How is factorising related to expanding brackets?

Factorising is the reverse of expanding: expanding removes brackets, and factorising puts them back in. That also gives a way of checking an answer, since expanding a factorised expression should return the expression you started with.

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Factorisation (SQA National 5 Maths): Flashcards

Exam code: X847 75

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Define a factorised expression.

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Define a factorised expression.
A factorised expression is one written as a product, meaning two or more terms multiplied together.
So $3 (x + 2)$ and $3 x y$ are factorised, but $3 x + 6$ is not.
How is factorising related to expanding brackets?
Factorising is the reverse of expanding: expanding removes brackets, and factorising puts them back in.
That also gives a way of checking an answer, since expanding a factorised expression should return the expression you started with.
In $12 x^{2} + 18 x$ how do you find the highest common factor of the two terms?
Find the highest common factor of the number parts and of the algebra parts separately, then multiply the two together.
Here that is $6$ from $12$ and $18$ , and $x$ from $x^{2}$ and $x$ , giving $6 x$ and the factorised form $6 x (2 x + 3)$ .
When two terms both contain powers of $a$ which power goes into the highest common factor?
The lowest power present goes into the highest common factor, because that is the largest power both terms can be divided by.
For $2 a^{3} b - 4 a^{2} b^{2}$ the highest common factor is $2 a^{2} b$ , giving $2 a^{2} b (a - 2 b)$ .
Complete the factorisation by filling in the missing factor and the missing number.
$30 x^{2} - 24 x =______(5 x -______)$
The completed factorisation is:
$30 x^{2} - 24 x = 6 x (5 x - 4)$
Checking each part, $6$ is the highest common factor of $30$ and $24$ , and $x$ is the highest power of $x$ that divides both terms.
True or False?
Once an expression has been written with a bracket, it has been factorised fully.
False.
An expression is only factorised fully when there is no common factor left to take out.
$x (6 x + 10)$ is only partly factorised, because $6 x + 10$ still has a factor of $2$ , and the full form is $2 x (3 x + 5)$ .
Why does $5 x + 15$ factorise to $5 (x + 3)$ rather than to something with an $x$ outside?
The second term $15$ contains no $x$ at all, so $x$ is not a common factor and cannot come outside the bracket.
Only the number $5$ divides both terms, which makes $5$ the highest common factor.
Define a difference of two squares.
A difference of two squares is one squared quantity subtracted from another, in the form $a^{2} - b^{2}$ .
Both $9 x^{2} - 16$ and ${(x + 1)}^{2} - {(x - 4)}^{2}$ have this form.
Complete the difference of two squares rule by filling in the two missing brackets.
$a^{2} - b^{2} =______\times______$
The completed rule is:
$a^{2} - b^{2} = (a + b) (a - b)$
Every difference of two squares factorises this way, whatever $a$ and $b$ stand for, so $9 x^{2} - 16$ becomes $(3 x + 4) (3 x - 4)$ .
Why does expanding $(a + b) (a - b)$ leave no $a b$ term behind?
The outer and inner products are $- a b$ and $+ b a$ , which are equal in size and opposite in sign, so they cancel each other out.
Only the squares of the two terms are left.
True or False?
$(3 + x) (3 - x)$ is another way of writing $x^{2} - 9$ .
False.
Expanding $(3 + x) (3 - x)$ gives $9 - x^{2}$ , which is the negative of $x^{2} - 9$ .
The two brackets may swap places with each other, but the terms inside a bracket may not swap.
How can you tell that $4 r^{2} - t^{4}$ is a difference of two squares?
Each term can be rewritten as something squared, since $4 r^{2} = {(2 r)}^{2}$ and $t^{4} = {(t^{2})}^{2}$ .
Any even power is a square, because its index can be halved.
The factorisation is therefore $(2 r + t^{2}) (2 r - t^{2})$ .
What should you do with $2 y^{2} - 50$ before using the difference of two squares?
Take out the common factor of $2$ first, which turns the expression into $2 (y^{2} - 25)$ .
The bracket is now a difference of two squares, giving the fully factorised answer $2 (y + 5) (y - 5)$ .
How can $(2 x + 3) (2 x - 3)$ be expanded in one step?
Brackets in the form $(a + b) (a - b)$ expand straight to $a^{2} - b^{2}$ , so the answer is $4 x^{2} - 9$ .
Spotting the pattern avoids writing out $4 x^{2} - 6 x + 6 x - 9$ and then collecting like terms.
Define a quadratic expression.
A quadratic expression is one in the form $a x^{2} + b x + c$ where $a$ is not zero.
It contains no power of $x$ higher than $2$ , and one with three terms like this is also called a trinomial.
When factorising $x^{2} + b x + c$ which pair of numbers do you look for?
Look for a pair that multiplies to give $c$ and adds to give $b$ .
For $x^{2} - 2 x - 8$ that pair is $+ 2$ and $- 4$ , since $2 \times (- 4) = - 8$ and $2 + (- 4) = - 2$ .
Complete the factorisation by filling in the two missing numbers.
$x^{2} - 4 x - 21 = (x +______) (x -______)$
The completed factorisation is:
$x^{2} - 4 x - 21 = (x + 3) (x - 7)$
Expanding the brackets returns $x^{2} - 7 x + 3 x - 21$ , which confirms the answer.
