Factorising

Introduction to Factorising

Factorising is a way of breaking down an algebraic expression into simpler parts (called factors) that, when multiplied together, give you the original expression. Think of it like "unpacking" a complicated expression into smaller, easier pieces.

We'll learn how to do this using four different methods:

Highest Common Factor (HCF)
Grouping
Difference of Two Squares
Quadratic Trinomials (Quadratic Equations)

1. Highest Common Factor (HCF)

What is it? The Highest Common Factor is the biggest thing (number, letter, or a combination) that can divide into all the terms in your expression.

How to do it:

Look for the biggest common factor in all the terms. This could be a number, a letter, or both.
Write down the HCF outside a set of brackets.
Inside the brackets, write what's left after you've divided each term by the HCF.

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Example: Factorise: $ab - 2a^2b + 3ab^2$ Step-by-Step Solution:

Identify the HCF: Look at each term:

$ab$ , $-2a^2b$ , and $3ab^2$
The common factors are $a$ and $b$ . So, the HCF is $ab$ .

Factor out the HCF: Write $( ab )$ outside the bracket:

$( ab( \ \ \ \ ) )$

Divide each term by the HCF:

$ab \div ab = 1$
$-2a^2b \div ab = -2a$
$3ab^2 \div ab = 3b$
So, inside the brackets, we write $1 - 2a + 3b$ .

Final answer: $ab(1 - 2a + 3b)$ This means the expression $( ab - 2a^2b + 3ab^2 )$ is factorised as $ab(1 - 2a + 3b)$ .

2. Factors by Grouping

What is it?

Factoring by grouping is used when an expression has four terms. The idea is to group the terms in pairs and factor out common factors within those pairs.

How to do it:

Group the terms into pairs that look like they have something in common.
Factor out the common factor from each pair.
Check if you have a common binomial (a binomial is two terms inside a bracket). If yes, factor that out.

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Example: Factorise: $2pr - 2ps + qr - qs$ Step-by-Step Solution:

Group the terms:

$2pr - 2ps$ and $qr - qs$

Factor out the common factor from each group:

From $2pr - 2ps$ , factor out $2p$ : $2p(r - s)$
From $qr - qs$ , factor out $q$ : $q(r - s)$

Now you have: $2p(r - s) + q(r - s)$

Notice $r - s$ is common in both, so factor that out:
$(2p + q)(r - s)$

Final answer: $(2p + q)(r - s)$ This means the expression $2pr - 2ps + qr - qs$ is factorised as $(2p + q)(r - s)$

3. Difference of Two Squares

What is it? This method is used when you have two perfect squares with a minus sign between them, like $( a^2 - b^2 )$ . It can be factorised as $(a - b)(a + b)$ .

How to do it:

Recognise the pattern: Check if both terms are squares and if there's a minus sign between them.
Write down the square roots of both terms.
Factorise as $(first square root - second square root)(first square root + second square root)$ .

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Example: Factorise: $9 - 4y^2$ Step-by-Step Solution:

Recognise the pattern:

$9$ is $3^2$ and $4y^2$ is $(2y)^2$ , and there's a minus sign.

Write the square roots:

The square root of $9$ is $3$ .
The square root of $4y^2$ is $2y$ .

Factorise:

$(3 - 2y)(3 + 2y)$

Final answer: $(3 - 2y)(3 + 2y)$ This means the expression $( 9 - 4y^2 )$ is factorised as $(3 - 2y)(3 + 2y)$ .

4. Quadratic Trinomials (Quadratic Equations)

What is it? Quadratic trinomials have three terms and look like $( ax^2 + bx + c )$ . The goal is to factorise it into two binomials, like $(x - m)(x - n)$ .

How to do it:

Find two numbers that multiply to give the last number $ac$ and add to give the middle number $b$ .
Split the middle term using these two numbers.
Factor by grouping.

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Example: Factorise: $x^2 - 10x + 21$ Step-by-Step Solution:

Identify $ac$ and $b$ :

$( a = 1 )$ , $( b = -10 )$ , $( c = 21 )$
Find two numbers that multiply to $21$ and add to $-10$ . These are $-7$ and $-3$ .

