Partial Fractions.
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Lesson Objectives
- Understand the concept of partial fraction decomposition.
- Apply the method of partial fractions to distinct linear denominators.
- Find partial fraction decompositions for repeated linear factors.
- Determine the decomposition for irreducible quadratic factors.
- Interpret and decompose improper rational fractions by polynomial division first.
Introduction to Partial Fractions
Partial fractions is a technique used to break apart a complicated rational expression into a sum of simpler fractions. This is the reverse process of combining fractions over a common denominator. It is widely used in calculus for integration, in engineering for Laplace transforms, and in control systems for system analysis. By mastering partial fractions, you will be able to simplify many algebraic problems and prepare for higher-level mathematics.
Key Idea:
A rational function $\frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials, with $\deg(P) < \deg(Q)$, can be expressed as a sum of simpler fractions whose denominators are the factors of $Q(x)$.
A rational function $\frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials, with $\deg(P) < \deg(Q)$, can be expressed as a sum of simpler fractions whose denominators are the factors of $Q(x)$.
Key Definitions:
• Rational Fraction: A ratio of two polynomials, e.g., $\frac{2x+1}{x^2-1}$.
• Proper Fraction: A rational fraction where the degree of the numerator is less than the degree of the denominator.
• Improper Fraction: A rational fraction where the degree of the numerator is greater than or equal to the degree of the denominator.
• Irreducible Quadratic: A quadratic factor that cannot be factored into linear factors over the reals, e.g., $x^2+1$.
• Rational Fraction: A ratio of two polynomials, e.g., $\frac{2x+1}{x^2-1}$.
• Proper Fraction: A rational fraction where the degree of the numerator is less than the degree of the denominator.
• Improper Fraction: A rational fraction where the degree of the numerator is greater than or equal to the degree of the denominator.
• Irreducible Quadratic: A quadratic factor that cannot be factored into linear factors over the reals, e.g., $x^2+1$.
Case 1: Distinct Linear Factors
If the denominator $Q(x)$ can be factored into distinct linear factors, i.e., $Q(x) = (a_1x+b_1)(a_2x+b_2)\cdots(a_nx+b_n)$, then the partial fraction decomposition takes the form: $$ \frac{P(x)}{Q(x)} = \frac{A_1}{a_1x+b_1} + \frac{A_2}{a_2x+b_2} + \cdots + \frac{A_n}{a_nx+b_n} $$ where $A_1, A_2, \dots, A_n$ are constants to be determined.
Step-by-Step Method:
1. Factor the denominator completely into distinct linear factors.
2. Set up the partial fraction decomposition with unknown constants for each factor.
3. Multiply both sides by the common denominator to clear fractions.
4. Expand and equate coefficients of like powers of $x$ (or substitute convenient values of $x$).
5. Solve the resulting system of equations for the constants.
1. Factor the denominator completely into distinct linear factors.
2. Set up the partial fraction decomposition with unknown constants for each factor.
3. Multiply both sides by the common denominator to clear fractions.
4. Expand and equate coefficients of like powers of $x$ (or substitute convenient values of $x$).
5. Solve the resulting system of equations for the constants.
Example 1: Distinct Linear Factors
Problem: Decompose $\frac{5x+1}{x^2-1}$ into partial fractions.
Formula: $\frac{5x+1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$
Solution:
Step 1: Multiply both sides by $(x-1)(x+1)$:
$5x+1 = A(x+1) + B(x-1)$
Step 2: Substitute $x = 1$: $5(1)+1 = A(2) + B(0) \Rightarrow 6 = 2A \Rightarrow A = 3$.
Step 3: Substitute $x = -1$: $5(-1)+1 = A(0) + B(-2) \Rightarrow -4 = -2B \Rightarrow B = 2$.
Answer: $\frac{5x+1}{x^2-1} = \frac{3}{x-1} + \frac{2}{x+1}$.
Problem: Decompose $\frac{5x+1}{x^2-1}$ into partial fractions.
Formula: $\frac{5x+1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$
Solution:
Step 1: Multiply both sides by $(x-1)(x+1)$:
$5x+1 = A(x+1) + B(x-1)$
Step 2: Substitute $x = 1$: $5(1)+1 = A(2) + B(0) \Rightarrow 6 = 2A \Rightarrow A = 3$.
