Algebraic Expressions Revision Notes for Grade 10 NSC Matric Mathematics

Factorisation

Factorisation is the reverse process of expanding algebraic expressions. While expanding involves multiplying out brackets to get a sum of terms, factorisation starts with a sum of terms and rewrites it as a product of factors.

When you expand $2(x + 1)$ , you get $2x + 2$ . Factorisation works in the opposite direction - starting with $2x + 2$ and writing it as $2(x + 1)$ . These two expressions are equivalent and have the same value for all values of $x$ .

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Understanding factorisation builds on your previous work with common factors and special patterns like the difference of squares. This skill is essential for solving equations, simplifying expressions, and working with more advanced algebraic concepts.

Common factors

Common factor factorisation relies on identifying factors that are shared by all terms in an expression. This is often the first step in any factorisation problem.

The process involves finding the highest common factor of all terms and writing it outside brackets. For instance, the expression $2x - 6x^2$ can be factorised by identifying that both terms contain the factor $2x$ :

$2x - 6x^2 = 2x(1 - 3x)$

When expressions have more complex patterns, you can use a "switch around" strategy. This involves recognising that subtraction can be rewritten to reveal common factors. For example, if you have $(2 - a)$ , you can rewrite this as $-(a - 2)$ .

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Worked Example: Factorising using the switch around strategy

Question: Factorise $5(a - 2) - b(2 - a)$

Solution:

The key is to notice that $(2 - a) = -(a - 2)$ . This allows us to rewrite the expression:

$5(a - 2) - b(2 - a) = 5(a - 2) - b[-(a - 2)]$

$= 5(a - 2) + b(a - 2)$

$= (a - 2)(5 + b)$

Always check your factorisation by expanding the result to verify it matches the original expression.

Difference of two squares

The difference of two squares is a special pattern that appears frequently in algebra. The key formula is:

$a^2 - b^2 = (a + b)(a - b)$

To identify a difference of two squares, look for expressions that have:

Exactly two terms
Terms with opposite signs (one positive, one negative)
Each term is a perfect square

Examples include $x^2 - 16$ , $4x^2 - y^2$ , and $49 - p^4$ .

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Worked Example: Factorising the difference of two squares

Question: Factorise $3a(a^2 - 4) - 7(a^2 - 4)$

Solution:

Step 1: Take out the common factor $(a^2 - 4)$

$3a(a^2 - 4) - 7(a^2 - 4) = (a^2 - 4)(3a - 7)$

Step 2: Factorise the difference of two squares $(a^2 - 4)$

$(a^2 - 4)(3a - 7) = (a - 2)(a + 2)(3a - 7)$

The final answer combines both the common factor method and the difference of squares pattern.

Factorising by grouping in pairs

When an expression has four terms with no common factor across all terms, you can often use grouping in pairs. This method involves grouping terms that do share common factors, then looking for a new common factor that emerges.

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The strategy works when you can group the terms so that each group has a common factor, and the remaining expressions after factoring are identical.

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Worked Example: Factorising by grouping in pairs

Question: Find the factors of $7x + 14y + bx + 2by$

Solution:

Step 1: Check for factors common to all terms - there aren't any.

Step 2: Group terms with common factors together

The first two terms share the factor 7, and the last two terms share the factor $b$ :

$7x + 14y + bx + 2by = (7x + 14y) + (bx + 2by)$

$= 7(x + 2y) + b(x + 2y)$

Step 3: Take out the new common factor $(x + 2y)$

$7(x + 2y) + b(x + 2y) = (x + 2y)(7 + b)$

The factors are $(7 + b)$ and $(x + 2y)$ .

Factorising quadratic trinomials

Quadratic trinomials have the form $ax^2 + bx + c$ and require a systematic approach to factorise. The goal is to find two binomial factors that multiply to give the original quadratic.

