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.

What ordinary arithmetic method is polynomial division modelled on?

Long division , the "bus stop" method used for dividing numbers. It splits a polynomial into a factor pair, with or without a remainder left at the end.

True or False? If a polynomial division leaves a remainder, the divisor is not a factor.

True. A factor divides exactly and leaves zero, so any non-zero remainder shows that the divisor is not a factor.

Once you have written a term on top of a polynomial division, what are the next two moves?

Multiply the whole divisor by that term, and subtract the result from what you are dividing. Then bring down the terms you have not yet used and repeat with the new leading term.

What is the factor theorem mainly used for in this course?

Factorising polynomials. It finds one linear factor quickly, with a single substitution and no division at all.

Which two techniques does factorising a cubic combine?

The factor theorem , to find one linear factor, and polynomial division , to get the rest. Neither is enough alone: the theorem gives a factor but not the quotient, and division needs a divisor before it can start.

Can the same method be used on a polynomial of degree higher than three?

Yes, because each linear factor found reduces the degree by one. A quartic simply needs the find-a-factor-then-divide cycle carried out twice before a quadratic is left.

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Polynomials (Edexcel A Level Maths: Pure): Flashcards

Exam code: 9MA0

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

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Cards in this collection (23)

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 polynomial.
A polynomial is an algebraic expression with a finite number of terms, in which every index is a non-negative integer.
So $3 x + 5$ and even the plain number $8$ are polynomials, but $\sqrt{x}$ , $5 x^{- 3}$ and $\frac{3}{x}$ are not.
What ordinary arithmetic method is polynomial division modelled on?
Long division, the "bus stop" method used for dividing numbers.
It splits a polynomial into a factor pair, with or without a remainder left at the end.
Dividing $x^{3} + 6 x^{2} - 9 x - 14$ by $x + 3$ gives a quotient of $x^{2} + 3 x - 18$ , so:
$x^{3} + 6 x^{2} - 9 x - 14 = (x + 3) (x^{2} + 3 x - 18) +______$
$x^{3} + 6 x^{2} - 9 x - 14 = (x + 3) (x^{2} + 3 x - 18) + 40$
Written this way the division shows the polynomial as a product plus a leftover, exactly as $17 = 5 \times 3 + 2$ does for numbers.
When dividing a polynomial, which term do you deal with first?
The highest power of the dividend, compared against the highest power of the divisor.
Dividing $x^{3} + 6 x^{2} - 9 x - 14$ by $x - 2$ , it takes $x^{2}$ to make $x^{3}$ , so $x^{2}$ goes on top.
True or False?
If a polynomial division leaves a remainder, the divisor is not a factor.
True.
A factor divides exactly and leaves zero, so any non-zero remainder shows that the divisor is not a factor.
Once you have written a term on top of a polynomial division, what are the next two moves?
Multiply the whole divisor by that term, and subtract the result from what you are dividing.
Then bring down the terms you have not yet used and repeat with the new leading term.
Define the factor theorem.
For a polynomial $f (x)$ : if $f (p) = 0$ then $(x - p)$ is a factor, and if $(x - p)$ is a factor then $f (p) = 0$ .
It holds in both directions, which is what makes it usable either way round.
If $h (- 2) = 0$ , which expression is a factor of $h (x)$ ?
$(x + 2)$ , because $x - (- 2) = x + 2$ .
The theorem is stated with a minus sign, so a negative value of $p$ produces a plus inside the bracket.
You are told that $(x + 2)$ is a factor of $h (x)$ . What can you write down straight away?
That $h (- 2) = 0$ .
Substituting $x = - 2$ must give zero, which can be used to confirm the factor or to find an unknown coefficient.
True or False?
If $f (2) = 5$ , then $(x - 2)$ is not a factor of $f (x)$ .
True.
The theorem requires $f (p)$ to be zero, so any non-zero value rules that bracket out.
What is the factor theorem mainly used for in this course?
Factorising polynomials.
It finds one linear factor quickly, with a single substitution and no division at all.
Which two techniques does factorising a cubic combine?
The factor theorem, to find one linear factor, and polynomial division, to get the rest.
Neither is enough alone: the theorem gives a factor but not the quotient, and division needs a divisor before it can start.
What is the goal when fully factorising a polynomial?
To write it as a product of linear factors, taken as far as it will go.
For example $x^{3} + 4 x^{2} - 11 x - 30 = (x + 2) (x - 3) (x + 5)$ .
What is the first move in factorising a cubic $f (x)$ ?
Find a value $p$ for which $f (p) = 0$ .
Until you have one factor there is nothing to divide by, so this has to come first.
True or False?
Every cubic can be written as a product of three linear factors.
False.
If the quadratic left after dividing does not factorise, the answer stops at one linear factor times a quadratic.
For example $2 x^{3} + 3 x^{2} + 10 x - 6 = (2 x - 1) (x^{2} + 2 x + 6)$ .
Factorising $x^{3} + 6 x^{2} - 9 x - 14$ : since $f (- 1) = 0$ , dividing by $(x + 1)$ leaves $x^{2} + 5 x - 14$ , which factorises as:
$(x +______) (x -______)$
$(x + 7) (x - 2)$
So the full factorisation is $x^{3} + 6 x^{2} - 9 x - 14 = (x + 1) (x + 7) (x - 2)$ .
Can the same method be used on a polynomial of degree higher than three?
Yes, because each linear factor found reduces the degree by one.
A quartic simply needs the find-a-factor-then-divide cycle carried out twice before a quadratic is left.

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