Step 3: Substitute $x = -1$: $5(-1)+1 = A(0) + B(-2) \Rightarrow -4 = -2B \Rightarrow B = 2$.
Answer: $\frac{5x+1}{x^2-1} = \frac{3}{x-1} + \frac{2}{x+1}$.
Watch Out!
Do not forget to factor the denominator completely. A common mistake is to stop after factoring partially, which leads to an incorrect decomposition.
Do not forget to factor the denominator completely. A common mistake is to stop after factoring partially, which leads to an incorrect decomposition.
Case 2: Repeated Linear Factors
If a linear factor $(ax+b)$ appears $k$ times in the denominator, then the partial fraction decomposition includes $k$ terms: $$ \frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \cdots + \frac{A_k}{(ax+b)^k} $$
Step-by-Step Method:
1. Factor the denominator, identifying any repeated linear factors.
2. Write the decomposition with a term for each power of the repeated factor.
3. Multiply through by the common denominator.
4. Solve for the constants by equating coefficients or substituting convenient $x$ values.
5. Write the final decomposition.
1. Factor the denominator, identifying any repeated linear factors.
2. Write the decomposition with a term for each power of the repeated factor.
3. Multiply through by the common denominator.
4. Solve for the constants by equating coefficients or substituting convenient $x$ values.
5. Write the final decomposition.
Example 2: Repeated Linear Factors
Problem: Decompose $\frac{2x^2+3x+1}{(x-1)^2(x+2)}$.
Formula: $\frac{2x^2+3x+1}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$
Solution:
Step 1: Multiply both sides by $(x-1)^2(x+2)$:
$2x^2+3x+1 = A(x-1)(x+2) + B(x+2) + C(x-1)^2$
Step 2: Expand: $2x^2+3x+1 = A(x^2+x-2) + B(x+2) + C(x^2-2x+1)$
$\Rightarrow 2x^2+3x+1 = (A+C)x^2 + (A+B-2C)x + (-2A+2B+C)$
Step 3: Equate coefficients:
(1) $A+C = 2$
(2) $A+B-2C = 3$
(3) $-2A+2B+C = 1$
Step 4: Solve: From (1), $C = 2-A$. Substitute into (2): $A+B-2(2-A)=3 \Rightarrow A+B-4+2A=3 \Rightarrow 3A+B=7 \Rightarrow B=7-3A$.
Substitute into (3): $-2A+2(7-3A)+(2-A)=1 \Rightarrow -2A+14-6A+2-A=1 \Rightarrow -9A+16=1 \Rightarrow -9A=-15 \Rightarrow A=\frac{5}{3}$.
Then $B = 7-3(\frac{5}{3}) = 7-5 = 2$, and $C = 2-\frac{5}{3} = \frac{1}{3}$.
Answer: $\frac{2x^2+3x+1}{(x-1)^2(x+2)} = \frac{5/3}{x-1} + \frac{2}{(x-1)^2} + \frac{1/3}{x+2}$.
Problem: Decompose $\frac{2x^2+3x+1}{(x-1)^2(x+2)}$.
Formula: $\frac{2x^2+3x+1}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$
Solution:
Step 1: Multiply both sides by $(x-1)^2(x+2)$:
$2x^2+3x+1 = A(x-1)(x+2) + B(x+2) + C(x-1)^2$
Step 2: Expand: $2x^2+3x+1 = A(x^2+x-2) + B(x+2) + C(x^2-2x+1)$
$\Rightarrow 2x^2+3x+1 = (A+C)x^2 + (A+B-2C)x + (-2A+2B+C)$
Step 3: Equate coefficients:
(1) $A+C = 2$
(2) $A+B-2C = 3$
(3) $-2A+2B+C = 1$
Step 4: Solve: From (1), $C = 2-A$. Substitute into (2): $A+B-2(2-A)=3 \Rightarrow A+B-4+2A=3 \Rightarrow 3A+B=7 \Rightarrow B=7-3A$.
Substitute into (3): $-2A+2(7-3A)+(2-A)=1 \Rightarrow -2A+14-6A+2-A=1 \Rightarrow -9A+16=1 \Rightarrow -9A=-15 \Rightarrow A=\frac{5}{3}$.
Then $B = 7-3(\frac{5}{3}) = 7-5 = 2$, and $C = 2-\frac{5}{3} = \frac{1}{3}$.
