Class 10 Math Chapter 4 Notes: Partial Fractions Resolution Rules & Solved Board Exam Questions PDF | FBISE & Punjab Board

Theoretical Foundations & Core Concepts

In matriculation mathematics (10th Class), an algebraic fraction expressed in the form $\frac{N(x)}{D(x)}$, where $N(x)$ and $D(x)$ are polynomials in $x$ with $D(x) \neq 0$, is called a rational fraction. Decomposing a single complex rational fraction into a sum of two or more simpler fractions is known as resolving into partial fractions.

1. Proper vs. Improper Rational Fractions

• Proper Rational Fraction: A rational fraction $\frac{N(x)}{D(x)}$ is proper if the degree of the numerator polynomial $N(x)$ is strictly less than the degree of the denominator polynomial $D(x)$. For example, $\frac{2x+3}{x^2+5x+6}$ (Degree of $N(x)=1$, Degree of $D(x)=2$).
• Improper Rational Fraction: A rational fraction $\frac{N(x)}{D(x)}$ is improper if the degree of $N(x)$ is greater than or equal to the degree of $D(x)$. For example, $\frac{x^3+1}{x^2-1}$. Before resolving into partial fractions, an improper fraction must be converted into a polynomial plus a proper rational fraction using polynomial long division: $$\frac{N(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}$$ where $Q(x)$ is the quotient and $R(x)$ is the remainder such that $\text{degree}(R(x)) < \text{degree}(D(x))$.

2. The Four Primary Cases of Partial Fractions

1. Case I: Non-Repeated Linear Factors
   When $D(x)$ contains distinct non-repeated linear factors of the form $(ax+b)$. Each factor produces a term of the form $\frac{A}{ax+b}$.
2. Case II: Repeated Linear Factors
   When $D(x)$ contains a repeated linear factor $(ax+b)^n$. This factor yields $n$ partial fractions: $$\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \dots + \frac{A_n}{(ax+b)^n}$$
3. Case III: Non-Repeated Irreducible Quadratic Factors
   An irreducible quadratic factor $ax^2+bx+c$ cannot be factored into real linear factors (i.e., discriminant $b^2-4ac < 0$). Each non-repeated factor produces a partial fraction of the form $\frac{Ax+B}{ax^2+bx+c}$.
4. Case IV: Repeated Irreducible Quadratic Factors
   When $D(x)$ contains a repeated irreducible quadratic factor $(ax^2+bx+c)^n$, it produces $n$ partial fractions of the form: $$\frac{A_1 x + B_1}{ax^2+bx+c} + \frac{A_2 x + B_2}{(ax^2+bx+c)^2} + \dots + \frac{A_n x + B_n}{(ax^2+bx+c)^n}$$

Formula Summary & Quick Reference

Denominator Condition ($D(x)$)	Algebraic Expression Form	Partial Fraction Decomposition Template
Distinct Linear Factors	$\frac{P(x)}{(x-a)(x-b)}$	$\frac{A}{x-a} + \frac{B}{x-b}$
Repeated Linear Factors	$\frac{P(x)}{(x-a)^2(x-b)}$	$\frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{x-b}$
Non-Repeated Irreducible Quadratic Factor	$\frac{P(x)}{(x-a)(x^2+bx+c)}$	$\frac{A}{x-a} + \frac{Bx+C}{x^2+bx+c}$
Repeated Irreducible Quadratic Factor	$\frac{P(x)}{(x-a)(x^2+c)^2}$	$\frac{A}{x-a} + \frac{Bx+C}{x^2+c} + \frac{Dx+E}{(x^2+c)^2}$

Step-by-Step Solved Board Exam Questions

Question 1 (Case I: Non-Repeated Linear Factors)

Problem: Resolve $\frac{7x-9}{(x+1)(x-3)}$ into partial fractions. (BISE Lahore 2022, FBISE 2021)

Solution:

Step 1: Set up the partial fraction decomposition.

$$\frac{7x-9}{(x+1)(x-3)} = \frac{A}{x+1} + \frac{B}{x-3} \quad \text{--- (Equation 1)}$$

