True or False? Factorisation can be thought of as the opposite of expanding brackets .

True. Factorisation can be thought of as the opposite of expanding brackets .

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Factorising (Edexcel IGCSE Maths A)

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FrontFactorising Out Terms
Describe how to factorise simple expressions such as $6 x + 8$ .
BackFactorising Out Terms
To factorise simple expressions like $6 x + 8$ :
Find the highest common factor of 6 and 8 (which is 2)
Write this factor outside a set of brackets
Write inside the brackets what you must multiply the factor by to get the original expression
So $6 x + 8$ becomes $2 (3 x + 4)$ .

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Cards in this collection (28)

Describe how to factorise simple expressions such as $6 x + 8$ .
To factorise simple expressions like $6 x + 8$ :
1. Find the highest common factor of 6 and 8 (which is 2)
2. Write this factor outside a set of brackets
3. Write inside the brackets what you must multiply the factor by to get the original expression
So $6 x + 8$ becomes $2 (3 x + 4)$ .
True or False?
$2 x$ is the highest common factor in the expression $4 x^{2} + 12 x$ .
False.
The highest common factor in $4 x^{2} + 12 x$ is $4 x$ , not $2 x$ .
True or False?
Factorisation can be thought of as the opposite of expanding brackets.
True.
Factorisation can be thought of as the opposite of expanding brackets.
An expression is factorised to get $5 (x - 1)$ .
How can this result be checked?
This result can be checked by expanding the expression.
So $5 (x - 1)$ expands to give $5 x - 5$ .
If that was the original question, then the factorised expression is correct.
True or False?
You can factorise out negative numbers.
E.g. $- 2$ can be factorised out from the expression $- 2 x - 4$ .
True.
You can factorise out negative numbers.
Just be very careful with the signs.
E.g. $- 2 x - 4$ factorises to $- 2 (x + 2)$ .
True or False?
The expression $3 x (5 x + 10)$ is factorised fully.
False.
The expression $3 x (5 x + 10)$ is not factorised fully.
You can still take out a 5 from inside the brackets.
This gives $15 x (x + 2)$ , which is now factorised fully.
True or False?
It is possible to factorise $(x - 3)$ out of the expression $2 (x - 3) + y (x - 3)$ .
True.
It is possible to factorise $(x - 3)$ out of the expression $2 (x - 3) + y (x - 3)$ . You can treat the $(x - 3)$ as if it were a single term.
This gives $(x - 3) (2 + y)$ .
Write down the first step when factorising the expression $x y + 3 y + 2 x + 6$ .
The first step when factorising $x y + 3 y + 2 x + 6$ is to fully factorise the first pair of terms and fully factorise the last pair of terms.
This gives $y (x + 3) + 2 (x + 3)$ .
True or False?
You get the same result when factorising $x y + 3 y + 2 x + 6$ as you do when swapping the middle terms and factorising $x y + 2 x + 3 y + 6$ .
True.
You get the same result when factorising $x y + 3 y + 2 x + 6$ as you do when swapping the middle terms and factorising $x y + 2 x + 3 y + 6$ .
For $x y + 3 y + 2 x + 6$ , you can factorise out $(x + 3)$ .
For $x y + 2 x + 3 y + 6$ , you can factorise out $(y + 2)$ .
Both end up with $(x + 3) (y + 2)$ .
Define a quadratic expression.
A quadratic expression is an expression of the form $a x^{2} + b x + c$ where $a \neq 0$ .
E.g. $x^{2} + 6 x - 2$ is a quadratic expression.
An expression $x^{2} + 8 x + 12$ factorises to $(x + p) (x + q)$ .
Explain how the numbers $p$ and $q$ relate to the numbers $8$ and $12$ .
If $x^{2} + 8 x + 12$ factorises to $(x + p) (x + q)$ , then:
- $p + q = 8$ (the numbers must add to give 8).
- $p q = 12$ (the numbers must multiply to give 12).
True or False?
If $x^{2} - 54 x + 288$ factorises to $(x + p) (x + q)$ then $p$ and $q$ must both be negative.
True.
If $x^{2} - 54 x + 288$ factorises to $(x + p) (x + q)$ , then $p$ and $q$ multiply to give 288, which is positive. That means that $p$ and $q$ could both be positive or both be negative.
But since $p$ and $q$ add to give -54 which is negative, then at least one of them is negative.
The two facts above mean that $p$ and $q$ are both negative.
True or False?
A quadratic expression will always factorise into double brackets.
False.
The quadratic expression $x^{2} + 5 x$ with no constant term factorises to $x (x + 5)$ which is not a double bracket expansion.
True or False?
To factorise $3 x^{2} - 5 x - 2$ , it could first be written in the form $3 x^{2} + x - 6 x - 2$ .
True.
To factorise harder quadratics like $3 x^{2} - 5 x - 2$ , you can:
1. Multiply the first and last numbers together, $3 \times (- 2) = - 6$
2. Find two numbers that add to the x coefficient, -5, and multiply to -6, i.e. 1 and -6
3. Split the middle term into $1 x$ and $- 6 x$
$3 x^{2} - 5 x - 2 = 3 x^{2} + x - 6 x - 2$ .
This then helps to factorise: $x (3 x + 1) - 2 (3 x + 1)$ giving $(3 x + 1) (x - 2)$ .
True or False?
$(5 x + 2) (3 - x)$ and $(2 - 5 x) (x - 3)$ are both correct factorised expressions of $- 5 x^{2} + 13 x + 6$ .