Answer: $\frac{2x^2+3x+1}{(x-1)^2(x+2)} = \frac{5/3}{x-1} + \frac{2}{(x-1)^2} + \frac{1/3}{x+2}$.
Watch Out!
For repeated factors, you must include all powers from 1 up to the multiplicity. Missing a term will give an incorrect decomposition.
For repeated factors, you must include all powers from 1 up to the multiplicity. Missing a term will give an incorrect decomposition.
Case 3: Quadratic Factors (Irreducible)
If the denominator contains an irreducible quadratic factor $ax^2+bx+c$ (which cannot be factored into real linear factors), then the partial fraction decomposition includes a term of the form: $$ \frac{Ax+B}{ax^2+bx+c} $$ If the quadratic factor is repeated $k$ times, then there will be $k$ such terms with increasing powers in the denominator.
Step-by-Step Method:
1. Factor the denominator, identifying irreducible quadratic factors.
2. Write the decomposition with linear numerator for each quadratic factor.
3. Multiply through by the common denominator.
4. Expand and equate coefficients or substitute values to solve for the constants.
5. State the final decomposition.
1. Factor the denominator, identifying irreducible quadratic factors.
2. Write the decomposition with linear numerator for each quadratic factor.
3. Multiply through by the common denominator.
4. Expand and equate coefficients or substitute values to solve for the constants.
5. State the final decomposition.
Example 3: Irreducible Quadratic Factor
Problem: Decompose $\frac{3x^2+2x+1}{(x^2+1)(x-2)}$.
Formula: $\frac{3x^2+2x+1}{(x^2+1)(x-2)} = \frac{Ax+B}{x^2+1} + \frac{C}{x-2}$
Solution:
Step 1: Multiply both sides by $(x^2+1)(x-2)$:
$3x^2+2x+1 = (Ax+B)(x-2) + C(x^2+1)$
Step 2: Expand: $3x^2+2x+1 = A(x^2-2x) + B(x-2) + Cx^2 + C$
$\Rightarrow 3x^2+2x+1 = (A+C)x^2 + (-2A+B)x + (-2B+C)$
Step 3: Equate coefficients:
(1) $A+C = 3$
(2) $-2A+B = 2$
(3) $-2B+C = 1$
Step 4: Solve: From (2), $B = 2+2A$. Substitute into (3): $-2(2+2A)+C = 1 \Rightarrow -4-4A+C=1 \Rightarrow C = 5+4A$.
Substitute into (1): $A + (5+4A) = 3 \Rightarrow 5A+5=3 \Rightarrow 5A=-2 \Rightarrow A=-\frac{2}{5}$.
Then $B = 2+2(-\frac{2}{5}) = 2-\frac{4}{5} = \frac{6}{5}$, and $C = 5+4(-\frac{2}{5}) = 5-\frac{8}{5} = \frac{17}{5}$.
Answer: $\frac{3x^2+2x+1}{(x^2+1)(x-2)} = \frac{-\frac{2}{5}x+\frac{6}{5}}{x^2+1} + \frac{17/5}{x-2}$.
Problem: Decompose $\frac{3x^2+2x+1}{(x^2+1)(x-2)}$.
Formula: $\frac{3x^2+2x+1}{(x^2+1)(x-2)} = \frac{Ax+B}{x^2+1} + \frac{C}{x-2}$
Solution:
Step 1: Multiply both sides by $(x^2+1)(x-2)$:
$3x^2+2x+1 = (Ax+B)(x-2) + C(x^2+1)$
Step 2: Expand: $3x^2+2x+1 = A(x^2-2x) + B(x-2) + Cx^2 + C$
$\Rightarrow 3x^2+2x+1 = (A+C)x^2 + (-2A+B)x + (-2B+C)$
Step 3: Equate coefficients:
(1) $A+C = 3$
(2) $-2A+B = 2$
(3) $-2B+C = 1$
Step 4: Solve: From (2), $B = 2+2A$. Substitute into (3): $-2(2+2A)+C = 1 \Rightarrow -4-4A+C=1 \Rightarrow C = 5+4A$.
Substitute into (1): $A + (5+4A) = 3 \Rightarrow 5A+5=3 \Rightarrow 5A=-2 \Rightarrow A=-\frac{2}{5}$.
Then $B = 2+2(-\frac{2}{5}) = 2-\frac{4}{5} = \frac{6}{5}$, and $C = 5+4(-\frac{2}{5}) = 5-\frac{8}{5} = \frac{17}{5}$.