How is $x^{2} - 5 x + 6$ factorised by splitting the middle term and grouping?
Split $- 5 x$ into $- 2 x - 3 x$ to give $x^{2} - 2 x - 3 x + 6$ , then factorise each pair to reach $x (x - 2) - 3 (x - 2)$ .
The shared bracket $(x - 2)$ is then taken out, leaving $(x - 2) (x - 3)$ .
For a quadratic whose $x^{2}$ coefficient is $1$ which method of factorising is quickest?
Inspection is quickest, because the pair of numbers is written straight into two brackets with no intermediate working.
Grouping and the grid reach the same answer, and grouping is worth learning because it also handles a quadratic whose $x^{2}$ coefficient is not $1$ .
True or False?
Factorising $x^{2} - 7 x + 12$ needs two negative numbers.
True.
Their product $12$ is positive, so the two numbers must have the same sign, and their sum $- 7$ is negative, so both of them are negative.
The pair is $- 3$ and $- 4$ , giving $(x - 3) (x - 4)$ .
How is a grid used to factorise a quadratic rather than to expand one?
Split the middle term and write the four resulting terms inside the grid, then work backwards to find the row and column headings.
Those headings are the two brackets, so for $x^{2} - 2 x - 24$ they come out as $(x + 4) (x - 6)$ .
In $a x^{2} + b x + c$ what changes about the pair of numbers you look for?
The pair must multiply to give $a c$ rather than just $c$ , while still adding to give $b$ .
For $4 x^{2} - 25 x - 21$ that means multiplying to $4 \times (- 21) = - 84$ and adding to $- 25$ , which gives $- 28$ and $+ 3$ .
Complete the split of the middle term, ready for factorising $6 x^{2} - 7 x - 3$ by grouping.
$6 x^{2} +______-______- 3$
The completed split is:
$6 x^{2} + 2 x - 9 x - 3$
Only $+ 2$ and $- 9$ multiply to $- 18$ and add to $- 7$ , so no other pair will do.
After splitting the middle term of $4 x^{2} - 25 x - 21$ how is the factorisation finished?
Factorise each pair of terms fully, which turns $4 x^{2} - 28 x + 3 x - 21$ into $4 x (x - 7) + 3 (x - 7)$ .
The identical bracket $(x - 7)$ is then taken out, giving $(x - 7) (4 x + 3)$ .
When the $x^{2}$ coefficient is not $1$ why can the two numbers not be written straight into two brackets?
The numbers found multiply to $a c$ rather than to $c$ , so they are not the numbers that end up inside the brackets.
They are used to split the middle term, and the brackets then come out of the grouping step instead.
True or False?
The two middle terms can be written in either order once the middle term has been split.
True.
Grouping them the other way round gives $6 x^{2} - 9 x + 2 x - 3$ , which factorises as $3 x (2 x - 3) + 1 (2 x - 3)$ .
That is the same pair of brackets $(2 x - 3) (3 x + 1)$ , simply found in the opposite order.
In a factorising grid where the $x^{2}$ coefficient is not $1$ what goes in the first row heading?
The highest common factor of the two terms in that row goes in the heading.
For a row containing $4 x^{2}$ and $- 28 x$ the heading is $4 x$ , and the two column headings then follow from it.
What should you always check for before choosing a method of factorising?
Check for a common factor of every term first, and take it out before anything else.
For $- 8 x^{2} + 100 x - 48$ taking out $- 4$ gives $- 4 (2 x^{2} - 25 x + 12)$ , which leaves a quadratic with a positive $x^{2}$ term to factorise.
How does the number of terms guide which factorising method to use?
With two terms, check first for a common factor, as in $x^{2} - 7 x = x (x - 7)$ , and then for a difference of two squares, as in $x^{2} - 9$ .
With three terms you have a quadratic, which factorises into two brackets using a pair of numbers.
For a three-term quadratic starting with $a x^{2}$ what decides which method to use?
Whether $a$ is a common factor of all three terms decides it.
In $3 x^{2} + 15 x + 18$ the $3$ can be taken out first, leaving an easier quadratic.
In $3 x^{2} - 5 x - 2$ it cannot, so that one needs grouping and gives $(3 x + 1) (x - 2)$ .
True or False?
The expression $2 x^{2} + 4 x + 2$ can be divided by $2$ to give $x^{2} + 2 x + 1$ .
False.
Dividing an expression by a number changes its value, so $2 x^{2} + 4 x + 2$ and $x^{2} + 2 x + 1$ are not the same thing.
An equation can be divided through, because both sides change together, so $2 x^{2} + 4 x + 2 = 0$ does become $x^{2} + 2 x + 1 = 0$ .
How can you tell in advance whether $a x^{2} + b x + c$ will factorise?
Look for a pair of whole numbers that multiply to $a c$ and add to $b$ , since finding one means the expression factorises.
Working out $b^{2} - 4 a c$ also settles it, because the expression factorises exactly when that value is a square number.
Complete the factorisation of this perfect square by filling in the missing number.
$x^{2} + 10 x + 25 = {(x +______)}^{2}$
The completed factorisation is:
$x^{2} + 10 x + 25 = {(x + 5)}^{2}$
Any quadratic in the form $x^{2} + 2 a x + a^{2}$ is a perfect square and factorises to ${(x + a)}^{2}$ .
After taking out $- 4$ from $- 8 x^{2} + 100 x - 48$ what is left to do?
Factorise the bracket $2 x^{2} - 25 x + 12$ , which needs a pair of numbers multiplying to $24$ and adding to $- 25$ .
That pair is $- 24$ and $- 1$ , giving $(2 x - 1) (x - 12)$ and the full answer $- 4 (2 x - 1) (x - 12)$ .

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