Step 2: Multiply both sides by the total denominator $(x+1)(x-3)$:

$$7x - 9 = A(x - 3) + B(x + 1) \quad \text{--- (Equation 2)}$$

Step 3: Find constant $A$ by substituting the root of $(x+1)$, which is $x = -1$ into Equation 2:

$$7(-1) - 9 = A(-1 - 3) + B(-1 + 1)$$ $$-7 - 9 = A(-4) + B(0)$$ $$-16 = -4A \implies A = 4$$

Step 4: Find constant $B$ by substituting the root of $(x-3)$, which is $x = 3$ into Equation 2:

$$7(3) - 9 = A(3 - 3) + B(3 + 1)$$ $$21 - 9 = A(0) + 4B$$ $$12 = 4B \implies B = 3$$

Step 5: Substitute $A = 4$ and $B = 3$ back into Equation 1:

$$\frac{7x-9}{(x+1)(x-3)} = \frac{4}{x+1} + \frac{3}{x-3}$$ ---

Question 2 (Case II: Repeated Linear Factors)

Problem: Resolve $\frac{x^2+1}{(x-1)^2(x+2)}$ into partial fractions. (BISE Rawalpindi 2023)

Solution:

Step 1: Write the form of partial fractions for repeated linear factors:

$$\frac{x^2+1}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2} \quad \text{--- (Equation 1)}$$

Step 2: Multiply both sides by $(x-1)^2(x+2)$:

$$x^2 + 1 = A(x - 1)(x + 2) + B(x + 2) + C(x - 1)^2 \quad \text{--- (Equation 2)}$$

Step 3: Substitute $x = 1$ into Equation 2 to find $B$: $$(1)^2 + 1 = A(0) + B(1 + 2) + C(0)$$ $$2 = 3B \implies B = \frac{2}{3}$$

Step 4: Substitute $x = -2$ into Equation 2 to find $C$:

$$(-2)^2 + 1 = A(0) + B(0) + C(-2 - 1)^2$$ $$4 + 1 = C(-3)^2$$ $$5 = 9C \implies C = \frac{5}{9}$$

Step 5: Expand Equation 2 and equate coefficients of $x^2$ to find $A$:

$$x^2 + 1 = A(x^2 + x - 2) + B(x + 2) + C(x^2 - 2x + 1)$$ $$x^2 + 1 = (A + C)x^2 + (A + B - 2C)x + (-2A + 2B + C)$$ Equating coefficients of $x^2$: $$1 = A + C$$ $$1 = A + \frac{5}{9} \implies A = 1 - \frac{5}{9} = \frac{4}{9}$$

Step 6: Substitute values of $A, B, C$ into Equation 1:

$$\frac{x^2+1}{(x-1)^2(x+2)} = \frac{4}{9(x-1)} + \frac{2}{3(x-1)^2} + \frac{5}{9(x+2)}$$ ---

Question 3 (Improper Fraction Pre-processing)

Problem: Resolve $\frac{x^3-2x^2-2}{x^2-1}$ into partial fractions. (FBISE 2019 Long Question)

Solution:

Step 1: Check degree. Degree of $N(x) = 3$, Degree of $D(x) = 2$. It is an improper fraction. Perform polynomial long division:

Dividing $x^3 - 2x^2 - 2$ by $x^2 - 1$: $$x^3 - 2x^2 - 2 = (x - 2)(x^2 - 1) + (x - 4)$$ $$\frac{x^3-2x^2-2}{x^2-1} = x - 2 + \frac{x - 4}{x^2 - 1} = x - 2 + \frac{x - 4}{(x-1)(x+1)} \quad \text{--- (Equation 1)}$$

Step 2: Decompose the proper rational part $\frac{x-4}{(x-1)(x+1)}$:

$$\frac{x-4}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$$ $$x - 4 = A(x + 1) + B(x - 1)$$

