More factorising - quadratics Revision Notes for OCR GCSE Maths

More factorising - quadratics

What is a Quadratic Expression?

A quadratic expression is any expression that includes a squared term, such as:

ax^2+bx+c

Where $a, b$ , and $c$ are constants.

Looking Back at Expanding Brackets

When we expanded the following expressions, we ended up with quadratic expressions:

$(a+6)(a+4)→a^2+10a+24$
$(p+10)(p−8)→p^2+2p−80$
$(t−9)(t+2)→t^2−7t−18$
$(m−7)(m−9)→m^2−16m+63$

Factorising Quadratics

Factorising quadratic expressions involves reversing the process of expanding brackets. This is a key skill in GCSE Maths and helps in solving quadratic equations by turning them back into their original binomial factors.

Factorising quadratics means you want to get from:

To do this you will need to solve a puzzle.

If you look back at the examples you will see that…

The Key to Factorising Quadratics

To successfully factorise a quadratic, you need to solve a little puzzle:

The numbers $m$ and $n$ in the brackets must:

Add together to give you the coefficient of $x$ (which is $b$ ).
Multiply together to give you the constant term (which is $c$ ).

Understanding the Process

When you factorise a quadratic expression, you're aiming to rewrite it from this form:

x^2+bx+c

into this form:

(x+m)(x+n)

Where $m$ and $n$ are numbers that meet specific conditions.

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Example: Factorise $x^2+7x+10$

Step-by-Step Process

Step 1: Identify the Coefficients

The quadratic expression is $x^2+7x+10.$
Here, $b=7$ and $c=10$ .

Step 2: Find Two Numbers that Work

We need to find two numbers that:
Add together to give $7$ (the coefficient of $x$ ).
Multiply together to give $10$ (the constant term).
The numbers $2$ and $5$ work because:

2+5=7

2×5=10

Step 3: Write the Factorised Form

We can now write the quadratic as:

(x+2)(x+5)

Final Answer:

x^2+7x+10=(x+2)(x+5)

Worked Examples

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Example 1: Factorise $x²+11x+24$

Step 1: Identify the Coefficients

The quadratic expression is $x^2+11x+24$ .
Here, $b=11$ and $c=24$ .

Step 2: Find Two Numbers that Work

We need to find two numbers that:
Multiply together to give $24$ (the constant term).
Add together to give $11$ (the coefficient of $x$ ). List the Possible Pairs:
$1$ and $24$ (because $1×24=24 and 1+24=25$ ) ← doesn't work
$2$ and $12$ (because $2×12=24 and 2+12=14$ ) ← doesn't work
$3$ and $8$ (because $3×8=24$ and $3+8=11 \ ✔$ )

Step 3: Write the Factorised Form

We can now write the quadratic as:

(x+3)(x+8)

Final Answer:

x^2+11x+24=(x+3)(x+8)

Double-check by expanding:

Expanding $(x+3)(x+8)$ gives:

x^2+8x+3x+24=x^2+11x+24

This confirms that our factorisation is correct.

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Example 2: Factorise $p^2+2p−15$

Step 1: Identify the Coefficients

The quadratic expression is $p^2+2p−15.$
Here, $b=2$ and $c=−15$ .

Step 2: Find Two Numbers that Work

We need to find two numbers that:
Multiply together to give $−15$ (the constant term).
Add together to give $2$ (the coefficient of $p$ ).

List the Possible Pairs:

$1$ and $-15$ (because $1×−15=−15$ and $1+(−15)=−14)$ ← doesn't work
- $1$ and $15$ (because $−1×15=−15$ and $−1+15=14$ ) ← doesn't work
$3$ and $-5$ (because $3×−5=−15$ and $3+(−5)=−2$ ) ← doesn't work
$-3$ and $5$ (because $−3×5=−15$ and $−3+5=2 \ ✔$ )

Step 3: Write the Factorised Form

We can now write the quadratic as:

(p−3)(p+5)

Final Answer:

p^2+2p−15=(p−3)(p+5)

Double-check by expanding:

Expanding $(p−3)(p+5)$ gives:

p^2+5p−3p−15=p^2+2p−15

This confirms that our factorisation is correct.

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Example 3: Factorise $k^2−13k−14$

Step 1: Identify the Coefficients

The quadratic expression is $k²−13k−14$ .
Here, $b=−13$ and $c=−14$ .

Step 2: Find Two Numbers that Work

We need to find two numbers that:
Multiply together to give $−14$ (the constant term).
Add together to give $−13$ (the coefficient of $k$ ).

List the Possible Pairs:

$−1$ and $14$ because $−1×14=−14$ and $−1+14=13$ ← doesn't work
$11$ and $−14$ because $1×−14=−14$ and $1+(−14)=−13 ✔$

Step 3: Write the Factorised Form

The factorised form of the quadratic is:

(k+1)(k−14)

or equivalently:

(k−14)(k+1)

Final Answer:

k^2−13k−14=(k+1)(k−14)

Double-check by expanding:

Expanding $(k+1)(k−14)$ gives:

k^2−14k+k−14=k^2−13k−14

This confirms that our factorisation is correct.

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Example 4: Factorise $v^2−9v+18$

Step 1: Identify the Coefficients

The quadratic expression is $v^2−9v+18.$
Here, $b=−9$ and $c=18$ .

Step 2: Find Two Numbers that Work

We need to find two numbers that:
Multiply together to give $18$ (the constant term).
Add together to give $−9$ (the coefficient of $v$ ).
Two negative numbers to get a positive $+18$ .

