Algebra: Expressions and Equations

What you'll learn

This revision guide covers algebraic expressions and equations as tested in the CXC CSEC Additional Mathematics syllabus. You will master techniques for manipulating algebraic fractions, expanding and factorising expressions, solving polynomial and simultaneous equations, and working with indices and surds. These foundational skills underpin success across all Additional Mathematics topics.

Key terms and definitions

Polynomial — an algebraic expression consisting of terms with non-negative integer powers of variables, such as 3x³ - 5x² + 2x - 7.

Factor theorem — states that (x - a) is a factor of polynomial f(x) if and only if f(a) = 0.

Remainder theorem — states that when polynomial f(x) is divided by (x - a), the remainder is f(a).

Partial fractions — the decomposition of a rational algebraic fraction into a sum of simpler fractions with linear or quadratic denominators.

Conjugate surd — a binomial expression that differs only in the sign between two terms involving surds, such as (a + √b) and (a - √b).

Discriminant — the expression b² - 4ac from the quadratic formula, which determines the nature of roots of a quadratic equation.

Simultaneous equations — two or more equations with multiple unknowns that must be solved together to find values satisfying all equations.

Rationalising the denominator — the process of eliminating surds from the denominator of a fraction by multiplying by an appropriate form of 1.

Core concepts

Algebraic fractions and simplification

Algebraic fractions follow the same arithmetic rules as numerical fractions. Always factorise numerators and denominators completely before simplifying.

Addition and subtraction:

• Find the lowest common denominator (LCD)
• Express each fraction with the LCD
• Combine numerators, keeping the common denominator
• Simplify by factorising and cancelling common factors

Example: Simplify (2x)/(x² - 4) + 3/(x + 2)

First factorise: x² - 4 = (x + 2)(x - 2)

LCD = (x + 2)(x - 2)

= 2x/[(x + 2)(x - 2)] + 3(x - 2)/[(x + 2)(x - 2)]

= [2x + 3(x - 2)]/[(x + 2)(x - 2)]

= (2x + 3x - 6)/[(x + 2)(x - 2)]

= (5x - 6)/[(x + 2)(x - 2)]

Multiplication and division:

• Factorise all expressions first
• For division, multiply by the reciprocal
• Cancel common factors across numerators and denominators
• Multiply remaining factors

Partial fractions

Partial fractions decompose complex rational expressions into simpler components. This technique is essential for integration in calculus.

Type 1: Distinct linear factors

For f(x)/[(ax + b)(cx + d)], write as A/(ax + b) + B/(cx + d)

Method:

1. Express as sum of partial fractions with unknown constants
2. Multiply through by the common denominator
3. Substitute convenient values of x to find constants, or equate coefficients

Type 2: Repeated linear factors

For f(x)/(ax + b)², write as A/(ax + b) + B/(ax + b)²

Type 3: Improper fractions

When the numerator degree ≥ denominator degree, use polynomial division first to obtain:

f(x)/g(x) = quotient + (remainder)/g(x)

Then decompose the proper fraction remainder/g(x).

Caribbean context example: A shipping company's cost function C(n) = (5n² + 13n + 6)/(n + 2) dollars for n containers can be expressed in partial fraction form for analysis of marginal costs.

Indices and surds

Laws of indices:

• aᵐ × aⁿ = aᵐ⁺ⁿ
• aᵐ ÷ aⁿ = aᵐ⁻ⁿ
• (aᵐ)ⁿ = aᵐⁿ
• a⁰ = 1 (a ≠ 0)
• a⁻ⁿ = 1/aⁿ
• a^(m/n) = ⁿ√(aᵐ) = (ⁿ√a)ᵐ

Surd manipulation:

• √a × √b = √(ab)
• √a ÷ √b = √(a/b)
• (√a)² = a
• Simplify by extracting perfect square factors: √32 = √(16 × 2) = 4√2

Rationalising denominators:

For a single surd: multiply numerator and denominator by the surd.

Example: 3/√5 = (3√5)/(√5 × √5) = (3√5)/5

For binomial with surds: multiply by the conjugate surd.

Example: 1/(2 + √3) = (2 - √3)/[(2 + √3)(2 - √3)] = (2 - √3)/(4 - 3) = 2 - √3

Polynomial division and factor theorem

Polynomial long division: Used to divide polynomial f(x) by polynomial g(x), yielding quotient Q(x) and remainder R(x):

f(x) = g(x) · Q(x) + R(x)

The degree of R(x) is less than the degree of g(x).

Applying the remainder theorem: To find the remainder when f(x) is divided by (x - a), simply calculate f(a).

Applying the factor theorem:

• If f(a) = 0, then (x - a) is a factor of f(x)
• Use this to factorise polynomials and solve polynomial equations
• Test possible factors systematically, often starting with factors of the constant term

Example: Show that (x - 2) is a factor of f(x) = 2x³ - 3x² - 5x + 6

f(2) = 2(2)³ - 3(2)² - 5(2) + 6 = 16 - 12 - 10 + 6 = 0

Since f(2) = 0, (x - 2) is a factor by the factor theorem.

Quadratic equations and inequalities

Factorisation method: Express as ax² + bx + c = 0, factorise, and apply the null factor law.

Quadratic formula: For ax² + bx + c = 0:

x = [-b ± √(b² - 4ac)]/(2a)

Nature of roots (using discriminant Δ = b² - 4ac):

• Δ > 0: two distinct real roots
• Δ = 0: one repeated real root (equal roots)
• Δ < 0: no real roots (two complex conjugate roots)

Completing the square: Useful for finding turning points and solving equations.

ax² + bx + c = a[x + (b/2a)]² + c - b²/(4a)

Quadratic inequalities:

1. Solve the corresponding equation to find critical values
2. Sketch the parabola or use a sign table
3. Identify regions satisfying the inequality
4. Express solution in inequality notation or set notation

Caribbean context: A mango farmer's profit P(x) = -2x² + 40x - 150 dollars depends on x baskets sold. Solving P(x) ≥ 0 determines profitable production ranges.

