Factorisation: Common Factors, Regrouping, and Algebraic Identities

In arithmetic, when we are given a composite number such as $30$, we can break it down into a product of its prime factors: $30 = 2 \times 3 \times 5$. Similarly, in algebra, when an algebraic expression is written as a product of simpler algebraic expressions, those simpler expressions are called the factors of the given expression, and the process is called factorisation.

Factorisation is essentially the inverse process of algebraic expansion. While expansion transforms a product of expressions into a sum or difference of terms (e.g., $2x(x + 3) = 2x^2 + 6x$), factorisation converts a sum or difference of terms back into a product of irreducible factors (e.g., $2x^2 + 6x = 2x(x + 3)$). Mastering factorisation is crucial for higher mathematics, as it provides the foundational toolkit for solving algebraic equations, simplifying rational expressions, studying polynomial functions, and analyzing real-world rates of change.

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1. In-Depth Conceptual Breakdown

To factorise algebraic expressions systematically, we categorize expressions based on their structural properties and apply specific mathematical techniques:

1. Finding common factors
2. Regrouping terms
3. Utilizing standard algebraic identities

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1.1 Understanding Factors and Irreducibility

An algebraic term is composed of numerical coefficients and literal (variable) factors. For example, in the term $12x^2y$:

• Numerical coefficient: $12 = 2 \times 2 \times 3$
• Variable factors: $x^2y = x \times x \times y$

Hence, the prime or irreducible factor form of $12x^2y$ is: $12x^2y = 2 \times 2 \times 3 \times x \times x \times y$

An irreducible factor is an expression that cannot be factorised further into simpler algebraic expressions. For instance, $(x + 2)$ is irreducible, whereas $(x^2 - 4)$ is reducible because it can be split into $(x - 2)(x + 2)$.

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1.2 Method 1: Factorisation by Common Factors

When all terms in a given algebraic expression share one or more common factors (numerical or variable), we extract the Highest Common Factor (HCF) using the distributive law: $p \cdot q + p \cdot r = p(q + r)$

Step-by-Step Procedure:

1. Find the HCF of the numerical coefficients of all terms.
2. Find the HCF of the variable parts by taking each common variable raised to its lowest exponent present in any term.
3. Multiply these individual HCFs to obtain the overall HCF of the expression.
4. Divide each original term by this HCF to determine the remaining expression inside the parentheses.

Walkthrough Example:

Factorise $14x^2y + 21xy^2 - 35xy$.

1. Numerical Coefficients: The coefficients are $14, 21, 35$. Their $\text{HCF}(14, 21, 35) = 7$.
2. Variable Parts:
   • For $x$: powers are $x^2, x^1, x^1$. Lowest power is $x^1$.
   • For $y$: powers are $y^1, y^2, y^1$. Lowest power is $y^1$.
   • Thus, the variable HCF is $xy$.
3. Overall HCF: $7xy$.
4. Extract HCF: $14x^2y + 21xy^2 - 35xy = 7xy\left(\frac{14x^2y}{7xy} + \frac{21xy^2}{7xy} - \frac{35xy}{7xy}\right) = 7xy(2x + 3y - 5)$

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1.3 Method 2: Factorisation by Regrouping Terms

In many expressions, there is no single common factor shared by all terms simultaneously. However, terms can often be arranged in groups such that each group shares a common binomial factor.

Step-by-Step Procedure:

1. Examine the terms and check if a global common factor exists.
2. Rearrange (regroup) the terms into pairs or groups that share common factors.
3. Factorise each group individually to expose a common binomial factor.
4. Factor out the common binomial factor to write the expression as a product of two factors.

Walkthrough Example:

Factorise $6xy - 4y + 6 - 9x$.

