In a polynomial division you have just written the next term of the answer above the line. What are the two things you do with it?

Multiply the whole divisor by it, then subtract that product from what is left of the polynomial. Anything the subtraction has not touched is carried straight down, ready for the next term to be found.

Apart from factorising, what else is polynomial division used for?

Handling improper algebraic fractions, the top-heavy ones whose numerator has a degree at least as large as the denominator's. The same long-division method applies to them unchanged.

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.

International A Level (IAL)MathsEdexcelPure 2FlashcardsAlgebra & FunctionsPolynomials

Polynomials (Edexcel International A Level (IAL) Maths: Pure 2): Flashcards

Exam code: YMA01

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FrontPolynomial Division
Define polynomial.

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Cards in this collection (22)

Define polynomial.
A polynomial is an algebraic expression made up of a finite number of terms, in which every index is a non-negative integer.
So $3 x + 5$ and $2 x^{2} y - 4 y + 6$ are polynomials, while $5 x^{- 3}$ and $\sqrt{x}$ are not.
What does polynomial division do to a polynomial?
It splits it into a factor pair, two expressions that multiply together to give it.
For example $x^{3} + x^{2} - x - 1 = (x + 1) (x^{2} - 1)$ , where $(x + 1)$ is the divisor and $(x^{2} - 1)$ is the result.
True or False?
Polynomial division can only be carried out when the divisor is a factor of the polynomial.
False.
The division works either way, and when the divisor is not a factor it simply finishes with a remainder, exactly as dividing $17$ by $5$ does.
When dividing a polynomial by $(x - p)$ , how do you decide the first term of the answer?
Compare the highest power term of the polynomial with the highest power term of the divisor.
Dividing $x^{3} + 6 x^{2} - 9 x - 14$ by $(x - 2)$ starts with $x^{2}$ , because $x^{2} \times x = x^{3}$ .
In a polynomial division you have just written the next term of the answer above the line. What are the two things you do with it?
Multiply the whole divisor by it, then subtract that product from what is left of the polynomial.
Anything the subtraction has not touched is carried straight down, ready for the next term to be found.
$x^{3} + 6 x^{2} - 9 x - 14$ is divided by $(x - 2)$ . Complete the result:
$x^{3} + 6 x^{2} - 9 x - 14 = (x - 2) (x^{2} +______x +______)$
The completed result is:
$x^{3} + 6 x^{2} - 9 x - 14 = (x - 2) (x^{2} + 8 x + 7)$
Multiplying the two brackets back out returns the original polynomial, which is how a division is checked.
Apart from factorising, what else is polynomial division used for?
Handling improper algebraic fractions, the top-heavy ones whose numerator has a degree at least as large as the denominator's.
The same long-division method applies to them unchanged.
Define the factor theorem.
For a polynomial $f (x)$ , if $f (p) = 0$ then $(x - p)$ is a factor of $f (x)$ , and if $(x - p)$ is a factor then $f (p) = 0$ .
It works in both directions, so it can be used either to find a factor or to deduce a value from one you are given.
True or False?
If $(x + 2)$ is a factor of $h (x)$ , then $h (2) = 0$ .
False.
The factor has to be read as $(x - (- 2))$ , so the value that matters is $- 2$ and it is $h (- 2)$ that is zero.
A factor written as $(x + p)$ always pairs with the value $- p$ .
What does the remainder theorem tell you?
That dividing a polynomial $f (x)$ by $(x - a)$ leaves a remainder of $f (a)$ .
Written formally, $f (x) = (x - a) Q (x) + f (a)$ .
Name the three parts of a polynomial division, where $f (x) = (x - a) Q (x) + R$ :
$(x - a)$ is the $______$ , $Q (x)$ is the $______$ , and $R$ is the $______$ .
The completed sentence is:
$(x - a)$ is the divisor, $Q (x)$ is the quotient, and $R$ is the remainder.
The quotient is the expression built up on top of the division as it is carried out.
How is the factor theorem a special case of the remainder theorem?
It is the case in which the remainder is zero, so the division comes out exactly.
Putting $f (a) = 0$ leaves $f (x) = (x - a) Q (x)$ , which says precisely that $(x - a)$ is a factor.
What is the remainder when $x^{2} - 2 x$ is divided by $(x - 3)$ , and how do you get it without dividing?
Substitute $x = 3$ into the polynomial: $3^{2} - 2 \times 3 = 3$ .
Whenever the remainder is all that is wanted, this replaces the entire division with a single substitution.
True or False?
When a polynomial is divided by $(x - a)$ , the remainder is always a number rather than an expression in $x$ .
True.
The division is written with everything involving $x$ collected into the bracketed quotient, leaving a single value outside it.
So $x^{3} + 2 x^{2} + 3 x + 4 = (x - 1) (x^{2} + 3 x + 6) + 10$ , and the remainder is simply $10$ .
The remainder when $x^{2} + p x$ is divided by $(x - 2)$ is $8$ . How do you find $p$ ?
Set the value of the polynomial at $x = 2$ equal to the remainder and solve: $2^{2} + 2 p = 8$ .
That gives $p = 2$ , and where a question carries more than one unknown the same move produces simultaneous equations.
What is the remainder when $f (x)$ is divided by $(a x - b)$ ?
$f (\frac{b}{a})$ , the value of the polynomial at the number that makes the divisor zero.
The shortcut is unchanged, but that number is no longer necessarily a whole one.
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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