Partial Fractions I

9709 Paper 3 · §3.1 Algebra · Lesson 4 of 6
💡Lead In

The Scenario

You know that $\dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$. Combining two fractions into one is easy — find a common denominator and add. But now the question turns around: if I hand you $\dfrac{5}{6}$ and ask "split this into two unit fractions," can you recover $\dfrac{1}{2} + \dfrac{1}{3}$?

The Question

You know $\dfrac{1}{x-2} + \dfrac{1}{x+3} = \dfrac{2x+1}{(x-2)(x+3)}$. But if I give you $\dfrac{5}{(x-2)(x+3)}$, how would you split it back into two simpler fractions — without guessing?

📖Key Concepts
Partial fractions = reverse of adding fractions. Given a fraction with a factorable denominator, split it into simpler fractions — one per factor of the denominator.
Key relation 1: Three distinct linear factors $\dfrac{\text{numerator}}{(ax+b)(cx+d)(ex+f)} = \dfrac{A}{ax+b} + \dfrac{B}{cx+d} + \dfrac{C}{ex+f}$
1
2
-1

Green (solid): target original $\dfrac{2x^2+5x-1}{(x-1)(x+1)(x+2)}$. Amber (dashed): current partial sum. Adjust $A$, $B$, $C$ until the curves match.

Key relation 2: Repeated linear factor $\dfrac{\text{numerator}}{(ax+b)(cx+d)^2} = \dfrac{A}{ax+b} + \dfrac{B}{cx+d} + \dfrac{C}{(cx+d)^2}$
A repeated factor $(cx+d)^2$ gets two partial fractions: one with denominator $(cx+d)$ and one with $(cx+d)^2$.
2
1
-1

Green (solid): target original $\dfrac{3x^2+3x+2}{(x-1)(x+1)^2}$. Amber (dashed): current partial sum. Adjust $A$, $B$, $C$ until the curves match.

Standard method — multiply through by the full denominator: After setting up the form, multiply both sides by the full denominator to clear all fractions: $$P(x) \equiv A(cx+d)(ex+f) + B(ax+b)(ex+f) + C(ax+b)(cx+d)$$ Then substitute well-chosen values of $x$ into this identity. The fastest choices are the roots of the linear factors ($x = -\tfrac{b}{a}$, $x = -\tfrac{d}{c}$, …) — each root makes the other terms vanish. For any constant that no root can isolate (e.g. the middle constant of a repeated factor), substitute a convenient value such as $x = 0$, or equate coefficients of matching powers of $x$.
Improper fractions — divide first: When the degree of the numerator is greater than or equal to the degree of the denominator, the fraction is improper. Perform polynomial division first to obtain $$\dfrac{P(x)}{D(x)} = Q(x) + \dfrac{R(x)}{D(x)}$$ where $\deg(R) < \deg(D)$. Then decompose the proper fraction $\dfrac{R(x)}{D(x)}$ as usual.
Shortcut for checking — the cover-up (Heaviside) 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. It is a fast mental shortcut, but it compresses the working — in written CIE solutions, show the full identity and your substitutions so the method is complete and rigorous. Use cover-up mainly to check answers quickly.
Activity 1 Matching Puzzle — Decomposition Form Race

Group work (5 min): In pairs, match each fraction to its correct decomposition form. Click a fraction card, then click the matching form card. Race to get all 4 correct!

Score: 0 / 4 ⏱ 0s
Fractions
$$\dfrac{2x^2+5x-1}{(x-1)(x+1)(x+2)}$$
$$\dfrac{4}{(1-3x)(1-x)^2}$$
$$\dfrac{6x^2+22x+18}{(x+1)(x+2)(x+3)}$$
$$\dfrac{3x^2+3x+2}{(x-1)(x+1)^2}$$
Decomposition Forms
$$\dfrac{A}{x+1} + \dfrac{B}{x+2} + \dfrac{C}{x+3}$$
$$\dfrac{A}{1-3x} + \dfrac{B}{1-x} + \dfrac{C}{(1-x)^2}$$
$$\dfrac{A}{x-1} + \dfrac{B}{x+1} + \dfrac{C}{x+2}$$
$$\dfrac{A}{x-1} + \dfrac{B}{x+1} + \dfrac{C}{(x+1)^2}$$
🧠 Think-First

If $\dfrac{2x^2+5x-1}{(x-1)(x+1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+1} + \dfrac{C}{x+2}$, what happens when you multiply both sides by the full denominator $(x-1)(x+1)(x+2)$ and then set $x = 1$? Think first.

