After subtracting, what do you do with the terms you have not yet used?

Bring them down and carry on, treating what is left as a new polynomial of lower degree and dividing that by the same divisor. The cycle of divide, multiply and subtract simply repeats, one power at a time, until nothing further can be divided.

True or False? Dividing a cubic by a linear divisor always leaves a quadratic quotient.

True. Multiplying a quadratic by a linear expression gives a cubic, so the quotient that has to pair with a linear divisor to rebuild a cubic must itself be a quadratic. Any remainder sits separately at the end and does not change that.

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.

Define rational expression .

A rational expression is an algebraic fraction: one polynomial divided by another. The name comes from ratio , in the same way that a rational number is a ratio of two integers.

How do you simplify a rational expression?

Factorise the numerator and the denominator , then cancel any factors common to both. Nothing can be cancelled until each is written as a product.

What happens if the numerator and denominator of a rational expression share no common factor?

It is already in its simplest form and cannot be reduced any further. Factorising both is still worth doing, because that is the only way to be certain nothing cancels.

How do you split an improper algebraic fraction into a quotient and a remainder?

Divide the numerator by the denominator , using algebraic division. The quotient becomes the whole part of the answer, and the remainder goes back over the original denominator.

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Polynomials (Cambridge (CIE) A Level Maths: Pure 3): Flashcards

Exam code: 9709

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FrontPolynomial Division
Define polynomial.

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Cards in this collection (30)

