AQA A-level Mathematics: Polynomial Manipulation, Algebraic Division and the Factor Theorem (7357) – Revision Notes

These notes condense content statement B6 of the AQA A-level Mathematics 7357 specification (version 1.3, 31 January 2018): polynomial manipulation, algebraic division by linear expressions, the factor theorem and rational expressions. For full explanations and longer worked examples, read the study guide. When you are ready to test yourself, use the practice questions, the course hub, the checklist and the free 10-minute diagnostics.

Key ideas at a glance

Skill	Core idea	Spec
Expand and collect	Multiply every term by every term, then group like powers	B6
Factorise	Common factor first, then pattern, grouping or the factor theorem	B6
Algebraic division	f(x) = (divisor)(quotient) + remainder; linear divisors only	B6
Factor theorem	f(a) = 0 exactly when (x − a) is a factor	B6
Remainder	Dividing by (x − a) leaves f(a)	B6
Rational expressions	Factorise, cancel whole factors, state excluded values	B6

Definitions

• Polynomial: a sum of terms axⁿ with n a non-negative integer.
• Degree: the highest power of x present.
• Root (or zero) of f: a value a with f(a) = 0.
• Quotient: the polynomial you get from dividing. Its degree is one less than f when the divisor is linear.
• Rational expression: one polynomial over another.

Expanding

Method in steps:

1. Expand two brackets first, then multiply by the third.
2. Write every product, including signs.
3. Collect like terms by power of x, highest first.

Reminder: (x + 2)(x − 4)(3x + 1). First (x + 2)(x − 4) = x² − 2x − 8. Then (x² − 2x − 8)(3x + 1) = 3x³ + x² − 6x² − 2x − 24x − 8 = 3x³ − 5x² − 26x − 8.

Check an expansion by substituting a simple value. At x = 1 the original gives 3 × (−3) × 4 = −36, and 3 − 5 − 26 − 8 = −36. The values agree.

Factorising

Type	Rule	Example
Common factor	Take out the largest factor	2x³ − 8x = 2x(x² − 4)
Difference of two squares	a² − b² = (a − b)(a + b)	4x² − 9 = (2x − 3)(2x + 3)
Quadratic, leading coefficient above 1	Find two numbers with product ac and sum b, split and group	12x² − 5x − 2 = (3x − 2)(4x + 1)
Four terms	Group in pairs	x³ − 5x² − 4x + 20 = x²(x − 5) − 4(x − 5) = (x − 5)(x − 2)(x + 2)
Cubic with no obvious pattern	Find a root by trial, divide, factorise the quotient	see below

For 12x² − 5x − 2: ac = −24, and the pair with sum −5 is −8 and 3. So 12x² − 8x + 3x − 2 = 4x(3x − 2) + (3x − 2) = (3x − 2)(4x + 1).

Algebraic division

Method in steps:

1. Write the polynomial in descending powers, inserting 0xⁿ for any missing power.
2. Divide the leading term of what remains by the leading term of the divisor.
3. Multiply the divisor by that result and subtract.
4. Repeat until the remainder has lower degree than the divisor.
5. State f(x) = (divisor)(quotient) + remainder.

Reminder: divide 2x³ + 5x² − 14x − 8 by (x − 2).

2x² × (x − 2) = 2x³ − 4x²      → remainder 9x² − 14x
9x  × (x − 2) = 9x² − 18x      → remainder 4x − 8
4   × (x − 2) = 4x − 8         → remainder 0

The quotient is 2x² + 9x + 4 = (2x + 1)(x + 4). So 2x³ + 5x² − 14x − 8 = (x − 2)(2x + 1)(x + 4).

Checking by multiplying back is a quick way to catch a sign slip.

Factor theorem and remainder

Divisor	Substitute	Meaning
x − a	x = a	f(a) is the remainder; zero means a factor
x + a	x = −a	f(−a) is the remainder
px + q	x = −q/p	f(−q/p) is the remainder

Method for unknown constants:

1. Substitute the value that matches each divisor.
2. Set the result equal to 0 for a factor, or to the stated remainder.
3. Solve the simultaneous equations.
4. Substitute back and check with one more evaluation.

Reminder: the remainder of 5x³ − 2x² + x − 7 on division by (x − 1) is f(1) = 5 − 2 + 1 − 7 = −3. On division by (2x − 1) it is f(1/2) = 5/8 − 1/2 + 1/2 − 7 = −51/8.

Candidate roots for integer coefficients: numerators dividing the constant term, denominators dividing the leading coefficient. Test the small ones first.

Rational expressions

Method in steps:

1. Factorise the numerator and the denominator fully.
2. Cancel factors that appear in both.
3. State any excluded values from the original denominator.
4. For a division, multiply by the reciprocal first, then cancel.
5. If the numerator has degree equal to or above the denominator, divide to get a polynomial plus a proper fraction.