Answer: $\frac{3x^2+2x+1}{(x^2+1)(x-2)} = \frac{-\frac{2}{5}x+\frac{6}{5}}{x^2+1} + \frac{17/5}{x-2}$.
Watch Out!
For an irreducible quadratic, the numerator must be a linear expression $Ax+B$, not just a constant. This is a common error.
For an irreducible quadratic, the numerator must be a linear expression $Ax+B$, not just a constant. This is a common error.
Case 4: Improper Rational Fractions
When the degree of the numerator is greater than or equal to the degree of the denominator, the fraction is improper. First, perform polynomial division to express it as a polynomial plus a proper fraction. Then apply partial fractions to the proper remainder.
Step-by-Step Method:
1. Divide the numerator by the denominator using polynomial long division.
2. Write the result as quotient + remainder/divisor.
3. Apply partial fraction decomposition to the proper fractional part.
4. Combine the quotient and the partial fractions for the final answer.
1. Divide the numerator by the denominator using polynomial long division.
2. Write the result as quotient + remainder/divisor.
3. Apply partial fraction decomposition to the proper fractional part.
4. Combine the quotient and the partial fractions for the final answer.
Example 4: Improper Fraction
Problem: Decompose $\frac{x^3+2x^2+3x+4}{x^2-1}$.
Formula: First divide, then partial fractions.
Solution:
Step 1: Divide $(x^3+2x^2+3x+4)$ by $(x^2-1)$:
$x^3+2x^2+3x+4 = (x+2)(x^2-1) + (4x+6)$ because $(x+2)(x^2-1) = x^3+2x^2 - x - 2$, subtract from original gives $4x+6$.
Step 2: So $\frac{x^3+2x^2+3x+4}{x^2-1} = x+2 + \frac{4x+6}{x^2-1}$.
Step 3: Decompose $\frac{4x+6}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$.
Multiply through: $4x+6 = A(x+1) + B(x-1)$.
Substitute $x=1$: $10 = A(2) \Rightarrow A=5$.
Substitute $x=-1$: $2 = B(-2) \Rightarrow B=-1$.
Answer: $\frac{x^3+2x^2+3x+4}{x^2-1} = x+2 + \frac{5}{x-1} - \frac{1}{x+1}$.
Problem: Decompose $\frac{x^3+2x^2+3x+4}{x^2-1}$.
Formula: First divide, then partial fractions.
Solution:
Step 1: Divide $(x^3+2x^2+3x+4)$ by $(x^2-1)$:
$x^3+2x^2+3x+4 = (x+2)(x^2-1) + (4x+6)$ because $(x+2)(x^2-1) = x^3+2x^2 - x - 2$, subtract from original gives $4x+6$.
Step 2: So $\frac{x^3+2x^2+3x+4}{x^2-1} = x+2 + \frac{4x+6}{x^2-1}$.
Step 3: Decompose $\frac{4x+6}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$.
Multiply through: $4x+6 = A(x+1) + B(x-1)$.
Substitute $x=1$: $10 = A(2) \Rightarrow A=5$.
Substitute $x=-1$: $2 = B(-2) \Rightarrow B=-1$.
Answer: $\frac{x^3+2x^2+3x+4}{x^2-1} = x+2 + \frac{5}{x-1} - \frac{1}{x+1}$.
Watch Out!
Do not attempt partial fractions directly on an improper fraction. Always divide first to get a proper fraction, or you will get the wrong decomposition.
Do not attempt partial fractions directly on an improper fraction. Always divide first to get a proper fraction, or you will get the wrong decomposition.
