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

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Which two techniques does factorising a cubic combine?

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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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