Activity 2 Guided Discovery — Build the Identity, Find the Constants

Think-Pair-Share (5 min): Follow the 4-step interactive guide to discover how multiplying through by the full denominator — then substituting a well-chosen root — isolates each constant. After isolating one constant, discuss with your partner how to find the remaining constants.

1 Multiply Through
→
2 Choose Root
→
3 Substitute
→
4 Read Constant
Target: $$\dfrac{2x^2+5x-1}{(x-1)(x+1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+1} + \dfrac{C}{x+2}$$
Step 1: Clear all the fractions in one move — multiply both sides of the equation by the full denominator. Which multiplier achieves this?
✏️Worked Examples
Exercise 7C Q13 Challenging

Write $\dfrac{6x^2 + 22x + 18}{(x + 1)(x + 2)(x + 3)}$ as a sum of partial fractions.

Hint

Three distinct linear factors, so use $\dfrac{A}{x+1} + \dfrac{B}{x+2} + \dfrac{C}{x+3}$. Multiply through by the full denominator to form an identity, then substitute the roots $x = -1$, $-2$, $-3$.

Set up the form

$$\dfrac{6x^2 + 22x + 18}{(x+1)(x+2)(x+3)} = \dfrac{A}{x+1} + \dfrac{B}{x+2} + \dfrac{C}{x+3}$$

Multiply through by the full denominator

Multiplying both sides by $(x+1)(x+2)(x+3)$ clears every fraction and gives a polynomial identity:

$$6x^2 + 22x + 18 \equiv A(x+2)(x+3) + B(x+1)(x+3) + C(x+1)(x+2)$$

Substitute the roots of the factors

Find $A$: set $x = -1$ (the root of $x+1$) — the $B$ and $C$ terms vanish:

$$6(-1)^2 + 22(-1) + 18 = A(-1+2)(-1+3) \;\Rightarrow\; 2 = 2A \;\Rightarrow\; A = 1$$

Find $B$: set $x = -2$ (the root of $x+2$):

$$6(-2)^2 + 22(-2) + 18 = B(-2+1)(-2+3) \;\Rightarrow\; -2 = -B \;\Rightarrow\; B = 2$$

Find $C$: set $x = -3$ (the root of $x+3$):

$$6(-3)^2 + 22(-3) + 18 = C(-3+1)(-3+2) \;\Rightarrow\; 6 = 2C \;\Rightarrow\; C = 3$$

Shortcut check — the cover-up idea

Once fluent, you can compress each substitution: "cover up" the factor $(x+1)$ in the original denominator and substitute $x = -1$ into what remains — $A = \dfrac{6(1) + 22(-1) + 18}{(-1+2)(-1+3)} = 1$, as above. Useful for checking, but in written CIE solutions show the full identity and your substitutions.

Check

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

Numerator $= (x^2+5x+6) + 2(x^2+4x+3) + 3(x^2+3x+2) = 6x^2 + 22x + 18$. ✓

Final Answer

$$\dfrac{6x^2 + 22x + 18}{(x+1)(x+2)(x+3)} = \dfrac{1}{x+1} + \dfrac{2}{x+2} + \dfrac{3}{x+3}$$

Exercise 7D Q1(i) Advanced

Express $\dfrac{4}{(1 - 3x)(1 - x)^2}$ as a sum of partial fractions.