Cumulative Practice Exercises
- $\frac{3x+2}{(x+1)(x-2)}$
- $\frac{2x-5}{(x-3)^2}$
- $\frac{4x^2+1}{(x^2+1)(x-1)}$
- $\frac{x^2+4x+5}{(x-2)(x+1)^2}$
- $\frac{2x^3-3x^2+4x-5}{x^2-2x+1}$
- $\frac{x+3}{(x^2+4)(x-1)}$
- $\frac{5x^2+3x+7}{(x-2)^3}$
- $\frac{3x^3+2x^2+x+1}{x^2+x-2}$
- $\frac{x^2+1}{(x+1)^2(x-1)}$
- $\frac{2x^4+5x^3+3x^2+2x+1}{x^2+1}$
Solutions to Cumulative Exercises
-
Step 1: $\frac{3x+2}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}$
Step 2: $3x+2 = A(x-2) + B(x+1)$
Step 3: $x=2 \Rightarrow 8 = B(3) \Rightarrow B = \frac{8}{3}$; $x=-1 \Rightarrow -1 = A(-3) \Rightarrow A = \frac{1}{3}$
Answer: $\frac{1/3}{x+1} + \frac{8/3}{x-2}$ -
Step 1: $\frac{2x-5}{(x-3)^2} = \frac{A}{x-3} + \frac{B}{(x-3)^2}$
Step 2: $2x-5 = A(x-3) + B$
Step 3: $x=3 \Rightarrow 1 = B$; equate coefficients: $A=2, -3A+B = -5 \Rightarrow -6+1=-5$ works.
Answer: $\frac{2}{x-3} + \frac{1}{(x-3)^2}$ -
Step 1: $\frac{4x^2+1}{(x^2+1)(x-1)} = \frac{Ax+B}{x^2+1} + \frac{C}{x-1}$
Step 2: $4x^2+1 = (Ax+B)(x-1) + C(x^2+1)$
Step 3: Expand: $(A+C)x^2 + (-A+B)x + (-B+C)$
Step 4: $A+C=4$, $-A+B=0 \Rightarrow B=A$, $-B+C=1$
Solve: $B=A$, $C=4-A$, then $-A+(4-A)=1 \Rightarrow -2A+4=1 \Rightarrow A=\frac{3}{2}$, $B=\frac{3}{2}$, $C=\frac{5}{2}$
Answer: $\frac{\frac{3}{2}x+\frac{3}{2}}{x^2+1} + \frac{5/2}{x-1}$ -
Step 1: $\frac{x^2+4x+5}{(x-2)(x+1)^2} = \frac{A}{x-2} + \frac{B}{x+1} + \frac{C}{(x+1)^2}$
Step 2: $x^2+4x+5 = A(x+1)^2 + B(x-2)(x+1) + C(x-2)$
Step 3: $x=2 \Rightarrow 17 = A(9) \Rightarrow A=\frac{17}{9}$; $x=-1 \Rightarrow 2 = C(-3) \Rightarrow C=-\frac{2}{3}$; $x=0 \Rightarrow 5 = A + B(-2) + C(-2) \Rightarrow 5 = \frac{17}{9} -2B + \frac{4}{3} \Rightarrow 5 = \frac{29}{9} - 2B \Rightarrow 2B = -\frac{16}{9} \Rightarrow B=-\frac{8}{9}$
Answer: $\frac{17/9}{x-2} + \frac{-8/9}{x+1} + \frac{-2/3}{(x+1)^2}$ -
Step 1: Divide: $\frac{2x^3-3x^2+4x-5}{x^2-2x+1}$. $x^2-2x+1 = (x-1)^2$.
Step 2: Polynomial division: quotient $2x+1$, remainder $x-6$ (verify: $(2x+1)(x-1)^2 = 2x^3-3x^2+1$, subtract gives $4x-5-1 = 4x-6$ wait: do carefully)
Actually: $(2x+1)(x^2-2x+1) = 2x^3-4x^2+2x + x^2-2x+1 = 2x^3-3x^2+0x+1$. Subtract from original: $(2x^3-3x^2+4x-5) - (2x^3-3x^2+0x+1) = 4x-6$. So quotient $2x+1$, remainder $4x-6$.
Step 3: Decompose $\frac{4x-6}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2}$; $4x-6 = A(x-1)+B$; $x=1 \Rightarrow -2=B$; $A=4$.
