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 .

GCSEMathsEdexcelFoundationFlashcardsAlgebraFactorising

Factorising (Edexcel GCSE Maths: Foundation): Flashcards

Exam code: 1MA1

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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 $2 (3 x + 4)$ .
BackFactorising Out Terms
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Cards in this collection (16)

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.
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.
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 $5 (x + 3) (x - 3)$ .

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