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Section B — Chapters 23–39

Solving Quadratic Equations by Factorisation

Junior Cycle — 3rd Year

  • ✓By the end of this lesson students will be able to identify and factorise algebraic expressions using the highest common factor (HCF).
  • ✓By the end of this lesson students will be able to identify and factorise expressions that are the difference of two squares.
  • ✓By the end of this lesson students will be able to factorise quadratic trinomials of the form x² + bx + c and ax² + bx + c.
  • ✓By the end of this lesson students will be able to solve quadratic equations by factorisation using the Null Factor Law.
  • ✓By the end of this lesson students will be able to formulate and solve quadratic equations from word problems.

Key concepts

Quadratic Equation

An equation of the form ax² + bx + c = 0, where a, b, and c are real numbers and a ≠ 0. The highest power of the variable is 2.

ax² + bx + c = 0
Factorisation

The process of writing an algebraic expression as a product of its factors.

Highest Common Factor (HCF)

The largest factor that divides two or more terms. Factorising by HCF involves taking out the HCF from all terms in an expression.

Difference of Two Squares (DOTS)

An expression of the form a² - b², which factorises to (a - b)(a + b).

a² - b² = (a - b)(a + b)
Quadratic Trinomial

An algebraic expression with three terms, where the highest power of the variable is 2, e.g., x² + bx + c or ax² + bx + c.

Null Factor Law (Zero Product Property)

If the product of two or more factors is zero, then at least one of the factors must be zero. If A × B = 0, then A = 0 or B = 0 (or both).

If A × B = 0, then A = 0 or B = 0

Key facts to remember

  • 1A quadratic equation has the form ax² + bx + c = 0, where a ≠ 0.
  • 2To solve a quadratic equation by factorisation, it must first be set equal to zero.
  • 3The Null Factor Law states that if A × B = 0, then A = 0 or B = 0.
  • 4Common factorisation methods include Highest Common Factor (HCF), Difference of Two Squares (DOTS), and Quadratic Trinomials.
  • 5Always check your solutions by substituting them back into the original equation.
  • 6A quadratic equation can have two, one, or no real solutions.

Worked examples

Example 1

Factorise 6x² + 9x.

IIdentify the HCF of 6x² and 9x. The HCF of 6 and 9 is 3. The HCF of x² and x is x. So, the HCF is 3x.
IIDivide each term by the HCF: 6x²/3x = 2x and 9x/3x = 3.
IIIWrite the expression as the product of the HCF and the remaining terms: 3x(2x + 3).

Answer

3x(2x + 3)

Always check your factorisation by expanding the brackets to ensure you get the original expression.

Example 2

Factorise x² - 49.

IRecognise the expression as the difference of two squares: x² is a square and 49 is 7².
IIApply the formula a² - b² = (a - b)(a + b), where a = x and b = 7.

Answer

(x - 7)(x + 7)

Remember that both terms must be perfect squares and separated by a minus sign.

Example 3

Factorise x² + 5x + 6.

ILook for two numbers that multiply to give the constant term (6) and add to give the coefficient of x (5).
IIThe numbers are 2 and 3 (since 2 × 3 = 6 and 2 + 3 = 5).
IIIWrite the trinomial as a product of two binomials: (x + 2)(x + 3).

Answer

(x + 2)(x + 3)

For trinomials of the form ax² + bx + c where a ≠ 1, more steps are involved, often using grouping or trial and error.

Example 4

Solve x² - 2x - 15 = 0.

IEnsure the equation is in the form ax² + bx + c = 0. (It is).
IIFactorise the quadratic trinomial: Find two numbers that multiply to -15 and add to -2. These are -5 and 3.
IIISo, (x - 5)(x + 3) = 0.
IVApply the Null Factor Law: x - 5 = 0 or x + 3 = 0.
VSolve each linear equation: x = 5 or x = -3.

Answer

x = 5, x = -3

Always show both solutions, as quadratic equations typically have two distinct solutions.

Example 5

The area of a rectangular garden is 40 m². The length of the garden is 3 m more than its width. Find the dimensions of the garden.

ILet the width of the garden be w metres.
IIThen the length is (w + 3) metres.
IIIArea = length × width, so w(w + 3) = 40.
IVExpand the equation: w² + 3w = 40.
VRearrange into the standard quadratic form: w² + 3w - 40 = 0.
VIFactorise the quadratic: Find two numbers that multiply to -40 and add to 3. These are 8 and -5.
VIISo, (w + 8)(w - 5) = 0.
VIIIApply the Null Factor Law: w + 8 = 0 or w - 5 = 0.
9Solve for w: w = -8 or w = 5.
10Since width cannot be negative, w = 5 m.
11Calculate the length: length = w + 3 = 5 + 3 = 8 m.

Answer

The width is 5 m and the length is 8 m.

Always check if your solutions make sense in the context of the word problem. Dimensions like length or width cannot be negative.

Common mistakes

  • ✗Forgetting to set the equation to zero before factorising and applying the Null Factor Law (e.g., solving x(x+3)=10 by setting x=10 or x+3=10).
  • ✗Incorrectly factorising the quadratic expression, leading to incorrect solutions.
  • ✗Making arithmetic errors when finding factors or solving the resulting linear equations.
  • ✗Not rejecting negative or impractical solutions in word problems (e.g., negative length or time).
  • ✗Confusing factorising an expression (writing it as a product of factors) with solving an equation (finding the values of the variable).

Exam tips

  • ★Always show all steps of your factorisation and solution process clearly, as marks are awarded for method.
  • ★If you are unsure of your factorisation, expand your factors to check if you get the original expression.
  • ★For word problems, define your variables clearly and ensure your final answer makes sense in the context of the problem.
  • ★Practise all types of factorisation regularly to build speed and accuracy.

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