List the Possible Pairs:

$-1$ and $-18$ because $-1×-18=18$ and $-1+(-18)=-19$ ← doesn't work
$-2$ and $-9$ because $-2×-9=18$ and $-2+(-9)=-11$ ← doesn't work
$−3$ and $−6$ because $−3×−6=18$ and $−3+(−6)=−9 ✔$

Step 3: Write the Factorised Form

The factorised form of the quadratic is:

(v−3)(v−6)

or equivalently:

(v−6)(v−3)

Final Answer:

v^2−9v+18=(v−3)(v−6)

Double-check by expanding:

Expanding $(v−3)(v−6)$ gives:

v^2−6v−3v+18=v^2−9v+18

This confirms that our factorisation is correct.

Understanding the Difference of Two Squares

A quadratic expression is a difference of two squares if it can be written in the form:

a^2−b^2

This expression can be factorised into:

(a+b)(a−b)

Here, $a$ and $b$ are the square roots of the terms in the original expression, and the signs in the factors are opposite.

Worked Examples

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Example: Factorise $x^2−16$

Step 1: Recognise the Structure

The expression $x^2−16$ fits the pattern of a difference of two squares because:

x^2=x×x\quad and \quad16=4×4

This works because $+4$ and $-4$ add together to $=0$ and cancel each other out.

You can think of the expression like this if it helps:

x^2+0x-16

Step 2: Apply the Difference of Two Squares Formula

Factorising $x^2−16$ gives:

(x+4)(x−4)

Final Answer:

x^2−16=(x+4)(x−4)

Double-check by expanding:

Expanding $(x+4)(x−4)$ gives:

x^2−4x+4x−16=x^2−16

This confirms that the factorisation is correct.

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Other examples are:

a^2 - 25 → (a+5)(a−5)

p^2 - 100 → (p+10)(p−10)

4t^2 - 49 → (2t+7)(2t−7)

Dealing with More Complex Quadratics

Now, let's consider quadratic expressions that are not immediately recognisable as the difference of two squares, but still can be factorised. These include expressions where the coefficient of the squared term is not $1$ .

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Example 1: Factorise $5g^2−9g−27$

Step 1: Identify the Coefficients

Here, the expression is $5g^2−9g−27$ .
We need to find two numbers that multiply to give $5×−27=−135$ and add to give $−9$ .

Step 2: Find Two Numbers

The two numbers that work are $15$ and $−9$ , because:

15×−9=−135\quad and \quad15−9=6

Step 3: Write the Factorised Form

We rewrite the expression by splitting the middle term:

5g^2+15g−9g−27

Then, factorise by grouping:

5g(g+3)−9(g+3)=(5g−9)(g+3)

Final Answer:

5g^2−9g−27=(5g−9)(g+3)

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Example 2: Factorise $6c^2−23c+20$

Step 1: Identify the Coefficients

The expression is $6c^2−23c+20$ .
We need to find two numbers that multiply to give $6×20=120$ and add to give $−23$ .

Step 2: Find Two Numbers

The two numbers that work are $−15$ and $−8$ , because:

−15×−8=120 \quad and \quad−15−8=−23

Step 3: Write the Factorised Form

Rewrite the expression by splitting the middle term:

6c^2−15c−8c+20

Then, factorise by grouping:

3c(2c−5)−4(2c−5)=(3c−4)(2c−5)

Final Answer:

6c^2−23c+20=(3c−4)(2c−5)

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Example 1: Factorise $2x^2−x−3$

Step 1: Set Up the Factor Pairs

We start by identifying the terms that will go into our brackets.
The first term in each bracket must multiply together to give $2x^2$ , so the first terms must be $2x$ and $x$ .

Step 2: Determine the Possible Factor Pairs for the Constant Term

Next, we look at the constant term, $−3$ . We need pairs of numbers that multiply to give $−3$ .
The pairs are $(−1,3), (3,−1), (1,−3),$ and $(−3,1).$

Step 3: Multiply Diagonally to Check Which Pair Works

We arrange the pairs in a table and perform diagonal multiplications to see which pair adds up to give the coefficient of $x$ in the original quadratic.
The middle term in our expression is $−x$ . We need a pair of numbers whose diagonal products add up to $−1x$ .

Step 4: Identify the Correct Pair

We test the pairs:
$2x×3+(−1)×x=6x−x=5x$ (No match)
$2x×−1+3×x=−2x+3x=1x$ (No match)
$2x×1+(−3)×x=2x−3x=−1x$ (This works!)

Step 5: Write the Factorised Form

Now that we have identified the correct pair, we place these numbers into the brackets:

(2x+1)(x−3)

Final Answer:

2x^2−x−3=(2x+1)(x−3)

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Example 2: Factorise $8x^2−2x−15$ Let's break down the process step by step.

Step 1: Identify Possible First Terms in the Brackets

The first terms in our brackets must multiply to give $8x^2$ . The possibilities could be:
$8x$ and $x$
$4x$ and $2x$

Step 2: Consider the Constant Term

Next, consider the constant term, $−15$ . We need pairs of numbers that multiply to give $−15$ . These pairs are:
$(−1,15)$
$(1,−15)$
$(3,−5)$
$(−3,5)$

Step 3: Use a Grid Method with Two Tables

To find the correct combination, we use two tables, testing the pairs of factors for each set of first terms:
Now, we multiply diagonally to find the pair of numbers that adds up to the coefficient of the middle term, $−2x:$
From the second table: $4x×−3+5×2x=−12x+10x=−2x$ (This is the correct pair!)

Step 4: Write Down the Factors

With the correct pair found, we can write the factors as:

(4x+5)(2x−3)

Step 5: Check Your Answer

It's essential to check by expanding the brackets:(

(4x+5)(2x−3)=8x^2−12x+10x−15=8x^2−2x−15

This confirms the factorisation is correct.

More factorising - quadratics (OCR GCSE Maths): Revision Notes