Hint

The factor $(1-x)^2$ is repeated, so it needs two terms: $\dfrac{B}{1-x} + \dfrac{C}{(1-x)^2}$. Multiply through by the full denominator to form an identity, substitute $x = \dfrac{1}{3}$ and $x = 1$ to find $A$ and $C$, then a convenient value (e.g. $x = 0$) for $B$.

Method 1 — Multiply through by the full denominator
Set up the form

$$\dfrac{4}{(1-3x)(1-x)^2} = \dfrac{A}{1-3x} + \dfrac{B}{1-x} + \dfrac{C}{(1-x)^2}$$

Multiply both sides by $(1-3x)(1-x)^2$ to clear all fractions:

$$4 \equiv A(1-x)^2 + B(1-3x)(1-x) + C(1-3x)$$

Find $A$: set $x = \dfrac{1}{3}$ (the root of $1-3x$) — the $B$ and $C$ terms vanish:

$$4 = A\left(1 - \dfrac{1}{3}\right)^2 = A\left(\dfrac{2}{3}\right)^2 = \dfrac{4A}{9} \;\Rightarrow\; A = 9$$

Find $C$: set $x = 1$ (the root of $1-x$):

$$4 = C(1 - 3) = -2C \;\Rightarrow\; C = -2$$

Find $B$: substitute $x = 0$ (a convenient value) into the identity:

$$4 = A(1) + B(1)(1) + C(1) = 9 + B - 2 \;\Rightarrow\; B = -3$$

Method 2 — Equating coefficients

Multiply both sides by $(1-3x)(1-x)^2$:

$$4 = A(1-x)^2 + B(1-3x)(1-x) + C(1-3x)$$

Expand:

$$\begin{aligned} 4 &= A(1 - 2x + x^2) + B(1 - 4x + 3x^2) + C(1 - 3x) \\ &= (A + 3B)x^2 + (-2A - 4B - 3C)x + (A + B + C) \end{aligned}$$

Equate coefficients of corresponding powers of $x$:

$$\begin{cases} x^2: & A + 3B = 0 \\ x^1: & -2A - 4B - 3C = 0 \\ x^0: & A + B + C = 4 \end{cases}$$

From $x^2$: $A = -3B$. Substitute into $x^0$: $-3B + B + C = 4 \;\Rightarrow\; C = 4 + 2B$.

Substitute into $x^1$: $-2(-3B) - 4B - 3(4+2B) = 0 \;\Rightarrow\; 6B - 4B - 12 - 6B = 0 \;\Rightarrow\; -4B = 12 \;\Rightarrow\; B = -3$.

Then $A = -3(-3) = 9$,   $C = 4 + 2(-3) = -2$.

Check

At $x = -1$: LHS $= \dfrac{4}{(4)(4)} = \dfrac{1}{4}$.   RHS $= \dfrac{9}{4} - \dfrac{3}{2} - \dfrac{2}{4} = \dfrac{9-6-2}{4} = \dfrac{1}{4}$. ✓

Final Answer

$$\dfrac{4}{(1-3x)(1-x)^2} = \dfrac{9}{1-3x} - \dfrac{3}{1-x} - \dfrac{2}{(1-x)^2}$$

Exercise 8E Q5 Challenging

Let $\text{f}(x) \equiv \dfrac{x^2 + 3x + 3}{(x + 1)(x + 3)}$. Express $\text{f}(x)$ in partial fractions.

The numerator and denominator both have degree 2. Expand the denominator, then perform polynomial division to separate the whole part from the proper fraction.

Step 1: Polynomial long division

Expand the denominator: $(x+1)(x+3) = x^2 + 4x + 3$.

Since $\deg(\text{numerator}) = \deg(\text{denominator}) = 2$, the fraction is improper. Perform long division:

100%
x² + 4x + 3 1 x² +3x +3 x² +4x +3 −x
(x² + 3x + 3) ÷ (x² + 4x + 3) = 1, remainder −x

So $\text{f}(x) = 1 + \dfrac{-x}{(x+1)(x+3)}$.

