2 exam-style questions plus unlimited generated ones on this topic, at up to four difficulty levels, with full mark schemes. Notes and worksheets are open to everyone. Practice needs a free account.
Or take a quick quiz: 10 questions on this subtopic in AS Maths (Edexcel), marked as you tap.
Working with polynomials: expanding, factorising, dividing by a linear expression and using the factor theorem to find the factors of a cubic. You also simplify algebraic fractions. A typical question says 'show that \((x - 2)\) is a factor of f(x), hence factorise f(x) completely' for 4 to 6 marks, often followed by a sketch or by solving \(f(x) = 0\)
Key facts
Factor theorem: \(f(a) = 0 \Leftrightarrow (x - a)\) is a factor
\(f\left(\dfrac{b}{a}\right) = 0 \Leftrightarrow (ax - b)\) is a factor
\(f(-3) = 0 \Leftrightarrow (x + 3)\) is a factor
First factor: try \(\pm\) the factors of the constant term
Alternatively compare coefficients: write \(f(x) \equiv (x - 3)(ax^2 + bx + c)\); the \(x^3\) terms give \(a\), the constants give \(c\), then find \(b\)
M = method mark, A = answer mark, B = a mark on its own.
Show a factor, then factorise completely
6 marksGrade C
Substitute the value into \(f(x)\), show it equals 0 and conclude.
Divide \(f(x)\) by the factor (or compare coefficients) to get a quadratic.
Factorise the quadratic.
Write \(f(x)\) as a product of three linear factors.
Example. \(f(x) = 2x^3 + x^2 - 13x + 6\) (a) Use the factor theorem to show that \((x - 2)\) is a factor of \(f(x)\). (b) Hence factorise \(f(x)\) completely.
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(a) \(f(2) = 2(8) + 4 - 26 + 6\)
(M1)
\(= 0\), so \((x - 2)\) is a factor.
(A1)
(b) Dividing by \((x - 2)\) or comparing coefficients
(M1)
\(f(x) = (x - 2)(2x^2 + 5x - 3)\)
(A1)
Factorising the quadratic
(M1)
\(f(x) = (x - 2)(2x - 1)(x + 3)\)
(A1)
Simplify an algebraic fraction
3 marksGrade C
Take out any common factor first.
Factorise the numerator and the denominator fully.
Cancel the common bracket.
Example. Simplify fully \(\dfrac{x^3 - 4x}{x^2 + x - 6}\)
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\(x^3 - 4x = x(x - 2)(x + 2)\)
(M1)
\(x^2 + x - 6 = (x + 3)(x - 2)\)
(B1)
\(\dfrac{x(x + 2)}{x + 3}\)
(A1)
Find unknown coefficients using the factor theorem
6 marksGrade A
For each factor \((ax - b)\), set \(f\left(\dfrac{b}{a}\right) = 0\)
Simplify each to a linear equation in the unknowns.
Solve the equations simultaneously.
Example.\(f(x) = 2x^3 + ax^2 + bx - 6\), where \(a\) and \(b\) are constants. Given that \((x + 1)\) and \((2x - 3)\) are both factors of \(f(x)\), find the values of \(a\) and \(b\)
Sign error in the factor theorem: to test \((x + 2)\), substitute \(x = -2\), not \(x = 2\)
No conclusion: '\(f(2) = 0\), so \((x - 2)\) is a factor' earns the accuracy mark.
Stopping at the quadratic: 'completely' means three linear factors when possible.
Missing out zero terms in long division, so the columns do not line up.
Cancelling terms instead of factors, e.g. cancelling the \(x^2\) in \(\dfrac{x^2 + 3}{x^2 + 5x}\)
More tips
Memory tricks
\(f(1)\) is the sum of the coefficients: if they add up to 0, \((x - 1)\) is a factor.
For \(ax^3 + bx^2 + cx + d\),\(f(-1) = -a + b - c + d\): quick to check in your head.
When comparing coefficients, the \(x^2\) coefficient of the quotient is the leading coefficient of \(f\) divided by the coefficient of \(x\) in the divisor.
Check a factorisation by expanding it, or by substituting \(x = 1\) into both forms.
Exam technique
'Use the factor theorem' means you must substitute and show the value is 0; dividing alone does not score that mark.
Show the substitution with numbers, e.g. \(f(2) = 2(8) + 4 - 26 + 6 = 0\), not just '\(f(2) = 0\)'.
'Hence' means use the factor you have just shown, not a new method.
The factorised form gives the roots and \(y\)-intercept you need for a sketch in the next part.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Expand brackets and collect like termsMultiply every term in one bracket by every term in the other, e.g. \((x + 2)(x^2 - 3x + 1) = x^3 - x^2 - 5x + 2\)
Grade D
Show a factor using the factor theoremShow \(f(a) = 0\) and conclude that \((x - a)\) is a factor of \(f(x)\)
Grade C
Divide a cubic by a linear factorUse long division or compare coefficients to find the quadratic factor.
