IGCSE Add Maths 0606 / Topic 3 / IGCSE

Find a remainder without dividing.

Dividing a cubic by \(x - a\) takes several lines of working, but the remainder is simply \(p(a)\). When that value is zero, \(x - a\) is a factor. Two substitutions give two equations, which is how you find unknown coefficients.

Topic 3All factors of polynomials lessons →

The key idea

Remainder theorem: when a polynomial \(p(x)\) is divided by \(x - a\), the remainder is \(p(a)\); when it is divided by \(ax - b\), the remainder is \(p\left(\dfrac{b}{a}\right)\). Factor theorem: \(x - a\) is a factor of \(p(x)\) exactly when \(p(a) = 0\) (and \(ax - b\) is a factor exactly when \(p\left(\dfrac{b}{a}\right) = 0\)). Each fact about a remainder or a factor gives one equation, so two facts let you find two unknown coefficients.

Fig. 1: the remainder p(a) is the height of the graph y = p(x) at x = a. Here p(2) = -4, the remainder on dividing by (x - 2). Wherever the graph cuts the x-axis, p(a) = 0 and (x - a) is a factor

Section 01

The remainder is just p(a).

Dividing a polynomial \(p(x)\) by \(x - a\) gives a quotient \(q(x)\) and a remainder \(R\), a number, so \(p(x) = (x - a)\,q(x) + R\) for every \(x\). Put \(x = a\): the bracket becomes zero and you are left with \(p(a) = R\). So you can find the remainder by substituting, with no long division at all (Fig. 1).

Divisor \(x - a\): remainder \(p(a)\). For \(x - 3\) substitute \(x = 3\).
Divisor \(x + a\): this is \(x - (-a)\), so the remainder is \(p(-a)\). For \(x + 2\) substitute \(x = -2\).
Divisor \(ax - b\): \(p(x) = (ax - b)\,q(x) + R\), and \(ax - b = 0\) when \(x = \dfrac{b}{a}\), so the remainder is \(p\left(\dfrac{b}{a}\right)\). For \(2x - 1\) substitute \(x = \dfrac{1}{2}\).
The rule of thumb: substitute the value of \(x\) that makes the divisor zero.

Worked example Dividing by x - a and by x + a

Find the remainder when \(2x^{3} - 3x^{2} + x - 5\) is divided by (a) \(x - 2\) (b) \(x + 1\).

Let \(p(x) = 2x^{3} - 3x^{2} + x - 5\).
(a) \(x - 2 = 0\) when \(x = 2\): \(p(2) = 2(8) - 3(4) + 2 - 5 = 16 - 12 + 2 - 5 = 1\).
(b) \(x + 1 = 0\) when \(x = -1\): \(p(-1) = 2(-1) - 3(1) + (-1) - 5 = -2 - 3 - 1 - 5 = -11\).

Answer: (a) \(1\) (b) \(-11\)

Worked example Dividing by ax - b

Find the remainder when \(4x^{3} + 2x^{2} - 3x + 1\) is divided by \(2x - 1\).

\(2x - 1 = 0\) when \(x = \dfrac{1}{2}\), so the remainder is \(p\left(\dfrac{1}{2}\right)\).
\(p\left(\dfrac{1}{2}\right) = 4\left(\dfrac{1}{8}\right) + 2\left(\dfrac{1}{4}\right) - 3\left(\dfrac{1}{2}\right) + 1 = \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{3}{2} + 1\).
\(= \dfrac{1}{2}\).

Answer: \(\dfrac{1}{2}\)

Your turn

Find the remainder when \(x^{3} + 4x^{2} - x + 2\) is divided by \(x + 3\).

Examiner trap

For \(x + 1\) substitute \(x = -1\), not \(x = 1\). For \(2x - 1\) substitute \(x = \frac{1}{2}\), not \(x = 1\) or \(x = 2\). And put negative values in brackets: \((-1)^{3} = -1\) but \((-1)^{2} = +1\), so \(-3x^{2}\) at \(x = -1\) is \(-3\), not \(+3\).

Section 02

A remainder of zero means a factor.

If the remainder is zero, then \(p(x) = (x - a)\,q(x)\) exactly, so \(x - a\) is a factor. It works the other way too: if \(x - a\) is a factor then \(p(a) = 0\). This is the factor theorem. To show that a given linear expression is a factor, substitute the value that makes it zero, show every term, and finish with a clear conclusion.

\(p(a) = 0\) exactly when \(x - a\) is a factor of \(p(x)\); then \(x = a\) is a root of \(p(x) = 0\).
\(p\left(\dfrac{b}{a}\right) = 0\) exactly when \(ax - b\) is a factor of \(p(x)\).
Looking for a factor of a cubic with integer coefficients and leading coefficient 1: try \(x = \pm 1, \pm 2, \dots\), the divisors of the constant term.
A 'show that' answer ends with a sentence: '\(p(3) = 0\), so \(x - 3\) is a factor'.

Worked example Show that x - 3 is a factor

Show that \(x - 3\) is a factor of \(p(x) = x^{3} - 4x^{2} + x + 6\).

\(x - 3 = 0\) when \(x = 3\).
\(p(3) = 27 - 4(9) + 3 + 6 = 27 - 36 + 3 + 6 = 0\).
The remainder is zero, so by the factor theorem \(x - 3\) is a factor of \(p(x)\).

Answer: \(p(3) = 0\), so \(x - 3\) is a factor

Worked example A factor of the form ax + b

Show that \(2x + 1\) is a factor of \(p(x) = 2x^{3} - x^{2} - 7x - 3\).

\(2x + 1 = 0\) when \(x = -\dfrac{1}{2}\).
\(p\left(-\dfrac{1}{2}\right) = 2\left(-\dfrac{1}{8}\right) - \dfrac{1}{4} - 7\left(-\dfrac{1}{2}\right) - 3 = -\dfrac{1}{4} - \dfrac{1}{4} + \dfrac{7}{2} - 3\).
\(= -\dfrac{1}{2} + \dfrac{1}{2} = 0\).
So \(2x + 1\) is a factor of \(p(x)\). (The next lesson shows how to find the quadratic factor that goes with it.)

Answer: \(p\left(-\frac{1}{2}\right) = 0\), so \(2x + 1\) is a factor

Your turn

Is \(x - 2\) a factor of \(x^{3} - 3x^{2} + 4\)?

Examiner trap

Do not just write '\(= 0\)'. Show the substitution term by term and state the conclusion, because a 'show that' question gives its marks for the working. And a factor \(x + 2\) means \(p(-2) = 0\): the sign changes when you go from the factor to the root.

Section 03

Two facts, two unknowns.

Many questions give a polynomial with unknown coefficients, such as \(a\) and \(b\), together with facts about its factors or remainders. Turn each fact into an equation: 'a factor' means the substitution gives \(0\); 'a remainder of \(R\)' means the substitution gives \(R\). One unknown needs one fact; two unknowns need two facts, which give a pair of simultaneous equations.

\(x - a\) is a factor: \(p(a) = 0\).
Remainder \(R\) on dividing by \(x - a\): \(p(a) = R\).
Same remainder for two divisors: set the two substitutions equal to each other.
Check: put your values back into the polynomial and test both facts.

Worked example A factor and a remainder

The polynomial \(p(x) = x^{3} + ax^{2} + bx - 6\) has a factor \(x - 1\). When \(p(x)\) is divided by \(x + 1\) the remainder is \(-8\). Find the values of \(a\) and \(b\).

Factor \(x - 1\): \(p(1) = 1 + a + b - 6 = 0\), so \(a + b = 5\).
Remainder \(-8\) for \(x + 1\): \(p(-1) = -1 + a - b - 6 = -8\), so \(a - b = -1\).
Add the equations: \(2a = 4\), so \(a = 2\); then \(b = 5 - 2 = 3\).
Check with \(p(x) = x^{3} + 2x^{2} + 3x - 6\): \(p(1) = 1 + 2 + 3 - 6 = 0\) and \(p(-1) = -1 + 2 - 3 - 6 = -8\).

Answer: \(a = 2,\ b = 3\)

Worked example A factor of the form ax - b and a remainder

The polynomial \(p(x) = 2x^{3} + ax^{2} + bx + 3\) has a factor \(2x - 1\), and leaves a remainder of \(15\) when divided by \(x - 2\). Find the values of \(a\) and \(b\).

Factor \(2x - 1\): \(p\left(\dfrac{1}{2}\right) = 2\left(\dfrac{1}{8}\right) + \dfrac{a}{4} + \dfrac{b}{2} + 3 = 0\). Multiply by 4: \(1 + a + 2b + 12 = 0\), so \(a + 2b = -13\).
Remainder \(15\): \(p(2) = 16 + 4a + 2b + 3 = 15\), so \(4a + 2b = -4\), that is \(2a + b = -2\).
From the second equation \(b = -2 - 2a\). Substitute: \(a - 4 - 4a = -13\), so \(-3a = -9\), giving \(a = 3\) and \(b = -8\).
Check with \(p(x) = 2x^{3} + 3x^{2} - 8x + 3\): \(p\left(\frac{1}{2}\right) = \frac{1}{4} + \frac{3}{4} - 4 + 3 = 0\) and \(p(2) = 16 + 12 - 16 + 3 = 15\).

Answer: \(a = 3,\ b = -8\)

Your turn

The polynomial \(x^{3} + kx^{2} - 2x + 1\) leaves the same remainder when divided by \(x - 1\) and by \(x + 2\). Find \(k\).

Examiner trap

A remainder is not a factor: 'remainder \(-8\)' means \(p(-1) = -8\), not \(p(-1) = 0\). Keep the constant term in every equation (the \(-6\) in \(p(x)\) does not disappear), and multiply through to clear fractions only after you have written every term.

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What is the remainder when \(x^{3} - 2x^{2} + 3x - 1\) is divided by \(x + 1\)?

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Which value gives the remainder when \(p(x)\) is divided by \(2x + 3\)?

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For a polynomial \(p(x)\), \(p(4) = 0\). Which statement must be true?

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\(x - 1\) is a factor of \(x^{3} + kx - 5\). What is \(k\)?

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Find the remainder when \(3x^{3} - x^{2} + 2\) is divided by \(3x - 1\).

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The polynomial \(x^{3} + ax^{2} + bx + 4\) has a factor \(x + 1\) and leaves a remainder of \(6\) when divided by \(x - 1\). Find \(a\) and \(b\).

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\(x - 2\) and \(x + 3\) are both factors of \(x^{3} + ax^{2} + bx - 6\). Find \(a\) and \(b\).

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The polynomial \(x^{3} + kx^{2} - x + 2\) leaves the same remainder when divided by \(x - 1\) and by \(x + 3\). Find \(k\).

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Remainders on dividing by \(x - a\) and by \(ax - b\), showing a factor, one unknown from a factor, and two unknowns from a factor and a remainder.