9709 P3 §3.1 — Partial Fractions II

💡Lead In

Quick Recall

1. From Lesson 4: what form does $\dfrac{P(x)}{(x-a)(x-b)}$ take?

2. Factorise $x^2 - 4$ and $x^2 + 4$. What do you notice about the second one?

3. For $x^2 + 1$, what is the discriminant $\Delta = b^2 - 4ac$? What does this tell you?

The Wall

Decompose $\dfrac{4 + 2x}{(2x - 1)(x^2 + 1)}$ into partial fractions. The factor $x^2 + 1$ does not factorise (its discriminant is $-4 < 0$). If you try $\dfrac{A}{2x-1} + \dfrac{B}{x^2+1}$, you only have two unknowns — but the numerator $4 + 2x$ has both an $x$-term and a constant. A single constant $B$ over $x^2+1$ cannot produce both. You need a linear numerator over the quadratic.

Left: y = x² - 1 crosses the x-axis at x = ±1, so it splits into (x-1)(x+1). Right: y = x² + 1 never reaches the x-axis (Δ < 0), so it cannot be split — its partial fraction needs a linear numerator (Bx + C).

The Need

We need to extend the partial fraction form: when a factor is an irreducible quadratic, its partial fraction gets a linear numerator $(Ax + B)$, not just a constant.

📖Key Concepts

Irreducible quadratic: A quadratic $cx^2 + dx + e$ is irreducible (cannot be factorised over the reals) if its discriminant is negative: $\Delta = d^2 - 4ce < 0$.
Examples: $x^2 + 1$ ($\Delta = -4$), $x^2 + 4$ ($\Delta = -16$), $2x^2 + 1$ ($\Delta = -8$).

The decomposition form: When the denominator contains an irreducible quadratic $(cx^2 + d)$ alongside a linear factor $(ax + b)$: $$\dfrac{P(x)}{(ax+b)(cx^2+d)} = \dfrac{A}{ax+b} + \dfrac{Bx + C}{cx^2+d}$$
The key difference from Lesson 4: the irreducible quadratic gets a linear numerator $(Bx + C)$, not just a constant.

Adjust A, B, C and watch the partial sum (dashed amber) overlay the original (solid green). Toggle individual curves below to explore how each partial fraction contributes.

$A$: 4

$B$: -2

$C$: 0

✓ Curves match! This is the correct decomposition.

Finding the constants:

Standard method — multiply through by the full denominator: clear all fractions to get a polynomial identity, e.g. $$P(x) \equiv A(ex^2+f) + (Bx+C)(ax+b)$$ Then substitute well-chosen values of $x$. The root of the linear factor ($x = -\tfrac{b}{a}$) isolates its constant in one step — every other term contains $(ax+b)$ and vanishes.
Then find $B$ and $C$ using either method:
- Method 1 — Substitute convenient values: Plug in simple $x$-values (e.g. $x = 0$, $x = 1$, $x = -1$) to get equations for $B$ and $C$. Quick and intuitive.
- Method 2 — Equate coefficients: Expand and match coefficients of $x^2$, $x$, and the constant term. Good for verification.
Shortcut for checking — the cover-up method: substituting the root of $(ax+b)$ into the identity is equivalent to "covering up" the $(ax+b)$ factor in the original denominator and substituting the root into what remains. Fast, but it compresses the working — in written CIE solutions, show the full identity and your substitutions.

Always check irreducibility first: Before deciding the form, test each quadratic factor with the discriminant. If $\Delta \geq 0$, factorise it and use Lesson 4 methods. If $\Delta < 0$, use the linear-numerator form above.

Activity 1 Factorability Detective — Sorting Game

Game-based sorting (5 min): You are a "Factorability Detective." Examine each quadratic card below and decide whether it is factorisable (can be split into linear factors with real coefficients) or irreducible (cannot be factorised over the reals). Click a card to flip it and reveal its category. Sort all 8 cards correctly to become a Master Detective!

Score 0 / 8

Correct 0

Wrong 0

✂️ Factorizable

🔒 Irreducible

Unsorted Cards — click a card to categorize it:

Activity 2 Decomposition Challenge — Guided Discovery

Guided discovery + peer discussion (8 min): Work with a partner to decompose this rational function step by step. Discuss each step before clicking to check. The goal is to find $A$, $B$, and $C$ so that the partial fractions exactly match the original function.