Decompose the proper fraction

$$\dfrac{-x}{(x+1)(x+3)} = \dfrac{A}{x+1} + \dfrac{B}{x+3}$$

Multiply through and substitute the roots

Multiplying both sides by $(x+1)(x+3)$ gives the identity $-x \equiv A(x+3) + B(x+1)$.

Find $A$: set $x = -1$ (the root of $x+1$):   $1 = 2A \;\Rightarrow\; A = \dfrac{1}{2}$.

Find $B$: set $x = -3$ (the root of $x+3$):   $3 = -2B \;\Rightarrow\; B = -\dfrac{3}{2}$.

Check

At $x = 0$: LHS $= \dfrac{3}{3} = 1$.   RHS $= 1 + \dfrac{1/2}{1} - \dfrac{3/2}{3} = 1 + \dfrac{1}{2} - \dfrac{1}{2} = 1$. ✓

Final Answer

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

🎯 Partial Fractions Concept Quiz — Quick Check

Test your understanding of all four lesson objectives. Each question targets one objective — try each before moving on!

Score: 0 / 5
Q1 (Objective 1) — Identifying the decomposition form

Which is the correct form for decomposing $\dfrac{P(x)}{(x-2)(x+1)^2}$?

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

A repeated factor $(x+1)^2$ needs two partial fractions: one with denominator $(x+1)$ and one with $(x+1)^2$. The form $\dfrac{A}{x-2} + \dfrac{B}{x+1} + \dfrac{C}{(x+1)^2}$ covers all factors of the denominator.

Q2 (Objective 2) — Finding $A$ from the identity

Multiplying $\dfrac{x+5}{(x-1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+2}$ through by $(x-1)(x+2)$ gives the identity $x + 5 \equiv A(x+2) + B(x-1)$. Which value of $x$ finds $A$ in one step?

$x = -2$
$x = 1$
$x = -1$
$x = 0$

Substituting the root of the factor $(x-1)$ — that is, $x = 1$ — makes the $B(x-1)$ term vanish: $1 + 5 = A(1+2)$, so $A = 2$. In general, substitute the root of each linear factor into the identity to isolate its constant. (The cover-up shortcut is a compressed mental version of exactly this substitution.)

Q3 (Objective 3) — Factor $(ax+b)$ with $a \neq 1$

In $\dfrac{3x}{(2x-1)(x+1)} = \dfrac{A}{2x-1} + \dfrac{B}{x+1}$, multiplying through gives the identity $3x \equiv A(x+1) + B(2x-1)$. What value of $x$ should you substitute to find $A$?

$x = 2$
$x = 1$
$x = \dfrac{1}{2}$
$x = -\dfrac{1}{2}$

The factor $(2x-1)$ has root $x = \dfrac{1}{2}$ (solving $2x - 1 = 0$). Substituting $x = \dfrac{1}{2}$ into the identity makes the $B(2x-1)$ term vanish, leaving $\dfrac{3}{2} = A \cdot \dfrac{3}{2}$, so $A = 1$. When the factor is $(ax+b)$ with $a \neq 1$, the root is $x = -\dfrac{b}{a}$, not $x = -b$.

Q4 (Objective 4) — Repeated factor $(cx+d)^2$

In $\dfrac{P(x)}{(x-1)(x+2)^2} = \dfrac{A}{x-1} + \dfrac{B}{x+2} + \dfrac{C}{(x+2)^2}$, how do you find $B$?

Substitute $x = 1$ into the identity
Substitute $x = -2$ into the identity
Set $B = C$
Once $A$ and $C$ are known, substitute a convenient value (e.g. $x=0$) into the identity and solve

Multiplying through gives $P(x) \equiv A(x+2)^2 + B(x-1)(x+2) + C(x-1)$. Substituting $x = 1$ isolates $A$; substituting $x = -2$ isolates $C$. But no single root isolates $B$ — its term $(x-1)(x+2)$ shares the factor $(x+2)$ with the $C$ term's $(x+2)^2$. So once $A$ and $C$ are known, substitute any other convenient value (e.g. $x = 0$) into the identity and solve for $B$.

