What does FOIL stand for, and when can you use it?

First, Outside, Inside, Last. It is the each-term-times-each-term rule carried out in a fixed order. It only applies when both brackets contain exactly two terms.

How do you expand three or more sets of brackets?

Two at a time. Expand and simplify the first pair, then multiply that result by the next bracket, and so on.

International A Level (IAL)MathsEdexcelPure 1FlashcardsAlgebra & FunctionsPolynomials

Polynomials (Edexcel International A Level (IAL) Maths: Pure 1): Flashcards

Exam code: YMA01

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FrontExpanding Brackets
What is the rule for expanding two brackets multiplied together?

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What is the rule for expanding two brackets multiplied together?
Every term in one bracket must be multiplied by every term in the other.
So $(a + b) (x + y + z)$ gives six products: $a x + a y + a z + b x + b y + b z$ .
${(a + b)}^{2}$ expands to $a^{2} +______+ b^{2}$ .
${(a + b)}^{2}$ expands to $a^{2} + 2 a b + b^{2}$ .
The middle term comes from the two cross-products, $a b$ and $b a$ , which is exactly what is lost if you square each term separately.
What does FOIL stand for, and when can you use it?
First, Outside, Inside, Last. It is the each-term-times-each-term rule carried out in a fixed order.
It only applies when both brackets contain exactly two terms.
True or False?
${(a + b)}^{3}$ can be expanded by writing the bracket out three times and multiplying.
True.
A cube is simply the bracket multiplied by itself three times, and at that size writing it out is perfectly practical.
How do you expand three or more sets of brackets?
Two at a time.
Expand and simplify the first pair, then multiply that result by the next bracket, and so on.
When is it worth using the binomial expansion instead of writing brackets out?
When the power is large, such as ${(a + b)}^{7}$ .
Writing out and multiplying that many brackets is impractical.
Define factorising.
Factorising is rewriting an expression that is a sum of terms as a product of factors.
It is the reverse of expanding: $x^{2} + 6 x - 16$ is a sum of three terms, and $(x + 8) (x - 2)$ is the product of two linear factors.
Whatever kind of expression you are factorising, what should you always check for first?
A factor common to every term, which may be a number, a letter, or both.
Taking it out first leaves a simpler expression to factorise: $10 x^{2} + 5 x = 5 x (2 x + 1)$ .
Complete this difference of two squares by filling in the two missing terms:
$9 x^{2} - 49 y^{2} = (______- 7 y) (3 x +______)$
The completed factorisation is:
$9 x^{2} - 49 y^{2} = (3 x - 7 y) (3 x + 7 y)$
Each bracket holds the square roots of the two terms, and the brackets differ only in the sign between them.
In the 'ac' method for factorising $a x^{2} + b x + c$ , which two numbers are you looking for?
Two numbers whose product is $a c$ and whose sum is $b$ .
For $6 x^{2} + 7 x - 3$ that means a product of $- 18$ and a sum of $7$ , so the numbers are $9$ and $- 2$ .
True or False?
The 'ac' method will factorise any quadratic expression that factorises at all, whatever the coefficient of $x^{2}$ .
True.
Most factorising shortcuts only apply under particular conditions, such as the coefficient of $x^{2}$ being $1$ , but the 'ac' method carries no such restriction.
It is most useful exactly where those shortcuts are hardest to use, when that coefficient is greater than $1$ and not prime.
You have found the two numbers $m$ and $n$ needed by the 'ac' method on $a x^{2} + b x + c$ . What do you do with them?
Split the middle term, writing $b x$ as $m x + n x$ .
Factorising the first two terms and the last two terms then leaves a common bracket, which comes out as one of the factors: $4 x^{2} + 20 x - x - 5 = 4 x (x + 5) - (x + 5) = (x + 5) (4 x - 1)$ .
Why can you always take out a factor of $x$ when factorising a cubic expression at this level?
Because a cubic at this level has no constant term, so every term contains at least one $x$ .
What is left inside the bracket is a quadratic, which may itself factorise again: $4 x^{3} + 22 x^{2} + 30 x = 2 x (2 x^{2} + 11 x + 15) = 2 x (2 x + 5) (x + 3)$ .

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