Target function:

$$\dfrac{4 + 2x}{(2x - 1)(x^2 + 1)}$$

Step 1

Step 2

Step 3

Step 4

Step 1: Choose the Decomposition Form

With your partner, decide which of the following is the correct form for the partial fractions. Remember: $x^2 + 1$ is irreducible.

$$\dfrac{A}{2x-1} + \dfrac{B}{x^2+1}$$

$$\dfrac{A}{2x-1} + \dfrac{Bx+C}{x^2+1}$$

$$\dfrac{Ax+B}{2x-1} + \dfrac{Cx+D}{x^2+1}$$

👥 Discuss with your partner: Why does the irreducible quadratic factor need a linear numerator instead of a constant? What would go wrong if you used a constant numerator?

Step 2: Find $A$ — Substitute the Root into the Identity

Multiplying through by the full denominator $(2x-1)(x^2+1)$ gives the identity $4 + 2x \equiv A(x^2+1) + (Bx+C)(2x-1)$. Substitute the root of the linear factor $(2x-1)$ to make the $(Bx+C)$ term vanish.

$A = $

Step 3: Find $B$ and $C$ — Equate Coefficients

Now find $B$ and $C$ by equating coefficients. Multiply through by the full denominator and match the coefficients of $x^2$, $x$, and the constant term.

$B = $

$C = $

Step 4: Verify — Graph Match

The graph below shows the original function (solid green) and your partial fraction decomposition (dashed purple). Enter values for $A$, $B$, $C$ and the graph updates instantly — when the curves perfectly overlap, your decomposition is correct!

$A = $

$B = $

$C = $

Adjust A, B, C to match the curves

Your final decomposition:

👥 Final discussion with partner: Explain to each other why the linear numerator is necessary. What happens to the graph if you set $B=0$? Why?

✏️Worked Examples

Exercise 7D Q3(i) Advanced

Write $\dfrac{6x - 8}{(x^2 + 1)(x + 1)}$ in the form $\dfrac{Ax + B}{x^2 + 1} + \dfrac{C}{x + 1}$.

Hint

Multiply through by the full denominator $(x^2+1)(x+1)$ to form an identity, then substitute $x = -1$ (the root of $x+1$) to find $C$. Then find $A$ and $B$ using either: (1) substituting convenient values like $x = 0$ and $x = 1$, or (2) equating coefficients.

Set up

$x^2 + 1$ is irreducible ($\Delta = -4 < 0$).

$$\dfrac{6x-8}{(x^2+1)(x+1)} = \dfrac{Ax+B}{x^2+1} + \dfrac{C}{x+1}$$

Multiply through by $(x^2+1)(x+1)$:

$$6x - 8 \equiv (Ax+B)(x+1) + C(x^2+1)$$

Find $C$ by substituting the root of $(x+1)$

Set $x = -1$ in the identity — the $(Ax+B)(x+1)$ term vanishes:

$$6(-1) - 8 = C\left((-1)^2+1\right) \;\Rightarrow\; -14 = 2C \;\Rightarrow\; C = -7$$

Method 1: Find $A$ and $B$ by substituting convenient values

We know $C = -7$. Substitute simple values of $x$ to get equations for $A$ and $B$.

Let $x = 0$:

$$6(0) - 8 = (A\cdot 0 + B)(0+1) + C(0+1)$$

$$-8 = B + C = B - 7 \;\Rightarrow\; B = -1$$

Let $x = 1$:

$$6(1) - 8 = (A\cdot 1 + B)(1+1) + C(1+1)$$

$$-2 = 2(A+B) + 2C = 2(A - 1) + 2(-7) = 2A - 2 - 14$$

$$-2 = 2A - 16 \;\Rightarrow\; 2A = 14 \;\Rightarrow\; A = 7$$

So $A = 7$, $B = -1$. ✓

Method 2: Find $A$ and $B$ by equating coefficients

Substitute $C = -7$ and expand:

$$6x - 8 = (Ax+B)(x+1) - 7(x^2+1)$$

$6x - 8 = Ax^2 + Ax + Bx + B - 7x^2 - 7$

$$6x - 8 = (A-7)x^2 + (A+B)x + (B-7)$$

Equating coefficients:

$x^2$: $0 = A - 7 \;\Rightarrow\; A = 7$.

$x^1$: $6 = A + B = 7 + B \;\Rightarrow\; B = -1$.

constant: $-8 = B - 7 = -1 - 7 = -8$. ✓

Check

At $x = 0$: LHS $= \dfrac{-8}{1 \cdot 1} = -8$. RHS $= \dfrac{-1}{1} - \dfrac{7}{1} = -8$. ✓

At $x = 2$: LHS $= \dfrac{4}{5 \cdot 3} = \dfrac{4}{15}$. RHS $= \dfrac{13}{5} - \dfrac{7}{3} = \dfrac{39-35}{15} = \dfrac{4}{15}$. ✓

Final Answer

$$\dfrac{6x-8}{(x^2+1)(x+1)} = \dfrac{7x-1}{x^2+1} - \dfrac{7}{x+1}$$

🧠 Concept Quiz

Four quick questions, one per learning objective. Click an option to lock in your answer.

