POLYNOMIALS: THE REMAINDER AND FACTOR THEOREMS
We name polynomials according to their highest degree of power. e.g. if a polynomial’s
highest degree was 𝑥 𝑛 then it is a polynomial of degree 𝒏. We always write polynomials in
descending order of power.
𝑥 4 − 5𝑥 2 + 6𝑥 − 3 is a polynomial of degree 4
A linear polynomial is a polynomial of degree 1 (𝑒. 𝑔. 5𝑥 − 6) while a quadratic polynomial is
a polynomial of degree 2 (𝑒. 𝑔. 𝑥 2 − 7𝑥 + 2).
A cubic polynomial is a polynomial of degree 3 (𝑒. 𝑔. 3𝑥 3 − 5𝑥 2 + 4𝑥 − 9).
AIM: In another few lessons we want to be able to sketch the graph of a cubic function. To
sketch any graph we almost invariably start with the 𝑥 − 𝑎𝑛𝑑 𝑦 − intercepts.
Consider the function 𝑦 = 3𝑥 3 − 5𝑥 2 + 4𝑥 − 9. If we wanted to find the 𝑦 − intercepts of
the function, we would set 𝑥 = 0. This would give us 𝑦 = −9. So far no problem.
However, if we want to find the 𝑥 − intercepts we need to set 𝑦 = 0. This gives us:
3𝑥 3 − 5𝑥 2 + 4𝑥 − 9 = 0 , to find the 𝑥 −values where 𝑦 = 0 we need to be able to factorise
3𝑥 3 − 5𝑥 2 + 4𝑥 − 9 . We can guess that this may be some binomial times a trinomial:
(𝑝𝑥 + 𝑞)(𝑎𝑥 2 + 𝑏𝑥 + 𝑐)
We would have little issue factorising the trinomial if only we could factorise our original
expression to mirror this form… and to do that we would need to know what factor “𝑝𝑥 + 𝑞”
to take out.
The objective of this lesson and the next is to be able to factorise cubic functions.
1
, Remainder Theorem:
When we take the quotient of two numbers, the denominator EITHER divides perfectly into
the numerator OR it does not divide perfectly and so we are left with a remainder. The
remainder is the part of the numerator which cannot ‘fit’ the denominator.
Consider the example 27 ÷ 6; 6 goes into 27, 4 times. This means that of the initial 27, 24 can
be made up by groupings of 6, after which there is a remainder of 3 which cannot be formed
into a group of 6.
The same rule applies to polynomials:
If a polynomial 𝑓(𝑥) is divided by a linear polynomial 𝒂𝑥 + 𝒃, then the remainder (𝑟) is:
𝒃
𝑟 = 𝑓 (− )
𝒂
𝒃
Where exactly does − come from?
𝒂
Glad you asked…set 𝑎𝑥 + 𝑏 = 𝟎. And solve for 𝑥.
𝒂𝑥 = −𝒃
𝒃
𝑥=−
𝒂
Example:
Use the Remainder Theorem to determine the remainder when
𝒇(𝒙) = 𝟓𝒙𝟑 − 𝟔𝒙𝟐 + 𝟐𝒙 − 𝟑 is divided by:
𝒂) 𝟐𝒙 − 𝟔
𝒃) 𝒙 − 𝟒
(a) 𝑓(𝑥) is being divided by 2𝑥 – 6
−6
so the remainder is equal to 𝑓 (− ) = 𝑓(3)
2
𝑓(𝑥) has become 𝑓(3); we have substituted
3 2
𝑓(3) = 5(3) − 6(3) + 2(3) − 3 3 in for 𝑥. We must therefore do the same
= 135 − 54 + 6 − 3 for the rest of the equation.
= 84
This means that when we divide 5𝑥 3 − 6𝑥 2 + 2𝑥 − 3 by 2𝑥 − 6, the remainder is the
constant value of 84
2
We name polynomials according to their highest degree of power. e.g. if a polynomial’s
highest degree was 𝑥 𝑛 then it is a polynomial of degree 𝒏. We always write polynomials in
descending order of power.
𝑥 4 − 5𝑥 2 + 6𝑥 − 3 is a polynomial of degree 4
A linear polynomial is a polynomial of degree 1 (𝑒. 𝑔. 5𝑥 − 6) while a quadratic polynomial is
a polynomial of degree 2 (𝑒. 𝑔. 𝑥 2 − 7𝑥 + 2).
A cubic polynomial is a polynomial of degree 3 (𝑒. 𝑔. 3𝑥 3 − 5𝑥 2 + 4𝑥 − 9).
AIM: In another few lessons we want to be able to sketch the graph of a cubic function. To
sketch any graph we almost invariably start with the 𝑥 − 𝑎𝑛𝑑 𝑦 − intercepts.
Consider the function 𝑦 = 3𝑥 3 − 5𝑥 2 + 4𝑥 − 9. If we wanted to find the 𝑦 − intercepts of
the function, we would set 𝑥 = 0. This would give us 𝑦 = −9. So far no problem.
However, if we want to find the 𝑥 − intercepts we need to set 𝑦 = 0. This gives us:
3𝑥 3 − 5𝑥 2 + 4𝑥 − 9 = 0 , to find the 𝑥 −values where 𝑦 = 0 we need to be able to factorise
3𝑥 3 − 5𝑥 2 + 4𝑥 − 9 . We can guess that this may be some binomial times a trinomial:
(𝑝𝑥 + 𝑞)(𝑎𝑥 2 + 𝑏𝑥 + 𝑐)
We would have little issue factorising the trinomial if only we could factorise our original
expression to mirror this form… and to do that we would need to know what factor “𝑝𝑥 + 𝑞”
to take out.
The objective of this lesson and the next is to be able to factorise cubic functions.
1
, Remainder Theorem:
When we take the quotient of two numbers, the denominator EITHER divides perfectly into
the numerator OR it does not divide perfectly and so we are left with a remainder. The
remainder is the part of the numerator which cannot ‘fit’ the denominator.
Consider the example 27 ÷ 6; 6 goes into 27, 4 times. This means that of the initial 27, 24 can
be made up by groupings of 6, after which there is a remainder of 3 which cannot be formed
into a group of 6.
The same rule applies to polynomials:
If a polynomial 𝑓(𝑥) is divided by a linear polynomial 𝒂𝑥 + 𝒃, then the remainder (𝑟) is:
𝒃
𝑟 = 𝑓 (− )
𝒂
𝒃
Where exactly does − come from?
𝒂
Glad you asked…set 𝑎𝑥 + 𝑏 = 𝟎. And solve for 𝑥.
𝒂𝑥 = −𝒃
𝒃
𝑥=−
𝒂
Example:
Use the Remainder Theorem to determine the remainder when
𝒇(𝒙) = 𝟓𝒙𝟑 − 𝟔𝒙𝟐 + 𝟐𝒙 − 𝟑 is divided by:
𝒂) 𝟐𝒙 − 𝟔
𝒃) 𝒙 − 𝟒
(a) 𝑓(𝑥) is being divided by 2𝑥 – 6
−6
so the remainder is equal to 𝑓 (− ) = 𝑓(3)
2
𝑓(𝑥) has become 𝑓(3); we have substituted
3 2
𝑓(3) = 5(3) − 6(3) + 2(3) − 3 3 in for 𝑥. We must therefore do the same
= 135 − 54 + 6 − 3 for the rest of the equation.
= 84
This means that when we divide 5𝑥 3 − 6𝑥 2 + 2𝑥 − 3 by 2𝑥 − 6, the remainder is the
constant value of 84
2
