What is 2.6: Polynomials and the factor theorem?

Polynomial manipulation uses expansion, collection and factorisation while preserving equality; choose the form that exposes the required structure. The factor theorem states that ax-b is a factor of f(x) exactly when f(b/a)=0; after finding a factor, divide by the linear expression to obtain the remaining polynomial.

How many marks are 2.6 questions worth?

2.6 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 2.6 appeared in?

We have not yet indexed a verified past-paper appearance for 2.6 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

2.6

Manipulate polynomials algebraically, including expanding brackets, collecting like terms, factorisation and simple algebraic division; use the factor theorem; simplify rational expressions by factorising, cancelling and algebraic division.

Draft — not yet indexed

Polynomials and the factor theorem

Worked answers and methods for 2.6 on Edexcel A-level Maths 9MA0.

Explanation

Polynomial manipulation uses expansion, collection and factorisation while preserving equality; choose the form that exposes the required structure. The factor theorem states that $ax-b$ is a factor of $f(x)$ exactly when $f(b/a)=0$ ; after finding a factor, divide by the linear expression to obtain the remaining polynomial.
For example, $f(2)=0$ for $f(x)=x^3-3x^2-4x+12$ , and division by $x-2$ gives $x^2-x-6$ . Cancel factors, not separate terms, in rational expressions, and retain exclusions from the original denominator even when a factor cancels.
Only linear divisors $ax+b$ or $ax-b$ are required for algebraic division.
When simplifying a rational expression, factor the complete numerator and denominator before cancelling.
A cancelled factor still records an excluded input from the original expression, so state that restriction with the simplified answer.

Worked example

Use the factor theorem to factorise $x^3-4x^2+x+6$ fully.

1.Let $f(x)=x^3-4x^2+x+6$ .
2.Since $f(2)=8-16+2+6=0$ , $x-2$ is a factor.
3.Division gives $x^2-2x-3$ , which factorises as $(x-3)(x+1)$ .
4.Hence $f(x)=(x-2)(x-3)(x+1)$ .

Answer: $(x-2)(x-3)(x+1)$

Common mistakes

Don't cancel separate terms across addition, such as cancelling the $x$ in $(x+2)/x$ .
Don't forget the excluded value from the original denominator after a common factor has cancelled.

Exam tip

Use the factor theorem by showing $f(a)=0$ , then divide by $x-a$ and factorise the quotient fully.

Worked practice

Tier 1 · Easy

Factorise

x^2-9x+20

completely.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$(x-4)(x-5)$	2
Notes
Find two numbers with product $20$ and sum $-9$ : they are $-4$ and $-5$ . Therefore $x^2-9x+20=(x-4)(x-5)$ .

(2 marks)

Tier 2 · Standard

The polynomial

p(x)=x^3+ax^2-4x-12

, where

a

is a constant, has a factor

x+2

. Find the value of

a

and hence factorise

p(x)

fully.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$a=3$ $p(x)=(x+2)(x-2)(x+3)$	4
Notes
By the factor theorem, $p(-2)=0$ . Hence $-8+4a+8-12=0$ , so $4a-12=0$ and $a=3$ . Then $p(x)=x^3+3x^2-4x-12=x^2(x+3)-4(x+3)=(x+3)(x^2-4)=(x+3)(x-2)(x+2)$ .

(4 marks)

Tier 3 · Hard

Simplify

\dfrac{x^3-4x^2+x+6}{x^2-4x+4}

as far as possible, then use algebraic division. State any excluded value.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$x-\dfrac{3}{x-2}$ $x\ne2$	5
Notes
Factor the numerator as $(x-2)(x-3)(x+1)$ and the denominator as $(x-2)^2$ . Cancelling one factor gives $(x^2-2x-3)/(x-2)$ , while the original expression still requires $x\ne2$ . Division gives $x^2-2x-3=x(x-2)-3$ , so the simplified form is $x-3/(x-2)$ .

(5 marks)

Tier 1 · Easy

Expand and simplify

(2x-3)(x^2+x-4)

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$2x^3-x^2-11x+12$	2
Notes
Expanding gives $2x^3+2x^2-8x-3x^2-3x+12$ . Collecting like terms produces $2x^3-x^2-11x+12$ .