Score: 0 / 4

Q1. Which of the following quadratics is irreducible over the reals? Objective 1 · Discriminant test

A. $x^2 - 9$

B. $2x^2 + 1$

C. $4x^2 - 1$

D. $x^2 - 5x + 6$

Answer: B. $2x^2 + 1$ has $\Delta = 0^2 - 4(2)(1) = -8 < 0$, so it has no real roots and cannot be factorised. The others all have $\Delta \geq 0$: $x^2-9$ ($\Delta=36$, factors as $(x-3)(x+3)$), $4x^2-1$ ($\Delta=16$, factors as $(2x-1)(2x+1)$), $x^2-5x+6$ ($\Delta=1$, factors as $(x-2)(x-3)$).

Q2. The correct partial fraction form for $\dfrac{P(x)}{(x+3)(x^2+2)}$ is: Objective 2 · Form selection

A. $\dfrac{A}{x+3} + \dfrac{B}{x^2+2}$

B. $\dfrac{Ax+B}{x+3} + \dfrac{C}{x^2+2}$

C. $\dfrac{A}{x+3} + \dfrac{B}{x} + \dfrac{C}{x^2+2}$

D. $\dfrac{A}{x+3} + \dfrac{Bx+C}{x^2+2}$

Answer: D. $x^2+2$ has $\Delta = -8 < 0$, so it is irreducible and must take a linear numerator $(Bx+C)$. Option A would only give a constant over the quadratic, which cannot match an arbitrary numerator. Option B puts the linear numerator on the wrong (linear) factor. Option C incorrectly splits the quadratic.

Q3. Multiplying $\dfrac{5x+1}{(x-1)(x^2+1)} = \dfrac{A}{x-1} + \dfrac{Bx+C}{x^2+1}$ through by the full denominator gives $5x+1 \equiv A(x^2+1) + (Bx+C)(x-1)$. Substituting $x = 1$ gives $A =$: Objective 3 · Substituting the root

A. $3$

B. $5$

C. $1$

D. $6$

Answer: A. Substituting $x = 1$ (the root of $x-1$) makes the $(Bx+C)(x-1)$ term vanish, leaving $5(1)+1 = A(1^2+1)$, so $A = \dfrac{6}{2} = 3$. (The cover-up shortcut — "cover $(x-1)$, substitute $x=1$ into what remains" — is a compressed mental version of this substitution.)

Q4. Given $\dfrac{4+2x}{(2x-1)(x^2+1)} = \dfrac{4}{2x-1} - \dfrac{2x}{x^2+1}$, which check correctly verifies the decomposition? Objective 4 · Verification

A. Substitute $x = \tfrac{1}{2}$: both sides equal $0$.

B. Compare degrees: LHS has degree $1$, RHS has degree $0$.

C. Substitute $x = 0$: LHS $= -4$, RHS $= -4 - 0 = -4$. ✓

D. Take $x \to \infty$: both sides approach $0$, so they are equal everywhere.

Answer: C. Substituting $x = 0$ gives LHS $= \dfrac{4}{(-1)(1)} = -4$ and RHS $= \dfrac{4}{-1} - \dfrac{0}{1} = -4$. ✓ Option A fails because $x = \tfrac{1}{2}$ makes both sides undefined (vertical asymptote), not zero. Option B is false — both are rational functions, not polynomials. Option D only checks the limit at infinity, which is necessary but not sufficient.

🎉

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✏️ More Practice

Exercise 7D Q6(i) Standard

Express $\dfrac{10}{(2 - x)(1 + x^2)}$ in partial fractions.

Solution

$1 + x^2$: $\Delta = -4 < 0$, irreducible. ✓ Note: $(2-x)$ is linear with root $x = 2$.

$$\dfrac{10}{(2-x)(1+x^2)} = \dfrac{A}{2-x} + \dfrac{Bx+C}{1+x^2}$$

Multiplying through by $(2-x)(1+x^2)$ gives the identity $10 \equiv A(1+x^2) + (Bx+C)(2-x)$.

Find $A$: set $x = 2$ (root of $2-x$) — the $(Bx+C)$ term vanishes: $10 = A(5) \;\Rightarrow\; A = 2$.

