Lesson Notes By Weeks and Term v3 - Senior Secondary 3
Partial fractions
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Partial fractions
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Subject: Further Mathematics
Class: Senior Secondary 3
Term: 1st Term
Week: 2
Theme: Pure Mathematics
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Students should be able to resolve rational functions in to partial fractions (degree of numerator less thanthat of numerator which is less than or equal to 4)
Check if proper: Degree of numerator (3) is less than the degree of the denominator (4). It is a proper fraction.
2. Identify factors: The denominator has a repeated irreducible quadratic factor $(x^2+1)^2$.
3. Set up the partial fraction form: $\frac{2x^3+x^2+4x-1}{(x^2+1)^2} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2}$
4. Clear the denominator: Multiply both sides by $(x^2+1)^2$: $2x^3+x^2+4x-1 = (Ax+B)(x^2+1) + (Cx+D)$
5. Determine constants A, B, C and D: Expand the right side: $2x^3+x^2+4x-1 = Ax^3+Ax+Bx^2+B + Cx+D$ Group terms by powers of x: $2x^3+x^2+4x-1 = Ax^3+Bx^2+(A+C)x+(B+D)$ Equate coefficients: Coefficient of $x^3$: $A=2$ Coefficient of $x^2$: $B=1$ Coefficient of $x$: $A+C=4$ Substitute $A=2$: $2+C=4 \implies C=2$ Constant term: $B+D=-1$ Substitute $B=1$: $1+D=-1 \implies D=-2$
6. Write the partial fraction decomposition: $\frac{2x^3+x^2+4x-1}{(x^2+1)^2} = \frac{2x+1}{x^2+1} + \frac{2x-2}{(x^2+1)^2}$ Summary of Forms for Partial Fractions: | Factor in Denominator $Q(x)$ | Corresponding Partial Fraction Term(s) | | :---------------------------------------------------------------- | :----------------------------------------------------------------------------------------------------------- | | Distinct Linear Factor: $(ax+b)$ | $\frac{A}{ax+b}$ | | Repeated Linear Factor: $(ax+b)^k$ | $\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + ... + \frac{A_k}{(ax+b)^k}$ | | Irreducible Quadratic Factor: $(ax^2+bx+c)$ | $\frac{Ax+B}{ax^2+bx+c}$ | | Repeated Irreducible Quadratic Factor: $(ax^2+bx+c)^k$ | $\frac{A_1x+B_1}{ax^2+bx+c} + \frac{A_2x+B_2}{(ax^2+bx+c)^2} + ... + \frac{A_kx+B_k}{(ax^2+bx+c)^k}$ | General Strategy for Solving:
1. Check if proper: If improper, perform polynomial long division first.
2. Factorize denominator $Q(x)$: Factorize $Q(x)$ completely into linear and/or irreducible quadratic factors.
3. Set up the partial fraction form: Based on the type of factors identified.
4. Clear denominators: Multiply both sides of the equation by $Q(x)$.
5. Determine constants: Use a combination of: Substitution: Choose values of $x$ that make linear factors zero to find some constants quickly. * Equating Coefficients: Expand the right side, group terms by powers of $x$, and equate coefficients of corresponding powers of $x$ on both sides to form a system of linear equations.
6. Write the final decomposition. This section provides a detailed breakdown of the theoretical underpinnings and practical methods required for resolving rational functions into partial fractions.
Definition of a Rational Function: A rational function is a function that can be written as the ratio of two polynomials, $P(x)$ and $Q(x)$, where $Q(x)$ is not the zero polynomial. That is, $f(x) = \frac{P(x)}{Q(x)}$.
Proper and Improper Rational Functions: Proper Rational Function: A rational function $\frac{P(x)}{Q(x)}$ is proper if the degree of the numerator polynomial $P(x)$ is strictly less than the degree of the denominator polynomial $Q(x)$. For example, $\frac{2x+1}{x^2+3x+2}$ is a proper rational function.
