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Algebra - Expansion and Factorisation

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Expansion: The process of removing brackets by multiplying terms using the distributive law.

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Factorisation: The inverse process of expansion, where an expression is written as a product of its factors.

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Common Factors: Identifying the Highest Common Factor (HCF) among all terms to pull outside a bracket.

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Expanding Binomials: Often referred to as the FOIL method (First, Outer, Inner, Last).

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Quadratic Factorisation: Factoring trinomials of the form x² + bx + c by finding two numbers that multiply to 'c' and add up to 'b'.

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Difference of Two Squares: A specific pattern for factorising expressions where two perfect squares are subtracted.

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Grouping: A method used for factorising expressions with four terms by grouping them into pairs with common factors.

📐Formulae

a(b+c)=ab+aca(b + c) = ab + ac

(x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a + b)x + ab

(a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

(a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2

a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b)

💡Examples

Problem 1:

Expand and simplify: (2x−3)(x+5)(2x - 3)(x + 5)

Solution:

2x2+7x−152x^2 + 7x - 15

Explanation:

Use the FOIL method: multiply the First terms (2x×x=2x22x \times x = 2x^2), Outer terms (2x×5=10x2x \times 5 = 10x), Inner terms (−3×x=−3x-3 \times x = -3x), and Last terms (−3×5=−15-3 \times 5 = -15). Combine like terms: 10x−3x=7x10x - 3x = 7x.

Problem 2:

Factorise completely: 12x2y−18xy212x^2y - 18xy^2

Solution:

6xy(2x−3y)6xy(2x - 3y)

Explanation:

Identify the Highest Common Factor (HCF) of the coefficients 12 and 18, which is 6. Identify the lowest power of xx and yy present in both terms (xx and yy). Divide each term by 6xy6xy to find the expression inside the bracket.

Problem 3:

Factorise: x2−5x−14x^2 - 5x - 14

Solution:

(x−7)(x+2)(x - 7)(x + 2)

Explanation:

Find two numbers that multiply to −14-14 and add up to −5-5. The factors of −14-14 are (−7,2)(-7, 2) and (7,−2)(7, -2). Since −7+2=−5-7 + 2 = -5, these are the correct numbers to place in the binomials.

Problem 4:

Factorise: 16m2−4916m^2 - 49

Solution:

(4m−7)(4m+7)(4m - 7)(4m + 7)

Explanation:

This is a Difference of Two Squares (a2−b2a^2 - b^2). Here, a=16m2=4ma = \sqrt{16m^2} = 4m and b=49=7b = \sqrt{49} = 7. Use the formula (a−b)(a+b)(a - b)(a + b).