Find $C$ by substituting $x = 0$:

$$10 = A(1) + C(2) = 2 + 2C \;\Rightarrow\; C = 4$$

Find $B$ by substituting $x = 1$:

$$10 = A(2) + (B+C)(1) = 4 + B + 4 \;\Rightarrow\; B = 2$$

Check

At $x = -1$: LHS $= \dfrac{10}{3 \cdot 2} = \dfrac{5}{3}$. RHS $= \dfrac{2}{3} + \dfrac{2}{2} = \dfrac{2}{3} + 1 = \dfrac{5}{3}$. ✓

Final Answer

$$\dfrac{10}{(2-x)(1+x^2)} = \dfrac{2}{2-x} + \dfrac{2x+4}{1+x^2}$$

Exercise 7D Q1(viii) Challenging

Express $\dfrac{2x^2 + x + 2}{(2x^2 + 1)(x + 1)}$ as a sum of partial fractions.

Hint 1

Check $2x^2+1$: $\Delta = 0 - 8 = -8 < 0$, irreducible. Use $\dfrac{Ax+B}{2x^2+1} + \dfrac{C}{x+1}$. Multiply through by the full denominator to form an identity, substitute $x = -1$ (the root of $x+1$) to find $C$, then convenient values like $x = 0$ and $x = 1$ to find $A$ and $B$.

Solution

$2x^2+1$: $\Delta = -8 < 0$, irreducible. ✓

$$\dfrac{2x^2+x+2}{(2x^2+1)(x+1)} = \dfrac{Ax+B}{2x^2+1} + \dfrac{C}{x+1}$$

Multiplying through by $(2x^2+1)(x+1)$ gives the identity $2x^2 + x + 2 \equiv (Ax+B)(x+1) + C(2x^2+1)$.

Find $C$: set $x = -1$ (root of $x+1$) — the $(Ax+B)$ term vanishes: $3 = 3C \;\Rightarrow\; C = 1$.

Find $B$ by substituting $x = 0$:

$$2 = B(1) + C(1) = B + 1 \;\Rightarrow\; B = 1$$

Find $A$ by substituting $x = 1$:

$$5 = (A+B)(2) + C(3) = 2(A+1) + 3 \;\Rightarrow\; 2A = 0 \;\Rightarrow\; A = 0$$

Check

At $x = 2$: LHS $= \dfrac{8+2+2}{9 \cdot 3} = \dfrac{12}{27} = \dfrac{4}{9}$. RHS $= \dfrac{1}{9} + \dfrac{1}{3} = \dfrac{1}{9} + \dfrac{3}{9} = \dfrac{4}{9}$. ✓

Final Answer

$$\dfrac{2x^2+x+2}{(2x^2+1)(x+1)} = \dfrac{1}{2x^2+1} + \dfrac{1}{x+1}$$

Note: $A = 0$, so the linear numerator collapses to a constant — the quadratic part only needed a constant numerator here.

Exercise 7D Q9(i) Advanced

Let $\mathrm{f}(x) = \dfrac{2x^2 - 7x - 1}{(x - 2)(x^2 + 3)}$. Express $\mathrm{f}(x)$ in partial fractions.

Hint 1

Check $x^2+3$: $\Delta = 0 - 12 = -12 < 0$, irreducible. Use $\dfrac{A}{x-2} + \dfrac{Bx+C}{x^2+3}$. Multiply through by the full denominator to form an identity, substitute $x = 2$ (the root of $x-2$) to find $A$, then convenient values like $x = 0$ and $x = 1$ to find $B$ and $C$.

Solution

$x^2+3$: $\Delta = -12 < 0$, irreducible. ✓

$$\dfrac{2x^2-7x-1}{(x-2)(x^2+3)} = \dfrac{A}{x-2} + \dfrac{Bx+C}{x^2+3}$$

Multiplying through by $(x-2)(x^2+3)$ gives the identity $2x^2 - 7x - 1 \equiv A(x^2+3) + (Bx+C)(x-2)$.

Find $A$: set $x = 2$ (root of $x-2$) — the $(Bx+C)$ term vanishes: $-7 = 7A \;\Rightarrow\; A = -1$.

