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

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

IGCSEMaths A: HigherEdexcelFlashcards

Factorising (Edexcel IGCSE Maths A)

Flashcards

1/27

FrontFactorising
Describe how to factorise simple expressions such as $6 x + 8$ .
BackFactorising
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 (27)

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?
The highest common factor in $4 x^{2} + 12 x$ is $2 x$ .
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 to expanding brackets.
True.
Factorisation can be thought of as the opposite to expanding brackets.
A student factorises an expression to get $5 (x - 1)$ , but they are not sure if they are correct.
What could they do to check?
To check if they are correct, they could expand their answer.
So $5 (x - 1)$ expands to give $5 x - 5$ . If that was the original question, then they are correct.
True or False?
You can factorise out negative numbers, such as a $- 2$ in the expression $- 2 x - 4$ .
True.
You can factorise out negative numbers, such as a $- 2$ in the expression $- 2 x - 4$ . Just be very careful with the signs.
$- 2 x - 4$ 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 $x y + 2 x + 3 y + 6$ .
The first way involves factorising out $(x + 3)$ and the second way involves factorising out $(y + 2)$ .
Both end up with $(x + 3) (y + 2)$ .
Define a quadratic expression.
A quadratic expression is one of the form $a x^{2} + b x + c$ where $a \neq 0$ .
If $x^{2} + 8 x + 12$ 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)
Explain how it is possible to see, without factorising, that if $x^{2} - 54 x + 288$ factorises to $(x + p) (x + q)$ then $p$ and $q$ must both be negative.
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 $p$ 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 $5 (x + 3) (x - 3)$ .
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 -5 (the x coefficient) 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 $(3 x + 1) (x - 2)$ .
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 equations 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 $2 x^{2} + 7 x + 3 = (2 x + 1) (x + 3)$ .
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 $25$ .
Explain why this tells you that $2 x^{2} + 7 x + 3$ can be factorised.
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.
The quadratic expression $x^{2} + 4$ cannot be factorised, for example.
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.
What should be the first step 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 divide expressions by numbers (you can only do that with equations).
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 $2 (3 x + 1) (x + 1)$ .
A calculator gives the solutions to $x^{2} - 4 x + 1 = 0$ as $2 + \sqrt{3}$ and $2 - \sqrt{3}$ .
Explain whether or not $x^{2} - 4 x + 1$ can be written as $(x - p) (x - q)$ where $p$ and $q$ are integers.
If the solutions to a quadratic equations 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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