Split the middle term:

Rewrite $-10x$ as $-7x - 3x$ :
$x^2 - 7x - 3x + 21$

Factor by grouping:

$x(x - 7) - 3(x - 7)$
Now, factor out $x - 7$ :
$(x - 7)(x - 3)$

Final answer: $(x - 7)(x - 3)$ This means the expression $x^2 - 10x + 21$ is factorised as $(x - 7)(x - 3)$ .

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Summary Tips for Factorising:

Always start by looking for a Highest Common Factor (HCF).
In Quadratic Trinomials, practice finding two numbers that work for the sum and product criteria.
For the Difference of Squares, remember it must be in the form $( a^2 - b^2 )$ . When factorising, the form is always (a+b)(a-b)

Further Explanation: Quadratic Trinomials (Quadratic Equations)

What is it? A quadratic trinomial is an expression with three terms that follows the form $ax^2 + bx + c$ , where:

$a$ is the coefficient of $x^2$ , (it might be $1$ , or another number)
$b$ is the coefficient of $x$ ,
$c$ is the constant (the number without a variable). Goal:

The goal is to factorise this expression into two binomials. In other words, we want to rewrite $( ax^2 + bx + c )$ as $(dx + e)(fx + g)$ .

Why is it tricky?

Quadratic trinomials can be challenging because you need to find two numbers that both add and multiply correctly to fit the expression. But don't worry! With practice, you can master it.

Steps to Factorise Quadratic Trinomials

Identify the coefficients $a$ , $b$ , and $c$ .
Multiply $a$ and $c$ to get a product. Let's call this product $ac$ .
Find two numbers that multiply to $ac$ and add to $b$ .
Rewrite the middle term using these two numbers.
Group the terms in pairs and factor out the common factor from each pair.
Factor out the common binomial. Let's go through this process with some examples.

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Example 1: Simple Case with $a = 1$ Factorise: $x^2 - 10x + 21$

Identify $a = 1$ , $b = -10$ , and $c = 21$ .
Multiply $a$ and $c$ :

$ac = 1 \times 21 = 21$ .

Find two numbers that multiply to $21$ and add to $-10$ :

The numbers are $-7$ and $-3$ .
$(-7) \times (-3) = 21$
$(-7) + (-3) = -10$

Rewrite the middle term:

Replace $-10x$ with $-7x - 3x$ :
$x^2 - 7x - 3x + 21$

Group the terms:

$(x^2 - 7x) + (-3x + 21)$

Factor each group:

$x(x - 7) - 3(x - 7)$

Factor out the common binomial:

$(x - 7)(x - 3)$ Final Answer: $( x^2 - 10x + 21 )$ factorises to $(x - 7)(x - 3)$ .

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Example 2: When $a$ is Not $1$ Factorise: $2x^2 - 5x + 3$

Identify $( a = 2 )$ , $( b = -5 )$ , and $( c = 3 )$ .
Multiply $a$ and $c$ :

$ac = 2 \times 3 = 6$ .

Find two numbers that multiply to $6$ and add to $-5$ :

The numbers are $-2$ and $-3$ .
$(-2) \times (-3) = 6$
$(-2) + (-3) = -5$

Rewrite the middle term:

Replace $-5x$ with $-2x - 3x$ :
$2x^2 - 2x - 3x + 3$

Group the terms:

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

Factor each group:

$( 2x(x - 1) - 3(x - 1) )$

Factor out the common binomial:

$(2x - 3)(x - 1)$ Final Answer: $( 2x^2 - 5x + 3 )$ factorises to $(2x - 3)(x - 1)$ .

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Example 3: Complex Case Factorise: $3x^2 + 14x + 8$

Identify $a = 3$ , $b = 14$ , and $c = 8$ .
Multiply $a$ and $c$ :

$( ac = 3 \times 8 = 24 )$ .

