 2(1) - 3(0) + D = 26$$ $$24 + 2 + D = 26$$ $$D = 0$$ Use either equation #4 or #5 to find E. Let's plug in for B, C, and D in equation #4: $$-6(0) - 3(1) + 2(0) + E = 5$$ $$-3 + E = 5$$ $$E = 8$$ A = -4, B = 0, C = 1, D = 0, E = 8. We can replace these in our model. $$\frac{-4x^4 + x^3 + 26x^2 + 5x - 26}{x^5 + 2x^4 - 6x^3 - 12x^2 + 9x + 18} = -\frac{4}{x + 2} + \frac{1}{x^2 - 3} + \frac{8}{(x^2 - 3)^2}$$ Since we performed a long division at the beginning of the problem, let's return to the original form. $$\frac{2x^5 - 11x^3 + 2x^2 + 23x + 10}{x^5 + 2x^4 - 6x^3 - 12x^2 + 9x + 18} = 2 + \frac{-4x^4 + x^3 + 26x^2 + 5x - 26}{x^5 + 2x^4 - 6x^3 - 12x^2 + 9x + 18}$$ Now we can replace the remainder fraction with our partial fraction decomposition and we are done. $$\frac{2x^5 - 11x^3 + 2x^2 + 23x + 10}{x^5 + 2x^4 - 6x^3 - 12x^2 + 9x + 18} = 2 -\frac{4}{x + 2} + \frac{1}{x^2 - 3} + \frac{8}{(x^2 - 3)^2}$$