Improper Rational Function: A rational function $\frac{P(x)}{Q(x)}$ is improper if the degree of the numerator $P(x)$ is greater than or equal to the degree of the denominator $Q(x)$. For example, $\frac{x^3+x^2-1}{x^2-4}$ is an improper rational function.
Note: The performance objective specifically targets cases where the degree of the numerator is less than that of the denominator (i.e., proper rational functions). If an improper fraction is encountered, polynomial long division must be performed first to express it as a sum of a polynomial and a proper rational fraction.
However, this lesson focuses on proper fractions as per the stated objective.
Partial Fraction Decomposition: The process of expressing a proper rational function as a sum of simpler fractions is called partial fraction decomposition. The form of the partial fractions depends on the nature of the factors in the denominator $Q(x)$. The denominator's degree is limited to a maximum of 4 in this topic. Cases for Denominator Factors (with degree of Q(x) ≤ 4): Case 1: Denominator contains distinct linear factors. If the denominator $Q(x)$ can be factored into distinct linear factors $(ax+b)(cx+d)...$, then the proper rational function $\frac{P(x)}{Q(x)}$ can be expressed as a sum of fractions of the form $\frac{A}{ax+b}$, $\frac{B}{cx+d}$, etc., where A, B, ... are constants to be determined. For example, if $Q(x) = (x-a)(x-b)(x-c)$, the decomposition form is: $\frac{P(x)}{(x-a)(x-b)(x-c)} = \frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c}$ Constraints: Degree of $P(x)$ < Degree of $Q(x)$. Degree of $Q(x)$ ≤
4. Worked Example 1 (Distinct Linear Factors): Resolve $\frac{3x+5}{x^2+x-2}$ into partial fractions.
Solution:
1. Check if proper: The degree of the numerator (1) is less than the degree of the denominator (2). It is a proper fraction.
2. Factorize the denominator: $x^2+x-2 = (x+2)(x-1)$.
3. Set up the partial fraction form: Since the factors are distinct linear factors, the form is: $\frac{3x+5}{(x+2)(x-1)} = \frac{A}{x+2} + \frac{B}{x-1}$
4. Clear the denominator: Multiply both sides by $(x+2)(x-1)$: $3x+5 = A(x-1) + B(x+2)$
5. Determine constants A and B: Method 1: Substitution (setting x to make factors zero) To find B, let $x=1$ (this makes the term with A zero): $3(1)+5 = A(1-1) + B(1+2)$ $8 = A(0) + B(3)$ $8 = 3B \implies B = \frac{8}{3}$ To find A, let $x=-2$ (this makes the term with B zero): $3(-2)+5 = A(-2-1) + B(-2+2)$ $-6+5 = A(-3) + B(0)$ $-1 = -3A \implies A = \frac{1}{3}$ Method 2: Equating Coefficients (alternative or supplementary)
Expand the right side: $3x+5 = Ax - A + Bx + 2B$ Group terms by powers of x: $3x+5 = (A+B)x + (-A+2B)$ Equate coefficients of x: $A+B = 3$ (Equation 1)
Equate constant terms: $-A+2B = 5$ (Equation 2)
Solve the simultaneous equations: Add (1) and (2): $(A+B) + (-A+2B) = 3+5 \implies 3B = 8 \implies B = \frac{8}{3}$ Substitute B into (1): $A + \frac{8}{3} = 3 \implies A = 3 - \frac{8}{3} = \frac{9-8}{3} = \frac{1}{3}$
6. Write the partial fraction decomposition: $\frac{3x+5}{x^2+x-2} = \frac{\frac{1}{3}}{x+2} + \frac{\frac{8}{3}}{x-1}$ or $\frac{1}{3(x+2)} + \frac{8}{3(x-1)}$ Case 2: Denominator contains repeated linear factors. If the denominator $Q(x)$ contains a repeated linear factor, such as $(ax+b)^k$, then for each such factor, there will be $k$ partial fractions of the form $\frac{A_1}{ax+b}, \frac{A_2}{(ax+b)^2}, ..., \frac{A_k}{(ax+b)^k}$. For example, if $Q(x) = (x-a)^3 (x-b)$, the decomposition form is: $\frac{P(x)}{(x-a)^3(x-b)} = \frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{(x-a)^3} + \frac{D}{x-b}$ Constraints: Degree of $P(x)$ < Degree of $Q(x)$. Degree of $Q(x)$ ≤
