Algebraic Expressions & Factorisation – Complete Chapter Guide

Algebra & Factorisation Grade 8-10 Guide

Algebraic Expressions & Factorisation

Comprehensive Mathematics Chapter Guide, Rules, Solved Examples & Self Practice

1. Definitions 2. Rules 3. Video Guide 4. 2 & 3 Terms 5. 4 Terms Grouping 6. Self Practice

Understanding Algebraic Expressions and Factorisation

An algebraic expression is formed by combining constants, variables, and mathematical operations (+, -, ×, ÷). Factorisation is the reverse process of expanding brackets—rewriting polynomial expressions as products of simpler linear or common factors using the Highest Common Factor (HCF).

1 Key Definitions

1. Algebraic Expression

Combination of variables (\(x, y\)), constants (\(2, 3\)), and operations (\(+, -\)).
Example: \(6x + 3\)

2. Factors

Numbers or variables multiplied together.
Example: In \(6x\), factors are \(2, 3,\) and \(x\).

3. Factorisation

Writing an expression as a product of its factors.
Example: \(6x + 3 = 3(2x + 1)\)

2 Rules for Taking Common Factors

A. For 2 Terms: Break down into prime factors, identify HCF common in both terms, and factor it out outside brackets.

B. For 3 Terms: Expand all three terms, identify factor present in all three terms simultaneously, and extract HCF.

C. For 4 Terms (Grouping): Group into pairs sharing common terms, factor each pair separately, then extract identical binomial brackets.

Factorisation of 4-Term Expressions (Short Video Guide):

Note:

प्लेयर कन्ट्रोलमा रहेको भोल्युम बटन थिचेर आवाज सुन्न र मिलाउन सक्नुहुन्छ।

4 Solved Questions (2 & 3 Terms)

Factorise: \(6x + 3\)

Solution:

\(= 6x + 3\)

\(= 2 \times \mathbf{3} \times x + \mathbf{3} \times 1\)

\(= \mathbf{3}(2x + 1)\) (Answer)

Rough Work:
• First Term: 6x = 2 × 3 × x
• Second Term: 3 = 3 × 1
• Highest Common Factor (HCF) = 3

Factorise: \(x^2 + 4x\)

Solution:

\(= x^2 + 4x\)

\(= \mathbf{x} \times x + 4 \times \mathbf{x}\)

\(= \mathbf{x}(x + 4)\) (Answer)

Rough Work:
• First Term: x² = x × x
• Second Term: 4x = 4 × x
• Common Variable = x

Factorise: \(12a + 3b\)

Solution:

\(= 12a + 3b\)

\(= 4 \times \mathbf{3} \times a + \mathbf{3} \times b\)

\(= \mathbf{3}(4a + b)\) (Answer)

Rough Work:
• First Term: 12a = 4 × 3 × a
• Second Term: 3b = 3 × b
• Common Factor = 3

Factorise: \(12p^2 + 6q^2\)

Solution:

\(= 12p^2 + 6q^2\)

\(= 2 \times \mathbf{6} \times p^2 + \mathbf{6} \times q^2\)

\(= \mathbf{6}(2p^2 + q^2)\) (Answer)

Rough Work:
• HCF of 12 and 6 is 6
• Term 1: 2 × 6 × p²
• Term 2: 6 × q²
• Common Factor = 6

Factorise: \(14xy + 7y\)

Solution:

\(= 14xy + 7y\)

\(= 2x \times \mathbf{7y} + 1 \times \mathbf{7y}\)

\(= \mathbf{7y}(2x + 1)\) (Answer)

Rough Work:
• Term 1: 2x × 7y
• Term 2: 1 × 7y
• Common Factor = 7y

Factorise: \(x + x^3\)

Solution:

\(= x + x^3\)

\(= \mathbf{x} \times 1 + \mathbf{x} \times x^2\)

\(= \mathbf{x}(1 + x^2)\) (Answer)

Rough Work:
• Term 1: x × 1
• Term 2: x × x²
• Common Factor = x

Factorise: \(12x^2 + xy + xz\)

Solution:

\(= 12x^2 + xy + xz\)

\(= 12x \times \mathbf{x} + y \times \mathbf{x} + z \times \mathbf{x}\)

\(= \mathbf{x}(12x + y + z)\) (Answer)

Rough Work:
• Term 1: 12x × x
• Term 2: y × x
• Term 3: z × x
• Common in all 3 terms = x

Factorise: \(x^3 + x^2 + x\)

Solution:

\(= x^3 + x^2 + x\)

\(= x^2 \times \mathbf{x} + x \times \mathbf{x} + 1 \times \mathbf{x}\)

\(= \mathbf{x}(x^2 + x + 1)\) (Answer)

Rough Work:
• Term 1: x² × x
• Term 2: x × x
• Term 3: 1 × x
• Smallest power of x is x¹ → Common Factor = x