Reminders:

(x² − 16)/(x² + x − 12) = (x − 4)(x + 4) / [(x − 3)(x + 4)] = (x − 4)/(x − 3),  x ≠ −4, 3
(3x² − x − 4)/(x² − 1)  = (3x − 4)(x + 1) / [(x − 1)(x + 1)] = (3x − 4)/(x − 1), x ≠ ±1
(x² + x − 3)/(x − 1) = x + 2 − 1/(x − 1)

For the last one, (x − 1)(x + 2) = x² + x − 2, so x² + x − 3 is one less, giving remainder −1.

Putting it together: a full cubic

Method in steps for “factorise f(x) completely and solve f(x) = 0”:

1. List candidate roots from the factors of the constant term and the leading coefficient.
2. Evaluate f at the smallest candidates until one gives 0.
3. Turn that root into a factor, using the table above.
4. Divide to find the quadratic quotient.
5. Factorise the quotient, or use the quadratic formula if it does not factorise.
6. List every root, then write the answer in the form the question asks for.

Reminder: f(x) = 2x³ − 5x² − x + 6.

f(1)  = 2 − 5 − 1 + 6 = 2     not a root
f(−1) = −2 − 5 + 1 + 6 = 0    so (x + 1) is a factor

Dividing by (x + 1) gives 2x² − 7x + 6 with remainder 0, and 2x² − 7x + 6 = (2x − 3)(x − 2). So f(x) = (x + 1)(2x − 3)(x − 2) and the roots are x = −1, 3/2 and 2.

Check using a different route: f(2) = 16 − 20 − 2 + 6 = 0, and f(3/2) = 27/4 − 45/4 − 3/2 + 6 = 0.

Deciding which tool to use

Question wording	Likely tool
“Show that (x − k) is a factor”	Evaluate f(k) and state it equals 0
“Find the remainder when … is divided by …”	Evaluate f at the root of the divisor
“Find the quotient” or “divide”	Long division or matching coefficients
“Factorise completely”	Factor theorem, division, then factorise the quotient
“Find the values of a and b”	Two substitutions and simultaneous equations
“Simplify the fraction”	Factorise, cancel, state exclusions
“Write in the form … + A/(x + k)”	Divide, then write quotient plus remainder over the divisor

Writing the tool you are using in a few words, such as “by the factor theorem”, makes your working easier to follow and shows the method clearly.

Must-know distinctions

• Factor versus term: (x + 3)/(x + 3y) cannot be cancelled to 1/y, but (x + 3)(x − 1)/(x + 3) can be cancelled.
• Root versus factor: a root of 2 gives the factor (x − 2). A root of −1/2 gives the factor (2x + 1).
• Remainder zero versus non-zero: only a zero remainder means a factor.
• Quotient versus remainder: the quotient is a polynomial in x. The remainder, for a linear divisor, is a constant.
• Simplify versus solve: simplifying a rational expression leaves an expression. Solving needs an equals sign.

Quick self-test

1. Expand (x − 3)(x + 3)(x + 1).
2. Factorise 12x² − 5x − 2.
3. Factorise 2x³ − 8x completely.
4. Find the remainder when 5x³ − 2x² + x − 7 is divided by (x − 1).
5. Show that (x + 1) is a factor of x³ + 3x² − x − 3 and factorise it completely.
6. Divide 2x³ + 5x² − 14x − 8 by (x − 2) and state the quotient.
7. Simplify (x² − 16)/(x² + x − 12).
8. Simplify (3x² − x − 4)/(x² − 1).
9. Write (x² + x − 3)/(x − 1) as a polynomial plus a fraction.
10. State the value of x you substitute to test whether (2x + 5) is a factor of a polynomial.

Answers

1. x³ + x² − 9x − 9
2. (3x − 2)(4x + 1)
3. 2x(x − 2)(x + 2)
4. −3
5. f(−1) = −1 + 3 + 1 − 3 = 0, so (x + 1) is a factor. Then x³ + 3x² − x − 3 = (x + 1)(x² + 2x − 3) = (x − 1)(x + 1)(x + 3)
6. 2x² + 9x + 4, with remainder 0
7. (x − 4)/(x − 3)
8. (3x − 4)/(x − 1)
9. x + 2 − 1/(x − 1)
10. x = −5/2

Where marks are usually lost

• Dropping or flipping a sign when subtracting a bracketed product.
• Leaving out the 0x² term and misaligning the long division.
• Testing the wrong value for a factor such as (x + 4), using 4 instead of −4.
• Substituting x = −1 for the divisor (2x + 1) rather than x = −1/2.
• Cancelling terms across a fraction instead of factors.
• Forgetting to state excluded values when the question asks for them.
• Finding one root and stopping when the question asks for all solutions.
• Giving a quadratic factor that can still be split, or splitting one that cannot (such as x² + 9).
• Not showing the substitution when the question says “show that”.

Official syllabus

AQA A-level Mathematics (7357) specification, for teaching from September 2017, version 1.3 dated 31 January 2018, published by AQA. These notes cover content statement B6.