1. Check for overall common factors: $\text{HCF}(6, 4, 6, 9) = 1$, and variable $x$ or $y$ is not present in all four terms.
2. Regroup: Group $6xy - 4y$ and $-9x + 6$. $6xy - 4y - 9x + 6 = (6xy - 4y) - (9x - 6)$
3. Factor each group:
   • From $(6xy - 4y)$, factor out $2y$: $2y(3x - 2)$
   • From $-(9x - 6)$, factor out $3$: $-3(3x - 2)$
4. Combine: $2y(3x - 2) - 3(3x - 2)$
5. Extract the common binomial factor $(3x - 2)$: $= (3x - 2)(2y - 3)$

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1.4 Method 3: Factorisation Using Standard Algebraic Identities

When expressions have specific geometric or structural patterns, we apply standard algebraic identities directly in reverse.

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Detailed Breakdown of Identities

Identity I & II: Perfect Square Trinomials

An expression of the form $a^2 \pm 2ab + b^2$ can be immediately written as $(a \pm b)^2$.

• Example: Factorise $9x^2 + 24xy + 16y^2$.
  • First term: $9x^2 = (3x)^2 \implies a = 3x$
  • Third term: $16y^2 = (4y)^2 \implies b = 4y$
  • Middle term check: $2ab = 2(3x)(4y) = 24xy$ (Matches!)
  • Result: $9x^2 + 24xy + 16y^2 = (3x + 4y)^2$

Identity III: Difference of Two Squares

An expression of the form $a^2 - b^2$ factors into the product of the sum and difference of the bases: $(a + b)(a - b)$.

• Example: Factorise $49x^2 - 64$.
  • First term: $49x^2 = (7x)^2 \implies a = 7x$
  • Second term: $64 = (8)^2 \implies b = 8$
  • Result: $49x^2 - 64 = (7x + 8)(7x - 8)$

Identity IV: Factorisation of Form $x^2 + px + q$ (Splitting the Middle Term)

To factorise a quadratic trinomial $x^2 + px + q$, we find two numbers $a$ and $b$ such that: $a + b = p \quad \text{and} \quad a \times b = q$ Once identified, the expression factorises into $(x + a)(x + b)$.

• Example: Factorise $x^2 - 7x + 12$.
  • Product needed ($ab$): $+12$
  • Sum needed ($a + b$): $-7$
  • Since the product is positive and the sum is negative, both numbers must be negative.
  • Factors of $12$: $(-1, -12)$, $(-2, -6)$, $(-3, -4)$.
  • Check sum: $(-3) + (-4) = -7$. Thus, $a = -3$ and $b = -4$.
  • Result: $x^2 - 7x + 12 = (x - 3)(x - 4)$

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2. Real-World Applications & Analogies

Analogy 1: Reversing a Manufacturing Process (Assembly vs. Disassembly)

Imagine an engineer building a compound machine out of standardized Lego components.

• Expansion is taking basic components, combining them, and expanding them into a complex physical model.
• Factorisation is taking that complex machine, identifying its fundamental structural units, and dismantling it back into its essential modular building blocks. In software engineering, this is identical to code refactoring, where complex redundant code is broken down into clean, reusable modular functions.

Application 1: Land Measurement and Architecture

Architects and civil engineers frequently compute physical areas using structural identities. Suppose a land developer owns a square plot of side length $x$ meters and needs to remove a smaller square section of side length $y$ meters for a utility installation. $\text{Remaining Area} = x^2 - y^2$ Using Identity III, the area is factorised into: $\text{Remaining Area} = (x + y)(x - y)$ If $x = 105\text{ m}$ and $y = 5\text{ m}$, instead of computing $105^2 - 5^2 = 11025 - 25 = 11000$, factorisation allows mental computation: $\text{Area} = (105 + 5)(105 - 5) = (110)(100) = 11000\text{ m}^2$

   +-----------------------+          +-----------------------+
   |                       |          |                       |
   |                       |          |                       |
   |       x * x           |  ===>    |      (x - y) * x      |
   |       Square          |          |       Rectangle       |
   |                       |          |                       |
   +---------------+-------+          +-----------------------+
   |  Cutout (y^2) |                  |     (x - y) * y       |
   +---------------+                  +-----------------------+
                                   Total Area = (x + y)(x - y)

Application 2: Cryptography and Digital Security

Modern digital security, including internet banking and data encryption (like RSA encryption), relies heavily on prime factorisation and polynomial factorisation. Encrypted messages are encoded using extremely large composite numbers or higher-degree algebraic polynomials. Decrypting the message requires finding the unique irreducible factors of these mathematical structures.