Q5 (Objective 5) — Improper fractions

What must you do first when decomposing $\dfrac{x^2 + 3x + 3}{(x+1)(x+3)}$ into partial fractions?

Multiply through by the denominator and solve for the constants directly
Perform polynomial division because $\deg(\text{numerator}) = \deg(\text{denominator})$
Factor the numerator first
Cross-multiply and equate coefficients

The numerator has degree 2 and the denominator expands to degree 2, so the fraction is improper. You must perform polynomial division first to obtain $Q(x) + \dfrac{R(x)}{D(x)}$ where $\deg(R) < \deg(D)$. Only then can you decompose the proper fraction.

🎉
You scored 0 / 5
✏️ More Practice
Exercise 7C Q4 Standard

Write $\dfrac{x + 5}{(x - 1)(x + 2)}$ as a sum of partial fractions.

Solution

$$\dfrac{x+5}{(x-1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+2}$$

Multiplying through by $(x-1)(x+2)$ gives the identity $x + 5 \equiv A(x+2) + B(x-1)$.

Find $A$: set $x = 1$ (root of $x-1$):   $6 = 3A \;\Rightarrow\; A = 2$.

Find $B$: set $x = -2$ (root of $x+2$):   $3 = -3B \;\Rightarrow\; B = -1$.

Check

$\dfrac{2}{x-1} - \dfrac{1}{x+2} = \dfrac{2(x+2)-(x-1)}{(x-1)(x+2)} = \dfrac{x+5}{(x-1)(x+2)}$. ✓

Final Answer

$$\dfrac{x+5}{(x-1)(x+2)} = \dfrac{2}{x-1} - \dfrac{1}{x+2}$$

Exercise 7D Q1(ix) Challenging

Express $\dfrac{4x^2 - 3}{x(2x - 1)^2}$ as a sum of partial fractions.

Hint 1

The denominator has a distinct factor $x$ and a repeated factor $(2x-1)^2$. The form is $\dfrac{A}{x} + \dfrac{B}{2x-1} + \dfrac{C}{(2x-1)^2}$. Multiply through by the full denominator to form an identity, substitute the roots $x = 0$ and $x = \dfrac{1}{2}$ to find $A$ and $C$, then a convenient value (e.g. $x = 1$) for $B$.

Solution

$$\dfrac{4x^2-3}{x(2x-1)^2} = \dfrac{A}{x} + \dfrac{B}{2x-1} + \dfrac{C}{(2x-1)^2}$$

Multiplying through by $x(2x-1)^2$ gives the identity $4x^2 - 3 \equiv A(2x-1)^2 + Bx(2x-1) + Cx$.

Find $A$: set $x = 0$ (root of $x$):   $-3 = A(-1)^2 \;\Rightarrow\; A = -3$.

Find $C$: set $x = \dfrac{1}{2}$ (root of $2x-1$):   $1 - 3 = \dfrac{C}{2} \;\Rightarrow\; C = -4$.

Find $B$: substitute $x = 1$ into the identity:

$$4 - 3 = A(1) + B(1)(1) + C(1) \;\Rightarrow\; 1 = -3 + B - 4 \;\Rightarrow\; B = 8$$

Check

At $x = 2$: LHS $= \dfrac{16-3}{2 \cdot 9} = \dfrac{13}{18}$.   RHS $= \dfrac{-3}{2} + \dfrac{8}{3} - \dfrac{4}{9} = \dfrac{-27+48-8}{18} = \dfrac{13}{18}$. ✓

Final Answer

$$\dfrac{4x^2-3}{x(2x-1)^2} = -\dfrac{3}{x} + \dfrac{8}{2x-1} - \dfrac{4}{(2x-1)^2}$$

Exercise 7C Q14 Advanced

Write $\dfrac{4x^2 - 25x - 3}{(2x + 1)(x - 1)(x - 3)}$ as a sum of partial fractions.

Hint 1

Three distinct linear factors including $(2x+1)$. The root of $(2x+1)$ is $x = -\frac{1}{2}$. Multiply through by the full denominator to form an identity, then substitute each root $x = -\frac{1}{2}$, $1$, $3$ — each substitution kills two terms at once.

