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Factorisation - Method of Common Factors

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 two or more expressions, called its factors. Visually, if an expression represents the area of a rectangle, factorisation is the process of finding the lengths of its sides.

The Method of Common Factors involves identifying the Highest Common Factor (HCF) of all the terms in an expression and then dividing each term by this HCF. Think of this as the 'reverse distributive law' where you move from ab+acab + ac to a(b+c)a(b + c).

To find the HCF of monomials, find the HCF of the numerical coefficients and multiply it by each common variable raised to its lowest power. For example, in the terms 8x3y28x^{3}y^{2} and 12x2y412x^{2}y^{4}, the numerical HCF is 44, and the variable HCF is x2y2x^{2}y^{2}, making the total HCF 4x2y24x^{2}y^{2}.

When factorising by grouping, terms of an expression are arranged into groups such that each group has a common factor. This is often used when there is no single factor common to all terms. For example, in ax+ay+bx+byax + ay + bx + by, the first two terms share aa and the last two share bb.

The sign of the common factor is crucial. If you extract a negative common factor, you must change the signs of all terms inside the brackets. For instance, 5x10-5x - 10 becomes 5(x+2)-5(x + 2). Visually, this ensures that if you were to expand the brackets again, you would return to the original expression.

A common binomial factor can also be extracted. If an expression looks like x(a+b)+y(a+b)x(a + b) + y(a + b), the entire bracket (a+b)(a + b) is treated as a single common factor, resulting in (a+b)(x+y)(a + b)(x + y). This looks like two rectangular blocks sharing a common side length (a+b)(a + b) placed side by side.

📐Formulae

ab+ac=a(b+c)ab + ac = a(b + c) (Distributive Law in reverse)

HCF(xm,xn)=xmin(m,n)HCF(x^{m}, x^{n}) = x^{min(m,n)}

ax+ay+bx+by=a(x+y)+b(x+y)=(x+y)(a+b)ax + ay + bx + by = a(x + y) + b(x + y) = (x + y)(a + b) (Grouping method)

a(xy)=a(yx)a(x - y) = -a(y - x)

💡Examples

Problem 1:

Factorise the expression: 15x2y20xy215x^{2}y - 20xy^{2}

Solution:

Step 1: Find the HCF of the numerical coefficients 1515 and 2020. The HCF is 55.\Step 2: Find the HCF of the literal (variable) parts. For x2x^{2} and xx, the HCF is xx. For yy and y2y^{2}, the HCF is yy.\Step 3: Combine them to get the overall HCF, which is 5xy5xy.\Step 4: Divide each term of the expression by the HCF: \15x2y5xy=3x\frac{15x^{2}y}{5xy} = 3x and 20xy25xy=4y\frac{-20xy^{2}}{5xy} = -4y.\Step 5: Write the expression as the product of the HCF and the remaining terms: 5xy(3x4y)5xy(3x - 4y).

Explanation:

This solution uses the basic method of identifying the HCF of coefficients and variables separately before combining them to factor out the expression.

Problem 2:

Factorise by grouping: z7+7xyxyzz - 7 + 7xy - xyz

Solution:

Step 1: Rearrange the terms if necessary to find common factors. Let's group the first two and last two: (z7)+(7xyxyz)(z - 7) + (7xy - xyz).\Step 2: In the first group (z7)(z - 7), the common factor is 11, so it remains 1(z7)1(z - 7).\Step 3: In the second group (7xyxyz)(7xy - xyz), the common factor is xyxy. Taking it out gives xy(7z)xy(7 - z).\Step 4: Notice that (z7)(z - 7) and (7z)(7 - z) are opposites. Rewrite (7z)(7 - z) as (z7)-(z - 7).\Step 5: The expression becomes 1(z7)xy(z7)1(z - 7) - xy(z - 7).\Step 6: Now, (z7)(z - 7) is a common binomial factor. Take it out to get (z7)(1xy)(z - 7)(1 - xy).

Explanation:

This example demonstrates how to group terms and handle sign changes when the binomial factors are additive inverses of each other.