(2 marks)

Tier 2 · Standard

Divide

2x^3-3x^2+4x-5

x-2

, giving the quotient and remainder.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Quotient $2x^2+x+6$ ; remainder $7$	4
Notes
Algebraic division gives leading terms $2x^2$ , then $x$ , then $6$ . Equivalently, $(x-2)(2x^2+x+6)=2x^3-3x^2+4x-12$ . Therefore $2x^3-3x^2+4x-5=(x-2)(2x^2+x+6)+7$ , so the quotient is $2x^2+x+6$ and the remainder is $7$ .

(4 marks)

Tier 3 · Hard

Both

x-1

and

x+2

divide

p(x)=x^3+ax^2+bx+6

. Determine

a

and

b

, and hence factorise

p(x)

fully.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$a=-2$ , $b=-5$ $p(x)=(x-1)(x+2)(x-3)$	5
Notes
The factor theorem gives $p(1)=0$ , so $a+b=-7$ . Also $p(-2)=0$ , so $-8+4a-2b+6=0$ , or $2a-b=1$ . Solving the two linear equations gives $a=-2$ and $b=-5$ . The polynomial is then $x^3-2x^2-5x+6$ ; division by the known factors, or inspection of the remaining root, gives $p(x)=(x-1)(x+2)(x-3)$ .

(5 marks)

Tier 2 · Standard

Find the remainder when

p(x)=4x^3-3x+7

is divided by

2x+1

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Remainder $8$	3
Notes
For a divisor $2x+1$ , the remainder is $p(-1/2)$ . Therefore $p(-1/2)=4(-1/2)^3-3(-1/2)+7=-1/2+3/2+7=8$ .

(3 marks)

Tier 3 · Hard

For

x\ne-1,4

, the rational function

F(x)=\dfrac{2x^3+ax^2+bx-4}{(x-4)(x+1)}

simplifies exactly to a linear polynomial. Determine

a

and

b

, and find the simplified polynomial.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$a=-5$ , $b=-11$ $F(x)=2x+1$ for $x\ne-1,4$	6
Notes
The quotient must have leading term $2x$ . Also $F(0)=(-4)/(-4)=1$ , so the linear quotient is $2x+1$ . Hence the numerator must equal $(2x+1)(x-4)(x+1)=(2x+1)(x^2-3x-4)=2x^3-5x^2-11x-4$ . Comparing coefficients gives $a=-5$ and $b=-11$ . The original excluded values $x=-1$ and $x=4$ remain excluded.

(6 marks)

Tier 3 · Hard

The polynomial

p(x)=2x^3+ax+b

leaves the same remainder when divided by

x-2

and

x+1

, and

x+3

is a factor. Determine

a

and

b

. Then divide

p(x)

x-1

, giving the quotient and remainder.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$a=-6$ , $b=36$ Quotient $2x^2+2x-4$ ; remainder $32$	6
Notes
Equal remainders give $p(2)=p(-1)$ , so $16+2a+b=-2-a+b$ and hence $a=-6$ . Since $x+3$ is a factor, $p(-3)=0$ . Thus $-54-3a+b=0$ ; using $a=-6$ gives $b=36$ . Therefore $p(x)=2x^3-6x+36$ . Algebraic division by $x-1$ gives $p(x)=(x-1)(2x^2+2x-4)+32$ , so the quotient is $2x^2+2x-4$ and the remainder is $32$ .

(6 marks)

Q10

Tier 3 · Hard

10.

A monic quartic polynomial

p(x)

has roots

-2

1

and a repeated root

r

. Its coefficient of

x^3

-5

. Find

r

, factorise the polynomial fully and state its constant term.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$r=3$ $p(x)=(x+2)(x-1)(x-3)^2$ The constant term is $-18$ .	5
Notes
The factorised polynomial is $(x+2)(x-1)(x-r)^2=(x^2+x-2)(x^2-2rx+r^2)$ . In the product, the $x^3$ terms are $-2rx^3$ from $x^2(-2rx)$ and $x^3$ from $x(x^2)$ , so the $x^3$ coefficient is $1-2r$ . Hence $1-2r=-5$ , giving $r=3$ . Therefore $p(x)=(x+2)(x-1)(x-3)^2$ . Setting $x=0$ gives the constant term $2(-1)(9)=-18$ .

(5 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for 2.6. Browse the Edexcel A-level Maths 9MA0 past papers directly.

Other points in 2 Algebra and functions

Want help turning this into marks?

Bring 2.6 or any tricky specification point, and we can work through the method and exam wording together.

Book a free intro call