Factorise: \(2x^2 – 2x^3 + 8x^4\)

Solution:

\(= 2x^2 – 2x^3 + 8x^4\)

\(= 1 \times \mathbf{2x^2} – x \times \mathbf{2x^2} + 4x^2 \times \mathbf{2x^2}\)

\(= \mathbf{2x^2}(1 – x + 4x^2)\) (Answer)

Rough Work:
• Numerical HCF of (2, 2, 8) = 2
• Variable HCF of (x², x³, x⁴) = x²
• Overall Common Factor = 2x²

5 Solved Questions (4 Terms Grouping)

Factorise: \(ax + by + ay + bx\)

Solution:

\(= ax + by + ay + bx\)

\(= (ax + ay) + (bx + by)\) (Rearranging terms)

\(= a\mathbf{(x + y)} + b\mathbf{(x + y)}\)

\(= \mathbf{(x + y)}(a + b)\) (Answer)

Rough Work:
• Group terms with ‘a’: ax + ay = a(x + y)
• Group terms with ‘b’: bx + by = b(x + y)
• Common Binomial Bracket = (x + y)

Factorise: \(2ab + a^2b – 2b – ab\)

Solution:

\(= 2ab + a^2b – 2b – ab\)

\(= (2ab – 2b) + (a^2b – ab)\)

\(= 2b\mathbf{(a – 1)} + ab\mathbf{(a – 1)}\)

\(= \mathbf{(a – 1)}(2b + ab)\)

\(= \mathbf{b(a – 1)(a + 2)}\) (Answer)

Rough Work:
• First pair: 2b(a – 1)
• Second pair: ab(a – 1)
• Factor out ‘b’ from (2b + ab) = b(2 + a)

Factorise: \(x^2y – xy + 2x^2y – 2xy\)

Solution:

\(= x^2y – xy + 2x^2y – 2xy\)

\(= (x^2y – xy) + (2x^2y – 2xy)\)

\(= xy\mathbf{(x – 1)} + 2xy\mathbf{(x – 1)}\)

\(= \mathbf{(x – 1)}(xy + 2xy)\)

\(= \mathbf{3xy(x – 1)}\) (Answer)

Rough Work:
• Alternative method: Combine like terms first:
x²y + 2x²y = 3x²y and -xy – 2xy = -3xy
3x²y – 3xy = 3xy(x – 1)

Factorise: \(x^2 + 3x + xy + 3y\)

Solution:

\(= x^2 + 3x + xy + 3y\)

\(= (x^2 + 3x) + (xy + 3y)\)

\(= x\mathbf{(x + 3)} + y\mathbf{(x + 3)}\)

\(= \mathbf{(x + 3)(x + y)}\) (Answer)

Rough Work:
• Group 1: x(x + 3)
• Group 2: y(x + 3)
• Common Binomial = (x + 3)

Factorise: \(2ab + 3a + 2b^2 + 3b\)

Solution:

\(= 2ab + 3a + 2b^2 + 3b\)

\(= (2ab + 2b^2) + (3a + 3b)\) (Rearranging)

\(= 2b\mathbf{(a + b)} + 3\mathbf{(a + b)}\)

\(= \mathbf{(a + b)(2b + 3)}\) (Answer)

Rough Work:
• Pair 2ab & 2b² → Common = 2b
• Pair 3a & 3b → Common = 3
• Common Binomial = (a + b)

Factorise: \(a – b + a^2 – ab\)

Solution:

\(= a – b + a^2 – ab\)

\(= (a – b) + (a^2 – ab)\)

\(= 1\mathbf{(a – b)} + a\mathbf{(a – b)}\)

\(= \mathbf{(a – b)(1 + a)}\) (Answer)

Rough Work:
• (a – b) = 1 × (a – b)
• (a² – ab) = a × (a – b)
• Common Binomial = (a – b)

Factorise: \(2a^2 + 5a – 6a – 15\)

Solution:

\(= 2a^2 + 5a – 6a – 15\)

\(= (2a^2 + 5a) – (6a + 15)\) (Note sign change inside)

\(= a\mathbf{(2a + 5)} – 3\mathbf{(2a + 5)}\)

\(= \mathbf{(2a + 5)(a – 3)}\) (Answer)

Rough Work:
• Factoring out -3 from (-6a – 15) turns inner signs positive: -3(2a + 5)

Factorise: \(2xa – x^2a + 2a – ax\)

Solution:

\(= 2xa – x^2a + 2a – ax\)

\(= (2xa + 2a) – (x^2a + ax)\)

\(= 2a\mathbf{(x + 1)} – ax\mathbf{(x + 1)}\)

\(= \mathbf{(x + 1)}(2a – ax)\)

\(= \mathbf{a(x + 1)(2 – x)}\) (Answer)

