Class 10 Math Chapter 4 Notes: Partial Fractions Resolution Rules & Solved Board Exam Questions PDF | FBISE & Punjab Board
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.
| 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 | $\frac{P(x)}{(x-a)(x^2+c)^2}$ | $\frac{A}{x-a} + \frac{Bx+C}{x^2+c} + \frac{Dx+E}{(x^2+c)^2}$ |
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:
$$\frac{7x-9}{(x+1)(x-3)} = \frac{4}{x+1} + \frac{3}{x-3}$$ ---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:
$$\frac{x^2+1}{(x-1)^2(x+2)} = \frac{4}{9(x-1)} + \frac{2}{3(x-1)^2} + \frac{5}{9(x+2)}$$ ---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:
$$\frac{x^3-2x^2-2}{x^2-1} = x - 2 - \frac{3}{2(x-1)} + \frac{5}{2(x+1)}$$ ---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:
$$\frac{3x+7}{(x^2+4)(x+1)} = \frac{-\frac{4}{5}x + \frac{19}{5}}{x^2+4} + \frac{\frac{4}{5}}{x+1} = \frac{-4x + 19}{5(x^2+4)} + \frac{4}{5(x+1)}$$ ---Q1: What type of fraction is $\frac{x^3 + 2}{x^2 + 3x + 2}$?
Q2: The partial fraction form of $\frac{1}{(x-1)(x+2)}$ is:
Q3: How many constant coefficients ($A, B, C, \dots$) are needed to decompose $\frac{x+5}{(x-2)^3}$?
Q4: For the partial fraction $\frac{x^2+1}{(x+1)(x^2+4)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+4}$, what is the value of $A$?
Q5: Which of the following quadratic expressions is an irreducible quadratic factor over the set of real numbers?
Q6: If $\frac{1}{x^2-1} = \frac{A}{x-1} + \frac{B}{x+1}$, then the values of $A$ and $B$ are respectively:
Q7: An equation which holds true for all values of the unkown variable involved is called an:
Q8: What is the partial fraction template for $\frac{x}{(x^2+1)^2}$?
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 $\text{degree}(N(x)) \ge \text{degree}(D(x))$.
Answer: Look strictly at the factor in the denominator:
Answer: You can verify your final partial fraction decomposition using the Value Test Method: