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Topic 3

Factors of polynomials

Syllabus 3.1–3.3 · Papers 1 and 2

🎯What you need to be able to do

  • State and use the remainder theorem and the factor theorem, including for divisors like \( 2x - 1 \).
  • Find unknown coefficients from given factors and remainders.
  • Write a cubic as a linear factor times a quadratic factor, by inspection or by algebraic long division.
  • Factorise completely, and solve cubic equations — including ones with surd roots, or only one real root.

📚The mathematics

The remainder theorem

When a polynomial \( p(x) \) is divided by \( x - a \), the remainder is \( p(a) \). More generally, dividing by \( ax - b \) leaves the remainder \( p\!\left(\tfrac{b}{a}\right) \) — substitute the value of \(x\) that makes the divisor zero.

divide by \( x - 3 \): remainder \( p(3) \)
divide by \( x + 2 \): remainder \( p(-2) \)
divide by \( 2x - 1 \): remainder \( p\!\left(\tfrac{1}{2}\right) \)

The sign is where marks go: \( x + 2 \) is \( x - (-2) \), so you substitute \(-2\).

The factor theorem

The special case of remainder zero: \( x - a \) is a factor of \( p(x) \) if and only if \( p(a) = 0 \). To show that something is a factor, substitute and write the conclusion: “\( p(2) = 0 \), so \( x - 2 \) is a factor.”

To find a factor of a cubic with integer coefficients, try the divisors of the constant term (and, if the leading coefficient is not 1, fractions such as \( \pm\tfrac{1}{2} \)). Start with \( \pm 1, \pm 2, \pm 3 \).

From one factor to all of them

Once you have a linear factor, the syllabus expects you to write the cubic as (linear) × (quadratic), then deal with the quadratic. Two methods:

  • Algebraic long division of \( p(x) \) by the factor.
  • Comparing coefficients (“by observation”): write \( 2x^{3} + x^{2} - 13x + 6 = (x - 2)(2x^{2} + cx - 3) \). The \( 2x^{2} \) and \( -3 \) are forced by the \( x^{3} \) and constant terms; matching the \( x^{2} \) terms gives \( c - 4 = 1 \), so \( c = 5 \).

Then factorise the quadratic, or use the formula. If its discriminant is negative, the cubic has only one real root — a common “show that” ending.

The cubic y = 2x cubed plus x squared minus 13x plus 6 crossing the x-axis at minus 3, one half and 2, crossing the y-axis at 6, and passing through (minus 1, 18), where a dashed line shows that the height of the curve is the remainder 18.
The worked example’s cubic. Its roots are its factors, and the remainder on dividing by \( x + 1 \) is simply the height of the curve at \( x = -1 \).

✏️Worked example

The polynomial \( p(x) = 2x^{3} + ax^{2} + bx + 6 \) has a factor \( x - 2 \). When \( p(x) \) is divided by \( x + 1 \) the remainder is 18. (a) Find the values of \(a\) and \(b\). [4] (b) Hence solve \( p(x) = 0 \). [3]

(a) Factor theorem: \( p(2) = 16 + 4a + 2b + 6 = 0 \), so \( 2a + b = -11 \). Remainder theorem: \( p(-1) = -2 + a - b + 6 = 18 \), so \( a - b = 14 \). Adding the two equations, \( 3a = 3 \), so \( a = 1 \) and \( b = -13 \).

(b) \( p(x) = 2x^{3} + x^{2} - 13x + 6 = (x - 2)(2x^{2} + 5x - 3) \), by comparing coefficients as above. Then \( 2x^{2} + 5x - 3 = (2x - 1)(x + 3) \), so

\[ p(x) = (x - 2)(2x - 1)(x + 3) = 0 \;\Longrightarrow\; x = 2,\ \tfrac{1}{2},\ -3 \]
Check it. The product of the roots of \( 2x^{3} + \ldots + 6 = 0 \) should be \( -\tfrac{6}{2} = -3 \), and \( 2 \times \tfrac{1}{2} \times (-3) = -3 \). Also \( p(-1) = -2 + 1 + 13 + 6 = 18 \), as given.
Stopping at the quadratic. “Solve” means every root. Writing \( (x - 2)(2x^{2} + 5x - 3) \) and giving only \( x = 2 \) loses the last two marks.

