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Algebra - Factorisation

Grade 9ICSE

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

🔑Concepts

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Definition of Factorisation: Factorisation is the process of expressing a given algebraic expression as a product of two or more expressions, which are called its factors. It is essentially the reverse of expansion. Visualise this as breaking down a large rectangular area into smaller rectangles whose side lengths are the factors.

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Factorisation by Taking Out Common Factors: This is the most basic method where we identify the Highest Common Factor (HCF) of all the terms in the expression and write it outside a bracket. For example, in 3x2+6x3x^2 + 6x, the HCF is 3x3x, so the expression becomes 3x(x+2)3x(x + 2). Visualise this as extracting a common 'block' from several different structures.

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Factorisation by Grouping: When an expression has four or more terms and no common factor exists for all of them, we group the terms into pairs that have a common factor. For example, in ax+ay+bx+byax + ay + bx + by, we group them as (ax+ay)+(bx+by)(ax + ay) + (bx + by), which becomes a(x+y)+b(x+y)a(x + y) + b(x + y), eventually leading to (a+b)(x+y)(a + b)(x + y).

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Trinomials (Splitting the Middle Term): For a quadratic trinomial of the form ax2+bx+cax^2 + bx + c, we find two numbers pp and qq such that p+q=bp + q = b and p×q=acp \times q = ac. The middle term bxbx is then replaced by px+qxpx + qx, and factorisation is completed by grouping. Think of this as adjusting the internal segments of a rectangle to find consistent side lengths.

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Difference of Two Squares: An expression in the form a2−b2a^2 - b^2 is factorised into (a+b)(a−b)(a + b)(a - b). Visually, this represents taking a large square of side aa and cutting out a smaller square of side bb from the corner; the remaining L-shaped area can be rearranged into a rectangle with dimensions (a+b)(a+b) and (a−b)(a-b).

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Sum and Difference of Two Cubes: These are specific identities used for third-degree polynomials. The sum a3+b3a^3 + b^3 factors into a linear part (a+b)(a+b) and a quadratic part (a2−ab+b2)(a^2 - ab + b^2), while the difference a3−b3a^3 - b^3 factors into (a−b)(a2+ab+b2)(a-b)(a^2 + ab + b^2). Notice the sign change in the middle term of the quadratic factor.

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Perfect Square Trinomials: Expressions like a2+2ab+b2a^2 + 2ab + b^2 and a2−2ab+b2a^2 - 2ab + b^2 are perfect squares of binomials (a+b)2(a+b)^2 and (a−b)2(a-b)^2 respectively. On a grid, these expressions would form a perfect square shape where the sides are of equal length (a+b)(a+b) or (a−b)(a-b).

📐Formulae

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

(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

a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2 - ab + b^2)

a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2)

x2+(p+q)x+pq=(x+p)(x+q)x^2 + (p + q)x + pq = (x + p)(x + q)

a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)

If a+b+c=0, then a3+b3+c3=3abc\text{If } a + b + c = 0, \text{ then } a^3 + b^3 + c^3 = 3abc

💡Examples

Problem 1:

Factorise the quadratic expression: 2x2+7x+62x^2 + 7x + 6

Solution:

Step 1: Identify a=2a = 2, b=7b = 7, and c=6c = 6. Calculate ac=2×6=12ac = 2 \times 6 = 12. Step 2: Find two numbers that multiply to 1212 and add to 77. These numbers are 33 and 44. Step 3: Split the middle term 7x7x into 4x+3x4x + 3x. 2x2+4x+3x+62x^2 + 4x + 3x + 6 Step 4: Factor by grouping. 2x(x+2)+3(x+2)2x(x + 2) + 3(x + 2) Step 5: Factor out the common binomial (x+2)(x + 2). (2x+3)(x+2)(2x + 3)(x + 2)

Explanation:

This problem uses the method of splitting the middle term. We converted a 3-term trinomial into a 4-term expression to enable factorisation by grouping.

Problem 2:

Factorise completely: 81x4−16y481x^4 - 16y^4

Solution:

Step 1: Express both terms as squares. (9x2)2−(4y2)2(9x^2)^2 - (4y^2)^2 Step 2: Apply the difference of two squares formula a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b). (9x2−4y2)(9x2+4y2)(9x^2 - 4y^2)(9x^2 + 4y^2) Step 3: Observe that the first factor (9x2−4y2)(9x^2 - 4y^2) is again a difference of two squares. ((3x)2−(2y)2)(9x2+4y2)((3x)^2 - (2y)^2)(9x^2 + 4y^2) Step 4: Factorise (3x)2−(2y)2(3x)^2 - (2y)^2 further. (3x−2y)(3x+2y)(9x2+4y2)(3x - 2y)(3x + 2y)(9x^2 + 4y^2)

Explanation:

This problem demonstrates repeated application of the difference of two squares identity. Note that the sum of two squares (9x2+4y2)(9x^2 + 4y^2) cannot be factorised further in real numbers.