Define polynomial.
A polynomial is an algebraic expression made up of a finite number of terms whose indices are all non-negative integers.
So $2 x^{2} y - 4 y + 6$ is a polynomial, but $\sqrt{x}$ and $\frac{3}{x}$ are not, because their indices are $\frac{1}{2}$ and $- 1$ .
When dividing one polynomial by another, which term of the answer comes first?
You find the highest power term first, by asking what the highest power of the divisor must be multiplied by to give the highest power of the dividend.
Dividing $x^{3} + 6 x^{2} - 9 x - 14$ by $x - 2$ starts with $x^{2}$ , because $x^{2} \times x = x^{3}$ .
Complete this division, which does not come out exactly:
$x^{3} + 2 x^{2} + 3 x + 4 = (x - 1) (x^{2} +______x + 6) +______$
The completed identity is:
$x^{3} + 2 x^{2} + 3 x + 4 = (x - 1) (x^{2} + 3 x + 6) + 10$
The $10$ is the remainder, and it is left over precisely because $(x - 1)$ is not a factor of the polynomial.
After subtracting, what do you do with the terms you have not yet used?
Bring them down and carry on, treating what is left as a new polynomial of lower degree and dividing that by the same divisor.
The cycle of divide, multiply and subtract simply repeats, one power at a time, until nothing further can be divided.
True or False?
Dividing a cubic by a linear divisor always leaves a quadratic quotient.
True.
Multiplying a quadratic by a linear expression gives a cubic, so the quotient that has to pair with a linear divisor to rebuild a cubic must itself be a quadratic.
Any remainder sits separately at the end and does not change that.
What degree is the quotient when a degree $n$ polynomial is divided by a quadratic?
The quotient has degree $n - 2$ , two lower than the polynomial you started with.
The degree drops by the degree of the divisor, so a quartic divided by a quadratic leaves a quadratic quotient.
What are the possible degrees of the remainder when dividing by a quadratic?
The remainder must have degree less than $2$ , so it is either linear, of the form $a x + b$ , or a constant.
If anything of degree $2$ or more were left, the divisor would still go into it, so the division would not yet be finished.
What does the factor theorem say about a polynomial $f$ and a number $p$ ?
It says that $f (p) = 0$ and $(x - p)$ being a factor of $f (x)$ are two ways of saying the same thing.
The link runs both ways: a zero tells you there is a factor, and a factor tells you there is a zero.
To test whether $(x + 2)$ is a factor of $f (x)$ , what do you work out?
Work out $f (- 2)$ , not $f (2)$ .
The theorem is stated for a bracket of the form $(x - p)$ , and $(x + 2)$ is $(x - (- 2))$ , so here $p = - 2$ .
If the result is zero, the bracket is a factor.
True or False?
The factor theorem is a special case of the remainder theorem.
True.
Whatever remainder the remainder theorem gives on division by $(x - a)$ , having a factor is exactly the case where that remainder comes out as zero.
Writing $f (x) = (x - a) Q (x) + remainder$ shows it: with a zero remainder, $(x - a)$ multiplies up to give $f (x)$ exactly.
What do you learn if $f (1)$ turns out not to be zero?
You learn that $(x - 1)$ is not a factor of $f (x)$ , and nothing more than that.
It rules out that single bracket and says nothing about any other value, which is why you simply move on and try the next one.
Complete the shortcut for the remainder when $f (x)$ is divided by $(2 x - 1)$ :
$remainder = f (______)$
The completed shortcut is:
$remainder = f (\frac{1}{2})$
You substitute the value of $x$ that makes the divisor zero, and $2 x - 1 = 0$ gives $x = \frac{1}{2}$ , which need not be a whole number.
Why is the remainder theorem quicker than carrying out the division?
Because it produces the remainder from a single substitution, with no division to set out at all.
Finding the remainder of $x^{2} - 2 x$ on division by $(x - 3)$ takes only $3^{2} - 2 \times 3 = 3$ .
A polynomial's remainder on division by $(x - 2)$ is known. How does that help you find an unknown coefficient?
Substituting $x = 2$ turns the known remainder into an equation in the unknown coefficient, which you can then solve.
If $(x^{2} + p x) \div (x - 2)$ leaves remainder $8$ , then $2^{2} + p (2) = 8$ , giving $p = 2$ .
With two unknowns you need two such remainders and solve the resulting pair simultaneously.
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.
Define rational expression.
A rational expression is an algebraic fraction: one polynomial divided by another.
The name comes from ratio, in the same way that a rational number is a ratio of two integers.
How do you simplify a rational expression?
Factorise the numerator and the denominator, then cancel any factors common to both.
Nothing can be cancelled until each is written as a product.
True or False?
$\frac{x^{2} + 3 x}{x^{2} + 5}$ simplifies to $\frac{3 x}{5}$ .
False.
Only common factors can be cancelled, and $x^{2}$ is a term in a sum here, not a factor.
The numerator does factorise, to $x (x + 3)$ , but the denominator has no matching factor, so nothing cancels at all.
Cancel the common factors:
$\frac{(x - 1) (x + 3) (x - 2)}{(x + 3) (x - 1)} =______$
$\frac{(x - 1) (x + 3) (x - 2)}{(x + 3) (x - 1)} = x - 2$
Both $(x + 3)$ and $(x - 1)$ appear top and bottom, so both go, leaving a single bracket.
What happens if the numerator and denominator of a rational expression share no common factor?
It is already in its simplest form and cannot be reduced any further.
Factorising both is still worth doing, because that is the only way to be certain nothing cancels.
Define improper algebraic fraction.
An algebraic fraction in which the degree of the numerator is greater than or equal to the degree of the denominator.
So $\frac{x^{3} + 2 x^{2} - x + 4}{x - 5}$ is improper, being degree $3$ over degree $1$ .
Any improper algebraic fraction can be written as:
$\frac{p (x)}{a x + b} \equiv______+ \frac{______}{a x + b}$
$\frac{p (x)}{a x + b} \equiv q (x) + \frac{r}{a x + b}$
Here $q (x)$ is the quotient and $r$ the remainder, and the remainder keeps the original denominator underneath it.
What is the arithmetic equivalent of writing an improper algebraic fraction as a quotient and a remainder?
Turning a top-heavy fraction into a mixed number.
$\frac{17}{5} = 3 \frac{2}{5}$ is exactly the same move: divide, then write whatever is left over as a fraction.
How do you split an improper algebraic fraction into a quotient and a remainder?
Divide the numerator by the denominator, using algebraic division.
The quotient becomes the whole part of the answer, and the remainder goes back over the original denominator.
True or False?
$\frac{x}{x + 1}$ is an improper algebraic fraction.
True.
Numerator and denominator are both degree 1, and the definition covers the case where the degrees are equal.
Cases like this are the easiest to miss, and this one can be rewritten as $1 - \frac{1}{x + 1}$ .

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