Find two numbers that multiply to $24$ and add to $14$ :

The numbers are $12$ and $2$ .
$12 \times 2 = 24$
$12 + 2 = 14$

Rewrite the middle term:

Replace $14x$ with $12x + 2x$ :
$3x^2 + 12x + 2x + 8$

Group the terms:

$(3x^2 + 12x) + (2x + 8)$

Factor each group:

$3x(x + 4) + 2(x + 4)$

Factor out the common binomial:

$(3x + 2)(x + 4)$ Final Answer: $( 3x^2 + 14x + 8 )$ factorises to $(3x + 2)(x + 4)$ .

Tips for Factorising Quadratic Trinomials 38. Check for a Common Factor First:

Always check if there's a common factor you can factor out from all three terms before you start.

If the Numbers Don't Work Out:

Go back and check your multiplication and addition. Consider if you might have missed an obvious pair.

Dealing with Negative Numbers:

Remember that two negatives multiply to a positive, but add to a negative.

Practice, Practice, Practice:

Factorising is a skill that improves with practice. The more you do it, the quicker you'll recognise patterns and pairs.

Problem Set 1: Highest Common Factor (HCF)

Problem 1:

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Question: Factorise the expression: $( 6x^2 - 9x )$

Step-by-Step Solution:

Identify the terms:

The expression has two terms: $6x^2$ and $-9x$ .

Find the Highest Common Factor (HCF):

First, look at the numbers: $6$ and $9$ . The largest number that can divide both $6$ and $9$ is $3$ .
Next, look at the variables: both terms have $x$ , with the lowest power being $x^1$ .
So, the HCF is $3x$ .

Factor out the HCF:

Divide each term by the HCF $3x$ .
$( 6x^2 \div 3x = 2x )$ (because $( 6 \div 3 = 2 )$ and $( x^2 \div x = x )$
$( -9x \div 3x = -3 )$ (because $( -9 \div 3 = -3 )$ and $( x \div x = 1 )$ , which is just -3)
Write the expression as $3x(2x - 3)$ .

Final Answer:

The factorised form is $3x(2x - 3)$ .

Explanation: We started by finding what was common in both terms, factored it out, and then rewrote the expression with the remaining terms inside the brackets.

Problem 2:

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Question: Factorise the expression: $( 12a^3b^2 - 8a^2b^3 + 4ab^4 )$

Step-by-Step Solution:

Identify the terms:

The expression has three terms: $12a^3b^2$ , $-8a^2b^3$ , and $4ab^4$ .

Find the HCF:

For the numbers: The HCF of $12$ , $8$ , and $4$ is $4$ .
For the variable $a$ : The lowest power of $a$ across all terms is $a^1$ (in the third term).
For the variable $b$ : The lowest power of $b$ across all terms is $b^2$ (in the first term).
So, the HCF is $4ab^2$ .

Factor out the HCF:

Divide each term by the HCF $4ab^2$ .
$12a^3b^2 \div 4ab^2 = 3a^2$ (because $( 12 \div 4 = 3 )$ , $( a^3 \div a = a^2 )$ , and $( b^2 \div b^2 = 1 )$ )
$-8a^2b^3 \div 4ab^2 = -2ab$ (because $( -8 \div 4 = -2 )$ , $( a^2 \div a = a )$ , and $( b^3 \div b^2 = b )$
$4ab^4 \div 4ab^2 = b^2$ (because $( 4 \div 4 = 1 )$ , $( a \div a = 1 )$ , and $( b^4 \div b^2 = b^2 )$
Write the expression as $4ab^2(3a^2 - 2ab + b^2)$ .

Final Answer:

The factorised form is $4ab^2(3a^2 - 2ab + b^2)$ .

Explanation: Each term was divided by the HCF to leave the simpler expression inside the brackets. The hardest part is finding the common factors for both numbers and variables.

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Exam Tip: When finding the HCF, always look carefully at both the numbers and the letters. It's easy to miss a common factor if you rush.