Rough Work:
• Grouping gives (x + 1)(2a – ax)
• Factor out ‘a’ from (2a – ax): a(2 – x)

Factorise: \(x^2y + 4xy – xy^2 – 4y^2\)

Solution:

\(= x^2y + 4xy – xy^2 – 4y^2\)

\(= (x^2y – xy^2) + (4xy – 4y^2)\)

\(= xy\mathbf{(x – y)} + 4y\mathbf{(x – y)}\)

\(= \mathbf{(x – y)}(xy + 4y)\)

\(= \mathbf{y(x – y)(x + 4)}\) (Answer)

Rough Work:
• Pair x²y & -xy² → xy(x – y)
• Pair 4xy & -4y² → 4y(x – y)
• Take out common ‘y’ from (xy + 4y)

Factorise: \(3x(x + y) + 3y(x + y)\)

Solution:

\(= 3x(x + y) + 3y(x + y)\)

\(= \mathbf{(x + y)}(3x + 3y)\)

\(= (x + y) \times 3(x + y)\)

\(= \mathbf{3(x + y)^2}\) (Answer)

Rough Work:
• Common binomial factor is (x + y)
• 3x + 3y = 3(x + y)
• (x + y)(x + y) = (x + y)²

Factorise: \(2x^2 + 3ax + 2ax + 3a^2\)

Solution:

\(= 2x^2 + 3ax + 2ax + 3a^2\)

\(= (2x^2 + 2ax) + (3ax + 3a^2)\)

\(= 2x\mathbf{(x + a)} + 3a\mathbf{(x + a)}\)

\(= \mathbf{(x + a)(2x + 3a)}\) (Answer)

Rough Work:
• Pair 2x² & 2ax → Common = 2x
• Pair 3ax & 3a² → Common = 3a
• Common Binomial = (x + a)

6 Self Practice Questions

Q1. Factorise: \(8x + 4\)

Show Answer & Hint

Common Factor = 4
\(= 4(2x + 1)\)

Q2. Factorise: \(y^2 + 5y\)

Show Answer & Hint

Common Factor = y
\(= y(y + 5)\)

Q3. Factorise: \(15p + 5q\)

Show Answer & Hint

Common Factor = 5
\(= 5(3p + q)\)

Q4. Factorise: \(18ab + 9b\)

Show Answer & Hint

Common Factor = 9b
\(= 9b(2a + 1)\)

Q5. Factorise: \(a + a^4\)

Show Answer & Hint

Common Factor = a
\(= a(1 + a^3)\)

Q6. Factorise: \(15x^2 + xy + xz\)

Show Answer & Hint

Common Factor = x
\(= x(15x + y + z)\)

Q7. Factorise: \(3x^2 – 3x^3 + 12x^4\)

Show Answer & Hint

HCF of 3, 3, 12 is 3. Smallest power of x is x² → Common Factor = 3x²
\(= 3x^2(1 – x + 4x^2)\)

7 Frequently Asked Questions (FAQs)

Q1. What is the main goal when we factorise an algebraic expression?

The primary goal is to transform an expression written as a sum or difference of terms into an equivalent product of simpler factors. This simplifies calculations and helps solve algebraic equations.

Q2. How do you find the Highest Common Factor (HCF) of algebraic terms?

First find the HCF of the numerical coefficients, then choose the lowest exponent for each variable present in all terms. Multiply the numerical HCF by the common variable factors.

Q3. Why does ‘1’ remain inside the bracket when a full term is factored out?

Because any quantity multiplied by 1 equals itself (multiplicative identity). For example, in \(x + x^3\), factoring out \(x\) leaves \(1 + x^2\) so that expanding \(x(1 + x^2)\) gives \(x + x^3\) back.

Q4. What is the difference between expanding and factorising?

Expanding removes brackets by multiplying terms (e.g., \(3(2x + 1) \rightarrow 6x + 3\)). Factorising does the exact opposite: writing an expanded expression as a product with brackets (e.g., \(6x + 3 \rightarrow 3(2x + 1)\)).

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Factorisation by taking common 10.1.1 [BLE]

Algebra & Factorisation Grade 8-10 Guide

Algebraic Expressions & Factorisation

Understanding Algebraic Expressions and Factorisation

1 Key Definitions

1. Algebraic Expression

2. Factors

3. Factorisation

2 Rules for Taking Common Factors

4 Solved Questions (2 & 3 Terms)

5 Solved Questions (4 Terms Grouping)

6 Self Practice Questions

7 Frequently Asked Questions (FAQs)

Q1. What is the main goal when we factorise an algebraic expression?

Q2. How do you find the Highest Common Factor (HCF) of algebraic terms?

Q3. Why does ‘1’ remain inside the bracket when a full term is factored out?

Q4. What is the difference between expanding and factorising?