Simultaneous equations

Linear simultaneous equations:

• Elimination method: multiply equations to create equal coefficients, then add or subtract
• Substitution method: solve one equation for a variable, substitute into the other

One linear, one non-linear:

1. Use the linear equation to express one variable in terms of the other
2. Substitute into the non-linear equation
3. Solve the resulting equation (often quadratic)
4. Substitute back to find the other variable
5. Check both solutions in both original equations

Two non-linear equations: May require strategic manipulation, substitution, or special techniques depending on form.

Worked examples

Example 1: Partial fractions

Question: Express (7x + 1)/[(x + 2)(x - 1)] in partial fractions.

Solution:

Let (7x + 1)/[(x + 2)(x - 1)] = A/(x + 2) + B/(x - 1)

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

7x + 1 = A(x - 1) + B(x + 2)

Method 1 - Substitution:

Let x = 1: 7(1) + 1 = A(0) + B(3) 8 = 3B B = 8/3

Let x = -2: 7(-2) + 1 = A(-3) + B(0) -13 = -3A A = 13/3

Answer: (7x + 1)/[(x + 2)(x - 1)] = (13/3)/(x + 2) + (8/3)/(x - 1)

or equivalently: 13/[3(x + 2)] + 8/[3(x - 1)]

Example 2: Factor theorem and polynomial factorisation

Question: The polynomial f(x) = 2x³ - 7x² - 17x + 10 has a factor (x - 5). Factorise f(x) completely and solve f(x) = 0.

Solution:

Since (x - 5) is given as a factor, divide f(x) by (x - 5):

Using polynomial division or synthetic division:

2x³ - 7x² - 17x + 10 = (x - 5)(2x² + 3x - 2)

Now factorise 2x² + 3x - 2:

2x² + 3x - 2 = (2x - 1)(x + 2)

Therefore: f(x) = (x - 5)(2x - 1)(x + 2)

Solving f(x) = 0:

(x - 5)(2x - 1)(x + 2) = 0

x = 5, x = 1/2, or x = -2

Answer: f(x) = (x - 5)(2x - 1)(x + 2); solutions are x = 5, x = 1/2, x = -2

Example 3: Simultaneous equations (one linear, one quadratic)

Question: Solve simultaneously: y = 2x - 1 ... (1) x² + y² = 13 ... (2)

Solution:

Substitute equation (1) into equation (2):

x² + (2x - 1)² = 13

x² + 4x² - 4x + 1 = 13

5x² - 4x + 1 = 13

5x² - 4x - 12 = 0

Using the quadratic formula:

x = [4 ± √(16 + 240)]/10 = [4 ± √256]/10 = [4 ± 16]/10

x = 20/10 = 2 or x = -12/10 = -6/5

When x = 2: y = 2(2) - 1 = 3

When x = -6/5: y = 2(-6/5) - 1 = -12/5 - 5/5 = -17/5

Answer: (2, 3) and (-6/5, -17/5)

Common mistakes and how to avoid them

• Cancelling terms instead of factors: You can only cancel common factors, not terms. For example, (x + 3)/(x + 5) cannot be simplified by cancelling x. Always factorise fully before cancelling.

• Errors with negative indices: Remember a⁻ⁿ = 1/aⁿ, not -aⁿ. When simplifying expressions like x⁻² ÷ x⁻⁵, apply index laws correctly: x⁻²⁻(⁻⁵) = x³, not x⁻⁷.

• Forgetting to multiply all terms when finding partial fractions: When you multiply through by the common denominator, ensure every term is multiplied. This is especially important with three or more partial fractions.

• Improper fractions in partial fractions: Always check if the numerator degree is greater than or equal to the denominator degree. If so, perform polynomial division first before decomposing.

• Sign errors when rationalising with conjugates: (a + √b)(a - √b) = a² - b, not a² + b. Be methodical when expanding brackets involving surds.

• Not checking all solutions in simultaneous equations: When solving one linear and one non-linear equation, you typically get two solution pairs. Verify both pairs satisfy both original equations.

Exam technique for "Algebra: Expressions and Equations"

• "Factorise completely" requires all factors reduced to irreducible form. For polynomials, this means expressing as a product of linear and irreducible quadratic factors. Show each factorisation step for method marks.

• "Show that" questions require complete working. For factor theorem questions, explicitly state f(a) = 0 and conclude that (x - a) is a factor. Don't skip logical steps even if the answer is given.

• Allocation of marks: Algebraic manipulation questions typically award 1 mark per significant step. In a 4-mark factorisation question, expect marks for initial factorisation, further factorisation, and final answer. Partial fractions questions reward the setup (1 mark), method to find constants (2 marks), and correct final form (1 mark).

• Time management: Multi-part algebra questions build in difficulty. Secure marks on early parts (factorisation, simplification) before attempting complex applications. If stuck on partial fractions, move on and return if time permits.

Quick revision summary

Master algebraic fractions by factorising before simplifying, always finding common denominators for addition. Apply the factor and remainder theorems systematically for polynomial problems. Decompose rational expressions into partial fractions using substitution or equating coefficients. Handle indices using the laws methodically and rationalise surds by multiplying by conjugates when necessary. Solve simultaneous equations by choosing elimination or substitution based on equation types, checking all solutions. Use the discriminant to analyse quadratic equations without solving.