Grade C
Factorise a cubic completelyFind one factor with the factor theorem, divide, then factorise the quadratic quotient.
Grade C
Simplify algebraic fractions by factorisingFactorise the top and bottom fully, then cancel common factors (whole brackets only).
Grade B
Use factors of the form ax − b\((ax - b)\) is a factor of \(f(x)\) if and only if \(f\left(\dfrac{b}{a}\right) = 0\)
Grade A
Find unknown coefficients from given factorsEach known factor gives an equation, e.g. \(f(-1) = 0\); solve the equations simultaneously.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy6 marks
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. \(\mathrm{f}(x) = x^{3} + 2x^{2} - 5x - 6\)
(a) Use the factor theorem to show that \((x - 2)\) is a factor of \(\mathrm{f}(x)\).[2]
(b) Hence fully factorise \(\mathrm{f}(x)\).[4]
Show the answer and mark scheme
(a)Answer: \(\mathrm{f}(2) = 0\), so \((x - 2)\) is a factor.
M1 for attempting \(\mathrm{f}(2)\)
A1* for \(\mathrm{f}(2) = 0\) with a conclusion, e.g. so \((x - 2)\) is a factor
Worked solution: \(\mathrm{f}(2) = 2^3 + 2(2)^2 - 5(2) - 6 = 8 + 8 - 10 - 6 = 0\). Since \(\mathrm{f}(2) = 0\), \((x - 2)\) is a factor of \(\mathrm{f}(x)\).
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. \(\mathrm{f}(x) = 2x^{3} - 13x^{2} + 10x + 25\)
(a) Use the factor theorem to show that \((2x - 5)\) is a factor of \(\mathrm{f}(x)\).[2]
(b) Hence fully factorise \(\mathrm{f}(x)\).[4]
(c) Hence solve \(\mathrm{f}(x) = 0\)[1]
Show the answer and mark scheme
(a)Answer: \(\mathrm{f}\left(\frac{5}{2}\right) = 0\), so \((2x - 5)\) is a factor.
M1 for attempting \(\mathrm{f}\left(\frac{5}{2}\right)\)
A1* for \(\mathrm{f}\left(\frac{5}{2}\right) = 0\) with a conclusion, e.g. so \((2x - 5)\) is a factor
Worked solution: \(\mathrm{f}\left(\frac{5}{2}\right) = 2\left(\frac{5}{2}\right)^3 - 13\left(\frac{5}{2}\right)^2 + 10\left(\frac{5}{2}\right) + 25 = \frac{125}{4} - \frac{325}{4} + 25 + 25 = 0\). Since \(\mathrm{f}\left(\frac{5}{2}\right) = 0\), \((2x - 5)\) is a factor of \(\mathrm{f}(x)\).
Worked solution: From the factorised form, \(x = -1, \ \frac{5}{2}, \ 5\).
Question 3Hard9 marks
(a) Explain why, when a polynomial is divided by a quadratic, the remainder can be written in the form \(ax + b\).[1]
(b) Find the remainder when \(x^{100}\) is divided by \(x^2 - 3x + 2\). Give your answer in the form \(ax + b\), where \(a\) and \(b\) are expressed in terms of powers of 2.[5]
(c) When the polynomial \(\mathrm{q}(x)\) is divided by \((x - 1)\) the remainder is 3, and when \(\mathrm{q}(x)\) is divided by \((x + 2)\) the remainder is \(-9\). Find the remainder when \(\mathrm{q}(x)\) is divided by \(x^2 + x - 2\).[3]
Show the answer and mark scheme
(a)Answer: The remainder has a lower degree than the divisor, so its degree is at most 1.
B1 for the degree of the remainder is less than the degree of the divisor (2), so it is linear or constant
Worked solution: In polynomial division the remainder always has a smaller degree than the divisor. The divisor has degree 2, so the remainder has degree at most 1 and can be written \(ax + b\) (with \(a\) possibly 0).
M1 for \(\mathrm{q}(x) = (x - 1)(x + 2)\mathrm{S}(x) + cx + d\) and using the remainder theorem: \(\mathrm{q}(1) = 3\), \(\mathrm{q}(-2) = -9\)
M1 for \(c + d = 3\) and \(-2c + d = -9\) solved
A1 for \(4x - 1\)
Worked solution: \(x^2 + x - 2 = (x - 1)(x + 2)\), so \(\mathrm{q}(x) = (x - 1)(x + 2)\mathrm{S}(x) + cx + d\). By the remainder theorem, \(\mathrm{q}(1) = c + d = 3\) and \(\mathrm{q}(-2) = -2c + d = -9\). So \(3c = 12\), \(c = 4\), \(d = -1\): the remainder is \(4x - 1\).