Solution

$$\dfrac{4x^2-25x-3}{(2x+1)(x-1)(x-3)} = \dfrac{A}{2x+1} + \dfrac{B}{x-1} + \dfrac{C}{x-3}$$

Multiplying through by $(2x+1)(x-1)(x-3)$ gives the identity:

$$4x^2 - 25x - 3 \equiv A(x-1)(x-3) + B(2x+1)(x-3) + C(2x+1)(x-1)$$

Find $A$: set $x = -\dfrac{1}{2}$ (root of $2x+1$):

$$4\cdot\tfrac{1}{4} + \tfrac{25}{2} - 3 = A\left(-\tfrac{3}{2}\right)\left(-\tfrac{7}{2}\right) \;\Rightarrow\; \tfrac{21}{2} = \tfrac{21}{4}A \;\Rightarrow\; A = 2$$

Find $B$: set $x = 1$ (root of $x-1$):   $4 - 25 - 3 = B(3)(-2) \;\Rightarrow\; -24 = -6B \;\Rightarrow\; B = 4$.

Find $C$: set $x = 3$ (root of $x-3$):   $36 - 75 - 3 = C(7)(2) \;\Rightarrow\; -42 = 14C \;\Rightarrow\; C = -3$.

Check

Numerator of RHS $= 2(x-1)(x-3) + 4(2x+1)(x-3) - 3(2x+1)(x-1)$

$= (2x^2-8x+6) + (8x^2-20x-12) - (6x^2-3x-3) = 4x^2 - 25x - 3$. ✓

Final Answer

$$\dfrac{4x^2-25x-3}{(2x+1)(x-1)(x-3)} = \dfrac{2}{2x+1} + \dfrac{4}{x-1} - \dfrac{3}{x-3}$$

Exercise 8E Q4(i) Advanced

Find the values of the constants $A$, $B$, $C$ and $D$ such that $$\dfrac{2x^3 - 1}{x^2(2x - 1)} \equiv A + \dfrac{B}{x} + \dfrac{C}{x^2} + \dfrac{D}{2x - 1}$$

Hint

The numerator and denominator both have degree 3. Perform polynomial division first to find $A$, then decompose the remaining proper fraction for $B$, $C$, and $D$.

Divide first

Expand denominator: $x^2(2x-1) = 2x^3 - x^2$.

Since $\deg(\text{numerator}) = \deg(\text{denominator}) = 3$, perform long division:

100%
2x³ − x² 1 2x³ +0x² +0x −1 2x³ −x² x² −1
(2x³ − 1) ÷ (2x³ − x²) = 1, remainder x² − 1

So $\dfrac{2x^3 - 1}{2x^3 - x^2} = 1 + \dfrac{x^2 - 1}{2x^3 - x^2}$.

Therefore $A = 1$ and the proper fraction is $\dfrac{x^2 - 1}{x^2(2x-1)}$.

Decompose the proper fraction

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

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

$$x^2 - 1 = Bx(2x-1) + C(2x-1) + Dx^2$$

Find $C$ and $D$ by substituting the roots

Find $C$: set $x = 0$ (root of $x$) in the identity $x^2 - 1 \equiv Bx(2x-1) + C(2x-1) + Dx^2$:

$$-1 = C(-1) \;\Rightarrow\; C = 1$$

Find $D$: set $x = \dfrac{1}{2}$ (root of $2x-1$):

$$\tfrac{1}{4} - 1 = D \cdot \tfrac{1}{4} \;\Rightarrow\; -\tfrac{3}{4} = \tfrac{D}{4} \;\Rightarrow\; D = -3$$

Find $B$ by substitution

Substitute $x = 1$ into the identity $x^2-1 = Bx(2x-1) + C(2x-1) + Dx^2$:

$$0 = B(1)(1) + 1(1) + (-3)(1) \;\Rightarrow\; 0 = B + 1 - 3 \;\Rightarrow\; B = 2$$

Final Answer

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

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