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General procedure for factorising trinomials:

Take out any common factor first to get the form $ax^2 + bx + c$ where $a$ , $b$ , and $c$ have no common factors
Write two brackets with $x$ in each: $(x\quad)(x\quad)$
Find the factors of the constant term $c$
Create options for possible factor combinations using these factors
Expand each option to find which gives the correct middle term

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Important tip: If $c$ is positive, both factors must have the same sign (both positive or both negative). If $c$ is negative, the factors must have opposite signs.

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Worked Example: Factorising a quadratic trinomial

Question: Factorise $3x^2 + 2x - 1$

Solution:

Step 1: Check the quadratic is in the form $ax^2 + bx + c$ ✓

Step 2: Write brackets and identify possible factors

$(x\quad)(x\quad)$

The possible factors for $a = 3$ are: 1 and 3 The possible factors for $c = -1$ are: -1 and 1

Step 3: Create options for factor combinations:

Option 1	Option 2
$(x - 1)(3x + 1)$	$(x + 1)(3x - 1)$
$3x^2 - 2x - 1$	$3x^2 + 2x - 1$

Step 4: Check by expanding

$(x + 1)(3x - 1) = 3x^2 - x + 3x - 1 = 3x^2 + 2x - 1$ ✓

Therefore: $3x^2 + 2x - 1 = (x + 1)(3x - 1)$

Sum and difference of two cubes

There are two special formulas for factorising cubic expressions:

Sum of two cubes: $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$

Difference of two cubes: $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$

These formulas are used when you can identify perfect cube terms in the expression.

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Worked Example: Factorising a difference of two cubes

Question: Factorise $a^3 - 1$

Solution:

Step 1: Identify the cube roots

We're working with the difference of two cubes: $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$

Here, $\sqrt[3]{a^3} = a$ and $\sqrt[3]{1} = 1$ , so $x = a$ and $y = 1$

Step 2: Apply the formula

$(a^3 - 1) = (a - 1)(a^2 + a + 1)$

Step 3: Verify by expanding

$(a - 1)(a^2 + a + 1) = a(a^2 + a + 1) - 1(a^2 + a + 1)$

$= a^3 + a^2 + a - a^2 - a - 1$

$= a^3 - 1$ ✓

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Worked Example: Factorising a sum of two cubes

Question: Factorise $x^3 + 8$

Solution:

Step 1: Identify the cube roots

We have $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$

Here, $\sqrt[3]{x^3} = x$ and $\sqrt[3]{8} = 2$ , so the terms are $x$ and $2$

Step 2: Apply the sum of cubes formula

$(x^3 + 8) = (x + 2)(x^2 - 2x + 4)$

Step 3: Verify by expanding

$(x + 2)(x^2 - 2x + 4) = x(x^2 - 2x + 4) + 2(x^2 - 2x + 4)$

$= x^3 - 2x^2 + 4x + 2x^2 - 4x + 8$

$= x^3 + 8$ ✓

Exam tips for factorisation

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Key Strategies for Success:

Always look for common factors first - this often simplifies the remaining work significantly
Check your work by expanding your factored form back to the original expression
For quadratics, remember that the signs of factors depend on the sign of the constant term
Learn the special patterns - difference of squares and sum/difference of cubes save time in exams
When grouping in pairs, try different groupings if the first attempt doesn't work
Use the switch around strategy when you see expressions like $(a - b)$ and $(b - a)$ - they're opposites

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Key Points to Remember:

Factorisation is the reverse process of expansion - use this relationship to check your answers
Common factors should always be taken out first before applying other factorisation methods
The difference of squares formula $a^2 - b^2 = (a + b)(a - b)$ is one of the most useful patterns in algebra
Grouping in pairs works well for expressions with four terms that don't have a single common factor
For quadratic trinomials, systematic trial and error with factor combinations will lead to the correct answer

Algebraic Expressions (Grade 10 NSC Matric Mathematics): Revision Notes

Factorisation

Common factors

Difference of two squares

Factorising by grouping in pairs

Factorising quadratic trinomials

Sum and difference of two cubes

Exam tips for factorisation

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