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3. Step-by-Step Solved Textbook Examples

Example 1: Regrouping Terms

Problem: Factorise the algebraic expression: $15xy - 6x + 5y - 2$

Solution:

Step 1: Observe all four terms to see if there is a common factor across all terms.

• Terms: $15xy, -6x, 5y, -2$.
• There is no factor common to all four terms.

Step 2: Group the terms into two pairs. $\text{Group 1}: (15xy - 6x) \quad \text{and} \quad \text{Group 2}: (5y - 2)$

Step 3: Take out the common factor from each group.

• From $(15xy - 6x)$, the common factor is $3x$: $15xy - 6x = 3x(5y - 2)$
• From $(5y - 2)$, the common factor is $1$: $5y - 2 = 1(5y - 2)$

Step 4: Re-write the combined expression: $3x(5y - 2) + 1(5y - 2)$

Step 5: Factor out the common binomial factor $(5y - 2)$: $(5y - 2)(3x + 1)$

Final Answer: $\mathbf{15xy - 6x + 5y - 2 = (5y - 2)(3x + 1)}$

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Example 2: Using Perfect Square Identities

Problem: Factorise $25x^2 - 40xy + 16y^2$.

Solution:

Step 1: Count the terms and analyze the signs.

• The expression has $3$ terms.
• The first and last terms are positive perfect squares.
• The middle term has a minus sign, suggesting Identity II: $a^2 - 2ab + b^2 = (a - b)^2$.

Step 2: Identify $a$ and $b$.

• First term: $25x^2 = (5x)^2 \implies a = 5x$
• Last term: $16y^2 = (4y)^2 \implies b = 4y$

Step 3: Verify the middle term $2ab$. $2ab = 2 \cdot (5x) \cdot (4y) = 40xy$ This matches the middle term $-40xy$ in the given expression.

Step 4: Apply the identity $a^2 - 2ab + b^2 = (a - b)^2$. $25x^2 - 40xy + 16y^2 = (5x - 4y)^2$

Final Answer: $\mathbf{25x^2 - 40xy + 16y^2 = (5x - 4y)(5x - 4y)}$

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Example 3: Repeated Application of Difference of Squares

Problem: Factorise $x^4 - 81$ completely.

Solution:

Step 1: Recognize the expression as a difference of two squares.

• $x^4 = (x^2)^2$
• $81 = 9^2$

Step 2: Apply Identity III ($a^2 - b^2 = (a + b)(a - b)$) with $a = x^2$ and $b = 9$. $x^4 - 81 = (x^2)^2 - 9^2 = (x^2 + 9)(x^2 - 9)$

Step 3: Inspect the resulting factors for further factorisation.

• Factor 1: $(x^2 + 9)$ is a sum of squares and cannot be factorised further over real numbers (it is irreducible).
• Factor 2: $(x^2 - 9)$ is again a difference of two squares!

Step 4: Apply Identity III again to $(x^2 - 9)$.

• Here $a = x$ and $b = 3$: $x^2 - 9 = x^2 - 3^2 = (x + 3)(x - 3)$

Step 5: Combine all irreducible factors together. $x^4 - 81 = (x^2 + 9)(x + 3)(x - 3)$

Final Answer: $\mathbf{x^4 - 81 = (x^2 + 9)(x + 3)(x - 3)}$

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Example 4: Form $x^2 + (a+b)x + ab$

Problem: Factorise $m^2 - 10m - 24$.

Solution:

Step 1: Compare with the standard quadratic form $m^2 + pm + q$.

• Here, $p = -10$ and $q = -24$.

Step 2: Find two integers $a$ and $b$ such that: $a + b = -10 \quad \text{and} \quad a \times b = -24$

Step 3: List factor pairs of $-24$ (since the product is negative, one factor must be positive and one negative):

• $1 \times (-24) \implies 1 + (-24) = -23$
• $2 \times (-12) \implies 2 + (-12) = -10$ $\quad \text{(This is the correct pair!)}$
• $3 \times (-8) \implies 3 + (-8) = -5$
• $4 \times (-6) \implies 4 + (-6) = -2$

So, $a = 2$ and $b = -12$.