Problem Set 2: Factors by Grouping

Problem 3:

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Question: Factorise the expression: $x^3 + 3x^2 + 2x + 6$

Step-by-Step Solution:

Group the terms:

Pair the terms into two groups: $(x^3 + 3x^2)$ and $(2x + 6)$ .

Factor out the common factor from each group:

From the first group $( x^3 + 3x^2 )$ :
The common factor is $x^2$ , so $x^3 + 3x^2$ becomes $x^2(x + 3)$ .
From the second group $2x + 6$ :
The common factor is $2$ , so $( 2x + 6 )$ becomes $2(x + 3)$ .

Check the binomials:

The binomials $(x + 3)$ in both groups are the same, so you can factor this out.

Factor out the common binomial:

Write the expression as $(x^2 + 2)(x + 3)$ .

Final Answer:

The factorised form is $(x^2 + 2)(x + 3)$ .

Explanation: We grouped the terms to simplify them, factored out what was common in each group, and then combined the common binomials.

Problem 4:

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Question: Factorise the expression: $3xy - 6x + 2y - 4$

Step-by-Step Solution:

Group the terms:

Pair the terms into two groups: $(3xy - 6x)$ and $(2y - 4)$ .

Factor out the common factor from each group:

From the first group $( 3xy - 6x )$ :
The common factor is $3x$ , so $( 3xy - 6x )$ becomes $3x(y - 2)$ .
From the second group $( 2y - 4 )$ :
The common factor is 2, so $( 2y - 4 )$ becomes $2(y - 2)$ .

Check the binomials:

The binomials $(y - 2)$ in both groups are the same, so you can factor this out.

Factor out the common binomial:

Write the expression as $(3x + 2)(y - 2)$ .

Final Answer:

The factorised form is $(3x + 2)(y - 2)$ .

Explanation: We grouped and factored each pair, then used the common binomial to simplify the expression further.

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Exam Tip: When grouping, it's important that the binomials match. If they don't, double-check your factoring or try rearranging the terms.

Problem Set 3: Difference of Two Squares

Problem 5:

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Question: Factorise the expression: $x^2 - 25$

Step-by-Step Solution:

Recognise the pattern:

Notice that $x^2$ is a perfect square, and $25$ is also a perfect square.
This is a difference of squares since there is a subtraction sign between them.

Write the expression in the form $( a^2 - b^2 )$ :

Here, $x^2 = (x)^2$ and $25 = (5)^2$ .
So, $x^2 - 25 = (x)^2 - (5)^2 .$

Factor using the difference of squares formula:

The difference of squares formula is $a^2 - b^2 = (a - b)(a + b)$ .
Apply it: $x^2 - 25 = (x - 5)(x + 5)$ .

Final Answer:

The factorised form is $(x - 5)(x + 5)$ .

Explanation: We identified that the expression was a difference of squares and applied the formula to factor it.

Problem 6:

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Question: Factorise the expression: $4x^2 - 9y^2$

Step-by-Step Solution:

Recognise the pattern:

Notice that $4x^2$ is a perfect square, and $9y^2$ is also a perfect square.
This is another example of a difference of squares.

Write the expression in the form $( a^2 - b^2 )$ :

Here, $( 4x^2 = (2x)^2 )$ and $( 9y^2 = (3y)^2 )$ .
So, $4x^2 - 9y^2 = (2x)^2 - (3y)^2 .$

Factor using the difference of squares formula:

Apply it: $4x^2 - 9y^2 = (2x - 3y)(2x + 3y)$ .

Final Answer:

The factorised form is $(2x - 3y)(2x + 3y)$ .

Explanation: We used the difference of squares formula to simplify the expression into two binomials.

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Exam Tip: Always make sure that both terms are perfect squares and that there's a subtraction sign between them. It's easy to confuse this with other forms of factorisation.

Problem Set 4: Quadratic Trinomials (Quadratic Equations)

Problem 7:

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Question: Factorise the expression: $x^2 + 7x + 12$

Step-by-Step Solution:

Identify the coefficients:

Here, $a = 1$ , $b = 7$ , and $c = 12$ .