False.
$(5 x + 2) (3 - x)$ is a correct factorised expression of $- 5 x^{2} + 13 x + 6$ .
However, $(2 - 5 x) (x - 3)$ expands to $- 5 x^{2} + 17 x - 6$ .
What should the first step be when factorising the expression $2 x^{2} + 8 x + 6$ ?
When factorising the expression $2 x^{2} + 8 x + 6$ , check whether each term in $2 x^{2} + 8 x + 6$ has a common factor. There is a common factor of 2.
So the first step when factorising $2 x^{2} + 8 x + 6$ should be to factorise out a 2, to get $2 (x^{2} + 4 x + 3)$ .
Do not divide by a 2 and get rid of it; you cannot get rid of numbers in expressions by dividing (you can only do that with equations).
Define the difference of two squares when talking about factorisation.
The difference of two squares says that $a^{2} - b^{2}$ factorises into the double brackets $(a + b) (a - b)$ .
True or False?
$x^{2} - 81$ is $(x + 9) (x - 9)$ but not $(x - 9) (x + 9)$ .
False.
$x^{2} - 81$ can be written as $(x + 9) (x - 9)$ or as $(x - 9) (x + 9)$ , because they both expand to give $x^{2} - 81$ .
True or False?
$x^{2} - 81$ is $(x + 9) (x - 9)$ but not $(9 + x) (9 - x)$ .
True.
$x^{2} - 81$ is $(x + 9) (x - 9)$ but not $(9 + x) (9 - x)$ . The second one expands to give $81 - x^{2}$ , not $x^{2} - 81$ .
True or False?
It is impossible to factorise a quadratic expression with no middle term in $x$ into double brackets.
False.
The quadratic expression $x^{2} - 9$ has no middle term in $x$ but factorises into $(x + 3) (x - 3)$ using the difference of two squares.
Explain how to use the difference of two squares to factorise $4 x^{2} - 25$ .
To factorise $4 x^{2} - 25$ using difference of two squares, the $4 x^{2}$ can be thought of as ${(2 x)}^{2}$ . So $4 x^{2} - 25$ is ${(2 x)}^{2} - 5^{2}$ .
The difference of two squares can then be used where $a = 2 x$ and $b = 5$ , giving $(2 x + 5) (2 x - 5)$ .
Explain how to use the difference of two squares to factorise $5 x^{2} - 45$ .
To factorise $5 x^{2} - 45$ using difference of two squares, first factorise out the 5 to get $5 (x^{2} - 9)$ .
Then use the difference of two squares for the $x^{2} - 9$ part.
This gives $5 (x + 3) (x - 3)$ .
A calculator gives the solutions to $2 x^{2} + 7 x + 3 = 0$ as $- 3$ and $- \frac{1}{2}$ .
Explain why this tells you that $2 x^{2} + 7 x + 3$ can be factorised into double brackets.
If the solutions to a quadratic equation are integers or (rational) fractions, then the quadratic factorises.
The solutions are $- 3$ and $- \frac{1}{2}$ which are integers or fractions, so it must factorise.
The factorisation is $2 x^{2} + 7 x + 3 = (2 x + 1) (x + 3)$ .
True or False?
The value of $b^{2} - 4 a c$ from the quadratic formula for the equation $2 x^{2} + 7 x + 3 = 0$ is equal to $25$ .
Therefore the expression $2 x^{2} + 7 x + 3$ can be factorised.
True.
If the value of $b^{2} - 4 a c$ from the quadratic formula is a positive square number, then the quadratic expression factorises.
As $b^{2} - 4 a c = 25$ and 25 is a square number, $2 x^{2} + 7 x + 3$ must factorise.
The factorisation is $2 x^{2} + 7 x + 3 = (2 x + 1) (x + 3)$ .
True or False?
Quadratic expressions with only two terms can always be factorised.
False.
Quadratic expressions with only two terms can not always be factorised.
E.g. the quadratic expression $x^{2} + 4$ cannot be factorised.
True or False?
The expression $x^{2} - 6 x$ can be simplified to $x - 6$ by dividing through by $x$ .
False.
$x^{2} - 6 x$ is an expression, not an equation, so you cannot divide both sides by $x$ (because there are not two sides!).
Instead, you can factorise out an $x$ to get $x (x - 6)$ .
You cannot simplify this any further.
True or False?
If $3 x^{2} + 4 x + 1$ factorises to $(3 x + 1) (x + 1)$ then multiplying by 2 means that $6 x^{2} + 8 x + 2$ factorises to $(6 x + 2) (2 x + 2)$ .
False.
If $3 x^{2} + 4 x + 1$ factorises to $(3 x + 1) (x + 1)$ then multiplying by 2 means that $6 x^{2} + 8 x + 2$ factorises to either $(6 x + 2) (x + 1)$ , where the 2 is taken into the first bracket, or $(3 x + 1) (2 x + 2)$ , where the 2 is taken into the second bracket.
You cannot put a 2 into both the first and second brackets to get $(6 x + 2) (2 x + 2)$ . That would be multiplying $3 x^{2} + 4 x + 1$ by 4.
Both correct versions factorise further to $2 (3 x + 1) (x + 1)$ .
True or False?
A calculator gives the solutions to $x^{2} - 4 x + 1 = 0$ as $2 + \sqrt{3}$ and $2 - \sqrt{3}$ .
This means that the expression $x^{2} - 4 x + 1$ can be written as $(x - p) (x - q)$ where $p$ and $q$ are integers.
False.
If the solutions to a quadratic equation are integers or (rational) fractions, then the quadratic factorises.
The solutions are $2 + \sqrt{3}$ and $2 - \sqrt{3}$ which are neither integers nor fractions, so it does not factorise.
$x^{2} - 4 x + 1$ cannot be written in the form $(x - p) (x - q)$ where $p$ and $q$ are integers.

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1. Topic 1
  Vectors
2. Topic 2
  Transformations

Factorising (Edexcel IGCSE Maths A)

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