4. Worked Example 2 (Repeated Linear Factors): Resolve $\frac{x^2+2x+3}{(x-1)^3}$ into partial fractions.
Solution:
1. Check if proper: \frac{\frac{8}{3}}{x-1}$ or $\frac{1}{3(x+2)} + \frac{8}{3(x-1)}$ Case 2: Denominator contains repeated linear factors. If the denominator $Q(x)$ contains a repeated linear factor, such as $(ax+b)^k$, then for each such factor, there will be $k$ partial fractions of the form $\frac{A_1}{ax+b}, \frac{A_2}{(ax+b)^2}, ..., \frac{A_k}{(ax+b)^k}$. For example, if $Q(x) = (x-a)^3 (x-b)$, the decomposition form is: $\frac{P(x)}{(x-a)^3(x-b)} = \frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{(x-a)^3} + \frac{D}{x-b}$ Constraints: Degree of $P(x)$ < Degree of $Q(x)$. Degree of $Q(x)$ ≤
4. Worked Example 2 (Repeated Linear Factors): Resolve $\frac{x^2+2x+3}{(x-1)^3}$ into partial fractions.
Solution:
1. Check if proper: The degree of the numerator (2) is less than the degree of the denominator (3). It is a proper fraction.
2. Set up the partial fraction form: The denominator has a repeated linear factor $(x-1)^3$. $\frac{x^2+2x+3}{(x-1)^3} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3}$
3. Clear the denominator: Multiply both sides by $(x-1)^3$: $x^2+2x+3 = A(x-1)^2 + B(x-1) + C$
4. Determine constants A, B and C: Substitution (for C): Let $x=1$: $(1)^2+2(1)+3 = A(1-1)^2 + B(1-1) + C$ $1+2+3 = A(0) + B(0) + C$ $6 = C$ Equating Coefficients (for A and B, or a mix of methods): Expand the right side: $x^2+2x+3 = A(x^2-2x+1) + B(x-1) + C$ $x^2+2x+3 = Ax^2 - 2Ax + A + Bx - B + C$ Group terms by powers of x: $x^2+2x+3 = Ax^2 + (-2A+B)x + (A-B+C)$ Equate coefficients of $x^2$: $A = 1$ Equate coefficients of $x$: $-2A+B = 2$ Substitute $A=1$: $-2(1)+B = 2 \implies -2+B=2 \implies B=4$ Equate constant terms: $A-B+C = 3$ Substitute $A=1, B=4, C=6$: $1-4+6 = 3 \implies 3=3$ (This checks our values)
5. Write the partial fraction decomposition: $\frac{x^2+2x+3}{(x-1)^3} = \frac{1}{x-1} + \frac{4}{(x-1)^2} + \frac{6}{(x-1)^3}$ Case 3: Denominator contains irreducible quadratic factors. An irreducible quadratic factor is a quadratic expression $ax^2+bx+c$ that cannot be factored into linear factors with real coefficients (i.e., its discriminant $b^2-4ac < 0$). If the denominator $Q(x)$ contains an irreducible quadratic factor $ax^2+bx+c$, then the corresponding partial fraction will be of the form $\frac{Ax+B}{ax^2+bx+c}$. For example, if $Q(x) = (x^2+a^2)(x-b)$, the decomposition form is: $\frac{P(x)}{(x^2+a^2)(x-b)} = \frac{Ax+B}{x^2+a^2} + \frac{C}{x-b}$ Constraints: Degree of $P(x)$ < Degree of $Q(x)$. Degree of $Q(x)$ ≤
4. Worked Example 3 (Irreducible Quadratic Factor): Resolve $\frac{2x^2-x+4}{x(x^2+4)}$ into partial fractions.
Solution:
1. Check if proper: Degree of numerator (2) < Degree of denominator (3). It is proper.
2. Factorize the denominator: $x(x^2+4)$. The factor $x^2+4$ is an irreducible quadratic since its discriminant is $0^2 - 4(1)(4) = -16 < 0$.