📝Practise

All of these are Paper 1 (no calculator) style: that is where polynomial work usually appears.

1. Find the remainder when \( 4x^{3} - 3x^{2} + 2x - 7 \) is divided by \( 2x - 1 \). [2]
Substitute \( x = \tfrac{1}{2} \): \( 4\left(\tfrac{1}{8}\right) - 3\left(\tfrac{1}{4}\right) + 2\left(\tfrac{1}{2}\right) - 7 = \tfrac{1}{2} - \tfrac{3}{4} + 1 - 7 = -\tfrac{25}{4} \).
2. Show that \( x + 2 \) is a factor of \( p(x) = x^{3} - 3x^{2} - 6x + 8 \), and hence factorise \( p(x) \) completely. [4]
\( p(-2) = -8 - 12 + 12 + 8 = 0 \), so \( x + 2 \) is a factor. Then \( p(x) = (x + 2)(x^{2} - 5x + 4) = (x + 2)(x - 1)(x - 4) \).
3. When \( x^{3} + kx^{2} - 5x + 6 \) is divided by \( x - 3 \) the remainder is 36. Find \(k\). [2]
\( 27 + 9k - 15 + 6 = 36 \Rightarrow 9k = 18 \Rightarrow k = 2 \).
4. Solve \( x^{3} - 7x - 6 = 0 \). [4]
Try \( x = -1 \): \( -1 + 7 - 6 = 0 \), so \( x + 1 \) is a factor. \( x^{3} - 7x - 6 = (x + 1)(x^{2} - x - 6) = (x + 1)(x + 2)(x - 3) \). So \( x = -1, -2, 3 \). Note the missing \( x^{2} \) term: in long division, write it as \( 0x^{2} \) to keep the columns straight.
5. Solve \( 2x^{3} - 5x^{2} - 4x + 3 = 0 \). [5]
\( p(3) = 54 - 45 - 12 + 3 = 0 \), so \( x - 3 \) is a factor. \( p(x) = (x - 3)(2x^{2} + x - 1) = (x - 3)(2x - 1)(x + 1) \). So \( x = 3,\ \tfrac{1}{2},\ -1 \).
6. Show that \( x = 1 \) is a root of \( x^{3} - 3x^{2} - 2x + 4 = 0 \), and find the other two roots in exact form. [5]
\( 1 - 3 - 2 + 4 = 0 \), so \( x = 1 \) is a root. Dividing, \( x^{3} - 3x^{2} - 2x + 4 = (x - 1)(x^{2} - 2x - 4) \). The quadratic gives \( x = \dfrac{2 \pm \sqrt{4 + 16}}{2} = 1 \pm \sqrt{5} \). The roots are \( 1,\ 1 + \sqrt{5},\ 1 - \sqrt{5} \).
7. \( p(x) = 2x^{3} + ax^{2} + bx + 4 \) has a factor \( x + 2 \), and leaves a remainder of 12 when divided by \( x - 1 \). (a) Find \(a\) and \(b\). (b) Show that \( p(x) = 0 \) has only one real root. [6]
(a) \( p(-2) = -16 + 4a - 2b + 4 = 0 \Rightarrow 2a - b = 6 \). \( p(1) = 2 + a + b + 4 = 12 \Rightarrow a + b = 6 \). So \( a = 4 \), \( b = 2 \). (b) \( p(x) = 2x^{3} + 4x^{2} + 2x + 4 = (x + 2)(2x^{2} + 2) \). The quadratic \( 2x^{2} + 2 \) has discriminant \( 0 - 16 < 0 \), so no real roots; the only real root of \( p(x) = 0 \) is \( x = -2 \).

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Desmos — graph your cubic to check the roots you found, and read off \( p(a) \) as a height
  • Khan Academy — polynomial long division, worked step by step