Step 4: Rewrite the middle term $-10m$ as $+2m - 12m$ and factorise by grouping: $m^2 - 10m - 24 = m^2 + 2m - 12m - 24$ $= m(m + 2) - 12(m + 2)$ $= (m + 2)(m - 12)$

Final Answer: $\mathbf{m^2 - 10m - 24 = (m + 2)(m - 12)}$

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4. Common Student Mistakes to Avoid

Mistake 1: Sign Error When Extracting a Negative Factor

• Incorrect: $a(x - y) - b(y - x) = (a - b)(x - y)$
• Correct Explanation: Notice that $(y - x) = -(x - y)$. When factoring out a negative sign, the sign inside the bracket must flip: $-b(y - x) = -b[-(x - y)] = +b(x - y)$ Therefore: $a(x - y) - b(y - x) = a(x - y) + b(x - y) = (x - y)(a + b)$

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Mistake 2: Stopping Factorisation Incompletely

• Incorrect: Leaving $x^4 - 16 = (x^2 + 4)(x^2 - 4)$ as the final answer.
• Correct Explanation: Always check if any generated factor can be factorised further. The term $(x^2 - 4)$ is itself a difference of squares and must be broken down to $(x + 2)(x - 2)$. The complete answer is $(x^2 + 4)(x + 2)(x - 2)$.

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Mistake 3: Confusing $(a - b)^2$ with $(a^2 - b^2)$

• Incorrect: Writing $x^2 - 9 = (x - 3)^2$.
• Correct Explanation:
  • $(x - 3)^2$ is a perfect square trinomial: $(x - 3)^2 = x^2 - 6x + 9$.
  • $x^2 - 9$ is a difference of two squares: $x^2 - 9 = (x - 3)(x + 3)$. These two expressions are not equal.

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Mistake 4: Incorrect Splitting of Middle Term Signs

• Incorrect: For $x^2 - 5x - 6$, choosing factors $+2$ and $+3$ because $2 + 3 = 5$.
• Correct Explanation: The product must equal $-6$ (negative) and the sum must equal $-5$.
  • If you choose $+2$ and $+3$, their product is $+6 \neq -6$.
  • The correct factors are $-6$ and $+1$, because $(-6) \times (+1) = -6$ and $(-6) + (+1) = -5$.
  • Thus, $x^2 - 5x - 6 = (x - 6)(x + 1)$.

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5. Practice Questions for Self-Assessment

Question 1

Factorise by regrouping terms: $a^2b + ab^2 + a + b$

<details> <summary><b>Click to View Complete Solution</b></summary>

Step 1: Group terms in pairs. $(a^2b + ab^2) + (a + b)$

Step 2: Extract common factor $ab$ from the first pair. $ab(a + b) + 1(a + b)$

Step 3: Factor out the common binomial $(a + b)$. $(a + b)(ab + 1)$

Answer: $(a + b)(ab + 1)$

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Question 2

Factorise using suitable identities: $16x^4 - (y + z)^4$

<details> <summary><b>Click to View Complete Solution</b></summary>

Step 1: Recognize the expression as a difference of squares: $A^2 - B^2$.

• $A = 4x^2$ because $(4x^2)^2 = 16x^4$
• $B = (y + z)^2$ because $[(y + z)^2]^2 = (y + z)^4$

Step 2: Apply $A^2 - B^2 = (A + B)(A - B)$. $16x^4 - (y + z)^4 = [4x^2 + (y + z)^2][4x^2 - (y + z)^2]$

Step 3: Observe the second factor $[4x^2 - (y + z)^2]$. It is again a difference of squares where $a = 2x$ and $b = (y + z)$. $4x^2 - (y + z)^2 = [2x + (y + z)][2x - (y + z)] = (2x + y + z)(2x - y - z)$

Step 4: Combine all factors together. $[4x^2 + (y + z)^2](2x + y + z)(2x - y - z)$

Answer: $[4x^2 + (y + z)^2](2x + y + z)(2x - y - z)$

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Question 3

Factorise the expression: $p^2 - 11p + 28$

<details> <summary><b>Click to View Complete Solution</b></summary>

Step 1: Identify coefficients: product $q = +28$, sum $p = -11$.

