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Factorisation - Factorisation using Identities

Grade 8ICSE

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

🔑Concepts

Factorisation is the process of expressing an algebraic expression as a product of its simplest factors. Think of it as the reverse of multiplication; if you have a product, you are looking for the original 'dimensions' that created that area.

The identity a2b2=(ab)(a+b)a^2 - b^2 = (a - b)(a + b) is used to factorise the 'Difference of Two Squares'. Visually, imagine a large square of area a2a^2 with a smaller square of area b2b^2 removed from its corner. The remaining area can be cut and rearranged into a single rectangle with dimensions (ab)(a-b) and (a+b)(a+b).

A Perfect Square Trinomial follows the pattern a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a + b)^2. This represents the area of a large square with side length (a+b)(a+b). Geometrically, this large square is made of one square of area a2a^2, another square of area b2b^2, and two identical rectangles each having an area of abab.

The identity a22ab+b2=(ab)2a^2 - 2ab + b^2 = (a - b)^2 is used when the middle term of a trinomial is negative. To identify this, check if the first and last terms are positive perfect squares and if the middle term is exactly twice the product of their square roots.

Always check for a Greatest Common Factor (GCF) before applying identities. For example, in 3x2273x^2 - 27, you should first factor out 33 to get 3(x29)3(x^2 - 9), which then allows you to use the difference of squares identity on the expression inside the bracket.

Identification of terms is key: Always rewrite the given terms as perfect squares before applying the identity. For instance, rewrite 16x216x^2 as (4x)2(4x)^2 and 9y29y^2 as (3y)2(3y)^2 to clearly see that a=4xa = 4x and b=3yb = 3y.

Multi-step Factorisation: Sometimes an expression requires using identities more than once. For example, x4y4x^4 - y^4 is initially (x2)2(y2)2=(x2y2)(x2+y2)(x^2)^2 - (y^2)^2 = (x^2 - y^2)(x^2 + y^2). The first factor (x2y2)(x^2 - y^2) can be factorised further into (xy)(x+y)(x - y)(x + y).

📐Formulae

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

(ab)2=a22ab+b2(a - b)^2 = a^2 - 2ab + b^2

a2b2=(a+b)(ab)a^2 - b^2 = (a + b)(a - b)

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

💡Examples

Problem 1:

Factorise: 25x249y225x^2 - 49y^2

Solution:

  1. Recognize the expression as a difference of two squares: a2b2a^2 - b^2. \ 2. Rewrite each term as a perfect square: (5x)2(7y)2(5x)^2 - (7y)^2. \ 3. Here, a=5xa = 5x and b=7yb = 7y. \ 4. Apply the identity a2b2=(a+b)(ab)a^2 - b^2 = (a + b)(a - b). \ 5. Result: (5x+7y)(5x7y)(5x + 7y)(5x - 7y).

Explanation:

Since the expression consists of two perfect squares separated by a minus sign, we use the Difference of Squares identity.

Problem 2:

Factorise: 9a2+30ab+25b29a^2 + 30ab + 25b^2

Solution:

  1. Check if the first and last terms are perfect squares: 9a2=(3a)29a^2 = (3a)^2 and 25b2=(5b)225b^2 = (5b)^2. \ 2. Identify a=3aa = 3a and b=5bb = 5b. \ 3. Verify the middle term: 2×(3a)×(5b)=30ab2 \times (3a) \times (5b) = 30ab. \ 4. Since the middle term matches, apply the identity a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a + b)^2. \ 5. Result: (3a+5b)2(3a + 5b)^2.

Explanation:

This is a perfect square trinomial where the middle term is positive, so we use the square of a sum identity.