Step 3: Find $A$ by setting $x = 1$:

$$1 - 4 = A(1 + 1) + B(0) \implies -3 = 2A \implies A = -\frac{3}{2}$$

Step 4: Find $B$ by setting $x = -1$:

$$-1 - 4 = A(0) + B(-1 - 1) \implies -5 = -2B \implies B = \frac{5}{2}$$

Step 5: Combine terms into final expression:

$$\frac{x^3-2x^2-2}{x^2-1} = x - 2 - \frac{3}{2(x-1)} + \frac{5}{2(x+1)}$$ ---

Question 4 (Case III: Non-Repeated Irreducible Quadratic Factor)

Problem: Resolve $\frac{3x+7}{(x^2+4)(x+1)}$ into partial fractions. (BISE Multan 2021, Gujranwala 2023)

Solution:

Step 1: Formulate the equation noting that $(x^2+4)$ is irreducible over real numbers:

$$\frac{3x+7}{(x^2+4)(x+1)} = \frac{Ax + B}{x^2+4} + \frac{C}{x+1} \quad \text{--- (Equation 1)}$$

Step 2: Multiply both sides by $(x^2+4)(x+1)$:

$$3x + 7 = (Ax + B)(x + 1) + C(x^2 + 4) \quad \text{--- (Equation 2)}$$

Step 3: Substitute $x = -1$ (root of linear factor $x+1$):

$$3(-1) + 7 = (A(-1) + B)(0) + C((-1)^2 + 4)$$ $$4 = C(1 + 4) \implies 4 = 5C \implies C = \frac{4}{5}$$

Step 4: Expand Equation 2 to compare coefficients:

$$3x + 7 = Ax^2 + Ax + Bx + B + Cx^2 + 4C$$ $$3x + 7 = (A + C)x^2 + (A + B)x + (B + 4C)$$ Equating coefficients of $x^2$: $$0 = A + C \implies A = -C = -\frac{4}{5}$$ Equating coefficients of $x$: $$3 = A + B \implies 3 = -\frac{4}{5} + B \implies B = 3 + \frac{4}{5} = \frac{19}{5}$$

Step 5: Substitute $A, B, C$ back into Equation 1:

$$\frac{3x+7}{(x^2+4)(x+1)} = \frac{-\frac{4}{5}x + \frac{19}{5}}{x^2+4} + \frac{\frac{4}{5}}{x+1} = \frac{-4x + 19}{5(x^2+4)} + \frac{4}{5(x+1)}$$ ---

Interactive Practice Quiz (Clickable MCQs)

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Frequently Asked Questions (FAQs)

Q1: Why must we convert improper rational fractions into proper ones before finding partial fractions?

Answer: Partial fraction decomposition techniques rely on polynomial division algebra, where the degree of the remainder fraction is strictly less than the denominator degree. If you attempt partial fraction decomposition on an improper fraction (where degree of $N(x) \ge D(x)$) without long division, the system of linear equations for the constants $(A, B, C, \dots)$ will lead to inconsistent or mathematically incorrect results. Always perform polynomial long division first when $\text{degree}(N(x)) \ge \text{degree}(D(x))$.

Q2: How do I know whether to use $A$ or $(Ax+B)$ in the numerator?

Answer: Look strictly at the factor in the denominator:

• If the denominator factor is linear (degree 1, like $x-a$ or $(ax+b)^n$), use a single constant numerator ($A$, $B$, $C$, etc.).
• If the denominator factor is an irreducible quadratic (degree 2, like $x^2+a^2$ or $ax^2+bx+c$), use a linear numerator expression ($Ax+B$, $Cx+D$, etc.).

Q3: What is the fastest method to verify if my partial fraction decomposition is correct during an exam?

Answer: You can verify your final partial fraction decomposition using the Value Test Method:

1. Pick a simple test number for $x$ (such as $x = 0$ or $x = 2$) that does not make any denominator zero.
2. Substitute this $x$ value into the original single fraction.
3. Substitute the exact same $x$ value into your final decomposed partial fractions.
4. If both numerical values are equal, your decomposition is 100% correct.