Step 2: Find two negative numbers whose product is $+28$ and sum is $-11$.

• Factors of $28$: $(-1, -28)$, $(-2, -14)$, $(-4, -7)$.
• Check sum: $(-4) + (-7) = -11$.

Step 3: Split the middle term and factorise: $p^2 - 4p - 7p + 28 = p(p - 4) - 7(p - 4) = (p - 4)(p - 7)$

Answer: $(p - 4)(p - 7)$

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Question 4

Factorise the four-term expression combining identities and common factors: $25a^2 - 4b^2 + 28bc - 49c^2$

<details> <summary><b>Click to View Complete Solution</b></summary>

Step 1: Notice that the last three terms contain $b^2$, $bc$, and $c^2$. Group them together by factoring out a negative sign: $25a^2 - (4b^2 - 28bc + 49c^2)$

Step 2: Analyze the expression inside the parenthesis $(4b^2 - 28bc + 49c^2)$:

• $4b^2 = (2b)^2$
• $49c^2 = (7c)^2$
• Middle term: $2(2b)(7c) = 28bc$
• So, $4b^2 - 28bc + 49c^2 = (2b - 7c)^2$

Step 3: Substitute back into the main expression: $25a^2 - (2b - 7c)^2 = (5a)^2 - (2b - 7c)^2$

Step 4: Apply Identity III ($A^2 - B^2 = (A + B)(A - B)$) where $A = 5a$ and $B = (2b - 7c)$: $= [5a + (2b - 7c)][5a - (2b - 7c)]$ $= (5a + 2b - 7c)(5a - 2b + 7c)$

Answer: $(5a + 2b - 7c)(5a - 2b + 7c)$

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6. Exam Revision & FAQs

FAQ 1: How do I choose between Regrouping and Algebraic Identities?

Answer:

1. Count the number of terms:
   • 2 terms: Look for a common factor or a Difference of Squares ($a^2 - b^2$).
   • 3 terms (Trinomial): Check for Perfect Square Identities ($(a+b)^2$ or $(a-b)^2$) or use the middle-term splitting method ($(x+a)(x+b)$).
   • 4 terms: Try Common Factor extraction first. If that fails, attempt Regrouping in pairs $(2+2)$ or as a group of three forming a perfect square minus a term $(3-1)$.

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FAQ 2: What is the primary difference between a "Prime Factor" in arithmetic and an "Irreducible Factor" in algebra?

Answer: A prime factor is a natural number greater than $1$ that has no positive divisors other than $1$ and itself (e.g., $2, 3, 5$). An irreducible factor is an algebraic expression that cannot be expressed as a product of two or more expressions of lower degree (e.g., $(x + 5)$, $(2x - 3)$, or $(x^2 + 4)$ over real numbers).

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FAQ 3: Why do we write $a^2 - b^2 = (a+b)(a-b)$, but we cannot factorise $a^2 + b^2$ using real numbers?

Answer: The sum of two squares $a^2 + b^2$ cannot be factored into real linear factors because there are no two real numbers whose product is $+b^2$ and whose sum is $0$ when cross-multiplied. Over the set of Real Numbers ($\mathbb{R}$), $a^2 + b^2$ is an irreducible polynomial.

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FAQ 4: How can I double-check my factorisation answers during an exam?

Answer: Use Expansion (Multiplication) or Substitution:

1. Algebraic Method: Multiply the factors back out. If your factorised answer is $(2x - 3)(x + 4)$, expand it: $2x^2 + 8x - 3x - 12 = 2x^2 + 5x - 12$. If it matches the original question, your factorisation is $100\%$ correct.
2. Numerical Substitution Method: Choose a small integer for the variable (e.g., $x = 2$). Evaluate the original expression and evaluate your factorised answer. Both must yield the exact same numerical value.