Find $C$ by substituting $x = 0$:

$$-1 = A(3) + C(-2) = -3 - 2C \;\Rightarrow\; C = -1$$

Find $B$ by substituting $x = 1$:

$$-6 = A(4) + (B+C)(-1) = -4 - (B-1) \;\Rightarrow\; B = 3$$

Check

At $x = -1$: LHS $= \dfrac{2+7-1}{(-3)(4)} = \dfrac{8}{-12} = -\dfrac{2}{3}$. RHS $= \dfrac{-1}{-3} + \dfrac{-4}{4} = \dfrac{1}{3} - 1 = -\dfrac{2}{3}$. ✓

At $x = 3$: LHS $= \dfrac{18-21-1}{(1)(12)} = \dfrac{-4}{12} = -\dfrac{1}{3}$. RHS $= \dfrac{-1}{1} + \dfrac{8}{12} = -1 + \dfrac{2}{3} = -\dfrac{1}{3}$. ✓

Final Answer

$$\dfrac{2x^2-7x-1}{(x-2)(x^2+3)} = -\dfrac{1}{x-2} + \dfrac{3x-1}{x^2+3}$$

Improper Fraction Challenge Advanced

Express $\dfrac{2x^3 - x^2 + 3x - 1}{(x - 1)(x^2 + 1)}$ in the form $A + \dfrac{B}{x-1} + \dfrac{Cx+D}{x^2+1}$.

Hint

The numerator has degree 3 and the denominator also has degree 3 — this is an improper fraction. First perform polynomial division to find the quotient $A$ (or compare coefficients of $x^3$). Then decompose the remaining proper fraction: multiply through by the full denominator to form an identity, substitute $x = 1$ (the root of $x-1$) to find $B$, and convenient values for $C$ and $D$.

Step 1: Recognise the improper fraction

Numerator: degree 3. Denominator: $(x-1)(x^2+1) = x^3 - x^2 + x - 1$, also degree 3.

Since $\deg(\text{numerator}) = \deg(\text{denominator})$, we write:

$$\dfrac{2x^3 - x^2 + 3x - 1}{(x-1)(x^2+1)} = A + \dfrac{B}{x-1} + \dfrac{Cx+D}{x^2+1}$$

Step 2: Find $A$ — two methods

Method 1: Polynomial division

Divide the numerator by the denominator (long division):

So $A = 2$ and the remainder is $x^2 + x + 1$.

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

Method 2: Compare $x^3$ coefficients

Multiply through by $(x-1)(x^2+1)$:

$$2x^3 - x^2 + 3x - 1 = A(x-1)(x^2+1) + B(x^2+1) + (Cx+D)(x-1)$$

The $x^3$ term on the right comes only from $A \cdot x \cdot x^2 = A x^3$.

Comparing coefficients of $x^3$: $2 = A \;\Rightarrow\; A = 2$. ✓

Step 3: Find $B$ by substituting the root

The remainder proper fraction has numerator:

$2x^3 - x^2 + 3x - 1 - 2(x^3 - x^2 + x - 1) = 2x^3 - x^2 + 3x - 1 - 2x^3 + 2x^2 - 2x + 2 = x^2 + x + 1$

So we decompose $\dfrac{x^2 + x + 1}{(x-1)(x^2+1)}$. Multiplying through by $(x-1)(x^2+1)$ gives the identity $x^2 + x + 1 \equiv B(x^2+1) + (Cx+D)(x-1)$.

Find $B$: set $x = 1$ (root of $x-1$) — the $(Cx+D)$ term vanishes: $3 = 2B \;\Rightarrow\; B = \dfrac{3}{2}$.

Step 4: Find $C$ and $D$ by substitution

Find $D$ using $x = 0$:

$$1 = B(1) + D(-1) = \tfrac{3}{2} - D \;\Rightarrow\; D = \tfrac{1}{2}$$

Find $C$ using $x = -1$:

$$1 = B(2) + (-C+D)(-2) = 3 + 2C - 2\left(\tfrac{1}{2}\right) \;\Rightarrow\; 1 = 2 + 2C \;\Rightarrow\; C = -\tfrac{1}{2}$$

Check

At $x = 2$: LHS $= \dfrac{16 - 4 + 6 - 1}{(1)(5)} = \dfrac{17}{5}$. RHS $= 2 + \dfrac{3/2}{1} + \dfrac{-1 + 1/2}{5} = 2 + \dfrac{3}{2} + \dfrac{-1/2}{5} = 2 + \dfrac{3}{2} - \dfrac{1}{10} = \dfrac{20+15-1}{10} = \dfrac{34}{10} = \dfrac{17}{5}$. ✓

Final Answer

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

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Exit Ticket

Check the learning objectives you feel confident about:

I can test whether a quadratic is irreducible using the discriminant ($\Delta < 0$). I know that an irreducible quadratic factor gets a linear numerator $(Ax + B)$ in partial fractions. I can multiply through by the full denominator to form an identity, then substitute the root of the linear factor plus convenient values (or equate coefficients) to find all the constants. I can verify a decomposition by substituting values or recombining the fractions.