3. Set up the partial fraction form: $\frac{2x^2-x+4}{x(x^2+4)} = \frac{A}{x} + \frac{Bx+C}{x^2+4}$
4. Clear the denominator: Multiply both sides by $x(x^2+4)$: $2x^2-x+4 = A(x^2+4) + (Bx+C)x$ $2x^2-x+4 = Ax^2+4A + Bx^2+Cx$
5. Determine constants A, B and C: Substitution (for A): Let $x=0$: $2(0)^2-0+4 = A(0^2+4) + (B(0)+C)(0)$ $4 = 4A \implies A=1$ Equating Coefficients: Group terms by powers of x: $2x^2-x+4 = (A+B)x^2 + Cx + 4A$ Equate coefficients of $x^2$: $A+B=2$ Substitute $A=1$: $1+B=2 \implies B=1$ Equate coefficients of $x$: $C=-1$ Equate constant terms: $4A=4$ Substitute $A=1$: $4(1)=4$ (Checks our value for A)
6. Write the partial fraction decomposition: $\frac{2x^2-x+4}{x(x^2+4)} = \frac{1}{x} + \frac{x-1}{x^2+4}$ Case 4: Denominator contains repeated irreducible quadratic factors. If the denominator $Q(x)$ contains a repeated irreducible quadratic factor, such as $(ax^2+bx+c)^k$, then for each such factor, there will be $k$ partial fractions of the form $\frac{A_1x+B_1}{ax^2+bx+c}, \frac{A_2x+B_2}{(ax^2+bx+c)^2}, ..., \frac{A_kx+B_k}{(ax^2+bx+c)^k}$. For example, if $Q(x) = (x^2+a^2)^2$, the decomposition form is: $\frac{P(x)}{(x^2+a^2)^2} = \frac{Ax+B}{x^2+a^2} + \frac{Cx+D}{(x^2+a^2)^2}$ Constraints: Degree of $P(x)$ < Degree of $Q(x)$. Degree of $Q(x)$ ≤
4. Worked Example 4 (Repeated Irreducible Quadratic Factors): Resolve $\frac{x^3+x^2+x+1}{(x^2+1)^2}$ into partial fractions.
Solution:
1. Check if proper: Degree of numerator (3) < Degree of denominator (4). It is proper.
2. Factorize the denominator: $(x^2+1)^2$. The factor $x^2+1$ is an irreducible quadratic.
3. Set up the partial fraction form: $\frac{x^3+x^2+x+1}{(x^2+1)^2} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2}$
4. Clear the denominator: Multiply both sides by $(x^2+1)^2$: $x^3+x^2+x+1 = (Ax+B)(x^2+1) + (Cx+D)$
5. Determine constants A, B, C and D: Expand the right side: $x^3+x^2+x+1 = Ax^3+Ax+Bx^2+B + Cx+D$ * Group terms by $Q(x)$ ≤
4. Worked Example 4 (Repeated Irreducible Quadratic Factors): Resolve $\frac{x^3+x^2+x+1}{(x^2+1)^2}$ into partial fractions.
Solution:
1. Check if proper: Degree of numerator (3) < Degree of denominator (4). It is proper.
2. Factorize the denominator: $(x^2+1)^2$. The factor $x^2+1$ is an irreducible quadratic.
3. Set up the partial fraction form: $\frac{x^3+x^2+x+1}{(x^2+1)^2} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2}$
4. Clear the denominator: Multiply both sides by $(x^2+1)^2$: $x^3+x^2+x+1 = (Ax+B)(x^2+1) + (Cx+D)$
5. Determine constants A, B, C and D: Expand the right side: $x^3+x^2+x+1 = Ax^3+Ax+Bx^2+B + Cx+D$ Group terms by powers of x: $x^3+x^2+x+1 = Ax^3+Bx^2+(A+C)x+(B+D)$ Equate coefficients: Coefficient of $x^3$: $A=1$ Coefficient of $x^2$: $B=1$ Coefficient of $x$: $A+C=1$ Substitute $A=1$: $1+C=1 \implies C=0$ Constant term: $B+D=1$ Substitute $B=1$: $1+D=1 \implies D=0$
6. Write the partial fraction decomposition: $\frac{x^3+x^2+x+1}{(x^2+1)^2} = \frac{1x+1}{x^2+1} + \frac{0x+0}{(x^2+1)^2} = \frac{x+1}{x^2+1}$ (This example simplifies neatly, which is good for illustration, but learners should be prepared for non-zero C and D).
Important Considerations for Teachers: Always ensure the fraction is proper before proceeding. Emphasize careful algebraic manipulation, especially when expanding and equating coefficients. Encourage students to check their answers by combining the partial fractions back into the original rational function. For irreducible quadratics, remind students how to check the discriminant ($b^2-4ac < 0$). The constraint of the denominator degree being less than or equal to 4 means the highest power of x in Q(x) will be $x^4$. This limits the number of partial fractions and constants to find, making problems manageable for SS3.
This section provides practical ways to connect the topic of Partial Fractions to real-life situations relevant to Nigeria, making the learning more tangible and relatable. Electrical Circuit Analysis in Nigerian Context: Application: In designing and troubleshooting electrical systems for homes, offices, or even local workshops (e.g., generator repair shops, electronics vendors in Alaba International Market, Lagos), engineers and technicians often encounter circuits whose behavior is described by rational functions in the frequency domain (Laplace transforms).
Integration: If a circuit's transfer function is given as $H(s) = \frac{V_{out}(s)}{V_{in}(s)} = \frac{s+3}{(s+1)(s+2)}$, resolving this into partial fractions ($\frac{A}{s+1} + \frac{B}{s+2}$) simplifies the process of finding the circuit's response in the time domain. This decomposition allows engineers to understand how different components (resistors, inductors, capacitors) influence the overall circuit behavior, which is critical for ensuring stable power supply or optimal performance of electronic gadgets. A local electronics technician, though perhaps not explicitly performing partial fraction calculations, relies on the principles derived from such analysis when selecting components for repairs. Pharmacokinetics (Drug Dosage and Efficacy): Application: In medical research and pharmaceutical manufacturing (e.g., at drug companies like Emzor or May & Baker in Nigeria), partial fractions can be implicitly used in pharmacokinetic models that describe how drugs are absorbed, distributed, metabolized, and excreted from the body. These processes are often modeled using differential equations whose solutions involve rational functions.
Integration: If the concentration of a drug in a patient's bloodstream over time is represented by a complex rational function, decomposing it into partial fractions can simplify the analysis. This helps determine optimal dosing regimens, understanding drug half-life, and predicting drug efficacy for patients suffering from diseases prevalent in Nigeria, such as malaria or typhoid, ensuring effective treatment while minimizing side effects. While direct calculation by students is unlikely, the teacher can explain that understanding rates of change (which integration helps) is foundational to determining safe and effective drug dosages. Control Systems for Local Manufacturing/Agriculture: Application: Automated control systems are increasingly used in Nigerian industries, from large factories to specialized agricultural equipment (e.g., automated irrigation systems, processing lines for agricultural produce like rice or yam). The dynamic behavior of these systems is mathematically represented using transfer functions, which are rational expressions.
Integration: Consider a system controlling the temperature in a kiln used for pottery in Abuja or drying crops in Kano. If the system's response to an input (e.g., adjusting heat) is modeled by a rational function like $G(s) = \frac{s+10}{s(s+1)(s+5)}$, partial fraction decomposition allows engineers to analyze the individual components of the system's response (e.g., steady-state response, transient response). This helps in designing robust control systems that maintain desired conditions (e.g., precise temperature, consistent flow rates), improving efficiency and product quality in local manufacturing or agricultural processes.