Find two numbers that multiply to $c = 12$ and add to $b = 7$ :

We need to find two numbers that multiply to $12$ and add to $7$ .
These numbers are $3$ and $4$ because:
$3 \times 4 = 12$
$3 + 4 = 7$

Rewrite the expression using these two numbers:

Rewrite the middle term $7x$ as $3x + 4x$ :
$x^2 + 3x + 4x + 12$ .

Group and factor:

Group the first two terms and the last two terms:
$(x^2 + 3x) + (4x + 12)$ .
Factor out the common factor from each group:
$x(x + 3) + 4(x + 3)$ .

Factor out the common binomial:

The expression becomes $(x + 3)(x + 4)$ .

Final Answer:

The factorised form is $(x + 3)(x + 4)$ .

Explanation: By finding two numbers that work with both the multiplication and addition rules, we could rewrite the expression in a simpler form.

Problem 8:

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Question: Factorise the expression: $2x^2 + 5x - 3$

Step-by-Step Solution:

Identify the coefficients:

Here, $a = 2$ , $b = 5$ , and $c = -3$ .

Multiply $a \times c = 2 \times -3 = -6$ .
Find two numbers that multiply to $-6$ and add to $5$ :

These numbers are $6$ and $-1$ because:
$6 \times -1 = -6$
$6 + (-1) = 5$

Rewrite the expression using these two numbers:

Rewrite the middle term $5x$ as $6x - x$ :
$2x^2 + 6x - x - 3$ .

Group and factor:

Group the first two terms and the last two terms:
$(2x^2 + 6x) - (x + 3)$ .
Factor out the common factor from each group:
$2x(x + 3) - 1(x + 3)$ .

Factor out the common binomial:

The expression becomes $(2x - 1)(x + 3)$ .

Final Answer:

The factorised form is $(2x - 1)(x + 3)$ .

Explanation: Here, multiplying $( a \times c )$ helped us find the right pair of numbers to split the middle term and proceed with grouping.

Problem 9:

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Question: Factorise the expression: $6x^2 + 11x + 3$

Step-by-Step Solution:

Identify the coefficients:

Here, $a = 6$ , $b = 11$ , and $c = 3$ .

Multiply $a \times c = 6 \times 3 = 18$ .
Find two numbers that multiply to $18$ and add to $11$ :

These numbers are $9$ and $2$ because:
$9 \times 2 = 18$
$9 + 2 = 11$

Rewrite the expression using these two numbers:

Rewrite the middle term $11x$ as $9x + 2x$ :
$6x^2 + 9x + 2x + 3$ .

Group and factor:

Group the first two terms and the last two terms:
$(6x^2 + 9x) + (2x + 3)$ .
Factor out the common factor from each group:
$3x(2x + 3) + 1(2x + 3)$ .

Factor out the common binomial:

The expression becomes $(3x + 1)(2x + 3)$ .

Final Answer:

The factorised form is $(3x + 1)(2x + 3)$ .

Explanation: By splitting the middle term and carefully grouping, we could factorise the quadratic trinomial completely.

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Exam Tip: When dealing with quadratic trinomials, especially when $( a )$ is not $1$ , always take your time finding the correct pair of numbers. It's crucial to ensure they multiply to $ac$ and add to $b$ . Double-checking this step can prevent errors later on.

Factorising Simplified Revision Notes for Junior Cycle Mathematics

Factorising

Introduction to Factorising

1. Highest Common Factor (HCF)

2. Factors by Grouping

3. Difference of Two Squares

4. Quadratic Trinomials (Quadratic Equations)

Further Explanation: Quadratic Trinomials (Quadratic Equations)

Steps to Factorise Quadratic Trinomials

Problem Set 1: Highest Common Factor (HCF)

Problem 1:

Problem 2:

Problem Set 2: Factors by Grouping

Problem 3:

Problem 4:

Problem 5:

Problem 6:

Problem Set 4: Quadratic Trinomials (Quadratic Equations)

Problem 7:

Problem 8:

Problem 9:

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