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 .

IGCSEMathsCambridge (CIE)International MathsCoreFlashcardsAlgebra & SequencesExpanding & Factorising Brackets

Expanding & Factorising Brackets (Cambridge (CIE) IGCSE International Maths: Core): Flashcards

Exam code: 0607

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FrontExpanding & Simplifying Single Brackets
Explain how to expand a single bracket.
E.g. $2 (x + 3)$ .

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Cards in this collection (17)

Explain how to expand a single bracket.
E.g. $2 (x + 3)$ .
Expanding a bracket means multiplying the term outside the bracket by each term inside the bracket.
So $2 (x + 3)$ becomes $2 x + 6$ from $2 \times x + 2 \times 3$ .
True or False?
$- 2 (1 - x)$ expands to $- 2 - 2 x$ .
False.
$- 2 (1 - x)$ can be thought of as $(- 2) \times 1 + (- 2) \times (- x)$ .
Multiplying two negatives gives a positive, so the answer is $- 2 + 2 x$ .
How can you simplify $3 (x + 5 + x)$ before expanding the brackets?
You can simplify $3 (x + 5 + x)$ before expanding by first collecting like terms inside the brackets to give $3 (2 x + 5)$ .
This can also be written as $3 (5 + 2 x)$ .
Explain how to simplify an expression with two sets of brackets
E.g. $2 (x + 1) + 3 (x - 1)$ .
To simplify expressions like $2 (x + 1) + 3 (x - 1)$ you expand the brackets then collect like terms.
So $2 (x + 1) + 3 (x - 1)$ expands to give $2 x + 2 + 3 x - 3$ .
The terms are then collected to give $5 x - 1$ .
True or False?
$3 (x + 2)$ expands to $3 x + 6$ , then the 3's cancel to give $x + 2$ .
False.
You cannot cancel the 3's because $3 (x + 2)$ is an expression, not an equation.
You can only cancel both sides by 3 if you had an equation, like $3 (x + 2) = 9$ .
How do you find the highest power of $x$ when expanding expressions?
E.g. $2 x^{2} (x + x^{2})$ .
The highest power of $x$ will come from multiplying the outside term by the inside term with the highest power.
The highest power of $x$ in $2 x^{2} (x + x^{2})$ is $x^{4}$ .
This can be found by multiplying $x^{2}$ outside the bracket by the highest power, $x^{2}$ , from inside the bracket.
How many multiplications are needed to expand $(x + 1) (x + 3)$ , and why?
Four, because every term in the first bracket must be multiplied by every term in the second bracket.
The four products here are $x \times x$ , $x \times 3$ , $1 \times x$ and $1 \times 3$ .
Complete the expansion by filling in the missing numbers:
$(2 x - 3) (x + 4) = 2 x^{2} +______x -______$
The completed expansion is:
$(2 x - 3) (x + 4) = 2 x^{2} + 5 x - 12$
The two middle products, $8 x$ and $- 3 x$ , are like terms, so they combine into the single term $5 x$ .
When expanding $(x - 3) (3 x - 5)$ using FOIL, which two products are the Outside and Inside ones?
Outside is $x \times (- 5)$ , the first term of the first bracket multiplied by the last term of the second bracket.
Inside is $(- 3) \times 3 x$ , the last term of the first bracket multiplied by the first term of the second bracket.
True or False?
${(x + y)}^{2}$ expands to $x^{2} + y^{2}$ .
False.
The expansion of ${(x + y)}^{2}$ is $x^{2} + x y + x y + y^{2}$ , which simplifies to $x^{2} + 2 x y + y^{2}$ .
The two $x y$ cross terms are exactly what $x^{2} + y^{2}$ leaves out.
What is the first thing to do when expanding ${(x + 3)}^{2}$ ?
Write the square as two brackets multiplied together, $(x + 3) (x + 3)$ .
It can then be expanded in exactly the same way as any other pair of brackets.
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.

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