Answer: $2x+1 + \frac{4}{x-1} - \frac{2}{(x-1)^2}$ -
Step 1: $\frac{x+3}{(x^2+4)(x-1)} = \frac{Ax+B}{x^2+4} + \frac{C}{x-1}$
Step 2: $x+3 = (Ax+B)(x-1) + C(x^2+4)$
Step 3: Expand: $(A+C)x^2 + (-A+B)x + (-B+4C)$
Step 4: $A+C=0$, $-A+B=1$, $-B+4C=3$; from $A=-C$, $B=1-C$; then $-(1-C)+4C=3 \Rightarrow -1+C+4C=3 \Rightarrow 5C=4 \Rightarrow C=4/5$, $A=-4/5$, $B=1-4/5=1/5$
Answer: $\frac{-\frac{4}{5}x+\frac{1}{5}}{x^2+4} + \frac{4/5}{x-1}$ -
Step 1: $\frac{5x^2+3x+7}{(x-2)^3} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)^3}$
Step 2: $5x^2+3x+7 = A(x-2)^2 + B(x-2) + C$
Step 3: Expand: $A(x^2-4x+4) + Bx-2B + C = Ax^2 + (-4A+B)x + (4A-2B+C)$
Step 4: $A=5$, $-4A+B=3 \Rightarrow -20+B=3 \Rightarrow B=23$, $4A-2B+C=7 \Rightarrow 20-46+C=7 \Rightarrow C=33$
Answer: $\frac{5}{x-2} + \frac{23}{(x-2)^2} + \frac{33}{(x-2)^3}$ -
Step 1: Divide: $\frac{3x^3+2x^2+x+1}{x^2+x-2}$. Factor denom: $(x+2)(x-1)$.
Step 2: Divide: $(3x-1)(x^2+x-2) = 3x^3+3x^2-6x - x^2 - x +2 = 3x^3+2x^2-7x+2$. Subtract: $(3x^3+2x^2+x+1)-(3x^3+2x^2-7x+2) = 8x-1$. So quotient $3x-1$, remainder $8x-1$.
Step 3: Decompose $\frac{8x-1}{(x+2)(x-1)} = \frac{A}{x+2} + \frac{B}{x-1}$; $8x-1 = A(x-1)+B(x+2)$; $x=1 \Rightarrow 7 = B(3) \Rightarrow B=7/3$; $x=-2 \Rightarrow -17 = A(-3) \Rightarrow A=17/3$
Answer: $3x-1 + \frac{17/3}{x+2} + \frac{7/3}{x-1}$ -
Step 1: $\frac{x^2+1}{(x+1)^2(x-1)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-1}$
Step 2: $x^2+1 = A(x+1)(x-1) + B(x-1) + C(x+1)^2$
Step 3: Expand: $A(x^2-1) + Bx - B + C(x^2+2x+1) = (A+C)x^2 + (B+2C)x + (-A-B+C)$
Step 4: $A+C=1$, $B+2C=0$, $-A-B+C=1$; from $A=1-C$, $B=-2C$; then $-(1-C)-(-2C)+C=1 \Rightarrow -1+C+2C+C=1 \Rightarrow 4C=2 \Rightarrow C=1/2$, $A=1/2$, $B=-1$
Answer: $\frac{1/2}{x+1} - \frac{1}{(x+1)^2} + \frac{1/2}{x-1}$ -
Step 1: Divide: $\frac{2x^4+5x^3+3x^2+2x+1}{x^2+1}$
Step 2: Polynomial division: quotient $2x^2+5x+1$, remainder $-3x$ (since $(2x^2+5x+1)(x^2+1) = 2x^4+5x^3+x^2+2x^2+5x+1 = 2x^4+5x^3+3x^2+5x+1$; subtract gives $2x+1 -5x = -3x$). So quotient $2x^2+5x+1$, remainder $-3x$.
Step 3: Decompose $\frac{-3x}{x^2+1} = \frac{Ax+B}{x^2+1}$ with $A=-3, B=0$.
Answer: $2x^2+5x+1 - \frac{3x}{x^2+1}$
Conclusion & Summary
In this lesson, we learned how to decompose rational functions into partial fractions for four cases: distinct linear factors, repeated linear factors, irreducible quadratic factors, and improper fractions. This skill is essential for simplifying complex algebraic expressions and is widely used in calculus and engineering.
Key Takeaways:
1. Distinct Linear: $\frac{P(x)}{(a_1x+b_1)\cdots(a_nx+b_n)} = \frac{A_1}{a_1x+b_1} + \cdots + \frac{A_n}{a_nx+b_n}$
2. Repeated Linear: $\frac{P(x)}{(ax+b)^k} = \frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \cdots + \frac{A_k}{(ax+b)^k}$
3. Quadratic (irreducible): $\frac{P(x)}{(ax^2+bx+c)} = \frac{Ax+B}{ax^2+bx+c}$
4. Improper: Divide first, then apply partial fractions to the proper remainder.
Keep practicing, and partial fractions will become a powerful tool in your algebra toolkit!
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