11.2 The Remainder and Factor Theorems
AHL 2.12
- the remainder theorem: how to find a remainder without long division
- how to use given remainders to find unknown coefficients
- the factor theorem: how to test whether \(x - a\) (or \(ax - b\)) is a factor
- how to use the factor theorem when a polynomial is divisible by a quadratic
The remainder theorem
In Section 11.1 we found remainders by long division. When the divisor is linear, there is a much faster way.
Divide \(f(x)\) by \(x - a\). The divisor has degree 1, so the remainder \(R\) has degree 0: it is a number.
\[ f(x) \equiv (x - a)\,Q(x) + R . \]
This is true for every \(x\), so put \(x = a\):
\[ f(a) = (a - a)\,Q(a) + R = 0 \times Q(a) + R = R . \]
In the same way, for a divisor \(ax - b\), write \(f(x) \equiv (ax - b)\,Q(x) + R\) and put \(x = \dfrac{b}{a}\) (the value that makes \(ax - b = 0\)):
\[ f\!\left(\frac{b}{a}\right) = \left(a \times \frac{b}{a} - b\right) Q\!\left(\frac{b}{a}\right) + R = 0 + R = R . \]
The remainder theorem Not in the formula booklet
When a polynomial \(f(x)\) is divided by \(x - a\), the remainder is \(f(a)\).
When \(f(x)\) is divided by \(ax - b\), the remainder is \(f\!\left(\dfrac{b}{a}\right)\).
- \(a\): a constant (in \(ax - b\), \(a \neq 0\))
- \(b\): a constant
The remainder theorem gives the remainder only. To find the quotient, you still need division.
Tip. To find which value to substitute, solve “divisor \(= 0\)”. For \(x + 3\), use \(x = -3\); for \(2x - 1\), use \(x = \frac{1}{2}\).
Example 1 Given that \(f(x) = 2x^3 + 3x^2 - 5\), find the remainder when \(f(x)\) is divided by
(a) \(x + 2\), \(\qquad\) (b) \(2x + 1\).
(a) \(x + 2 = 0\) when \(x = -2\):
\[ f(-2) = 2(-8) + 3(4) - 5 = -16 + 12 - 5 = -9 . \]
(b) \(2x + 1 = 0\) when \(x = -\dfrac{1}{2}\):
\[ f\!\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{8}\right) + 3\left(\frac{1}{4}\right) - 5 = -\frac{1}{4} + \frac{3}{4} - 5 = -\frac{9}{2} . \]
Example 2 The expression \(3x^2 + px - 5\), where \(p\) is a constant, leaves a remainder of \(9\) when divided by \(x + 2\). Find the value of \(p\).
By the remainder theorem, the value at \(x = -2\) is \(9\):
\[ 3(4) + p(-2) - 5 = 9 \;\Longrightarrow\; 7 - 2p = 9 \;\Longrightarrow\; p = -1 . \]
Example 3 The expression \(x^3 + ax^2 + bx + 4\) leaves the same remainder when it is divided by \(x - 1\) and when it is divided by \(x + 3\). Show that \(2a - b = 7\).
Let \(f(x) = x^3 + ax^2 + bx + 4\).
\[ f(1) = 1 + a + b + 4 = a + b + 5 \]
\[ f(-3) = -27 + 9a - 3b + 4 = 9a - 3b - 23 \]
The remainders are equal, so \(f(1) = f(-3)\):
\[ a + b + 5 = 9a - 3b - 23 \;\Longrightarrow\; 28 = 8a - 4b \;\Longrightarrow\; 2a - b = 7 . \quad\blacksquare \]
The factor theorem
The factor theorem is the special case of the remainder theorem where the remainder is zero. If the remainder is \(0\), then \(f(x) \equiv (x - a)\,Q(x)\), so \(x - a\) is a factor.
The factor theorem Not in the formula booklet
\(x - a\) is a factor of the polynomial \(f(x)\) if and only if \(f(a) = 0\).
More generally, \(ax - b\) is a factor of \(f(x)\) if and only if \(f\!\left(\dfrac{b}{a}\right) = 0\).
- \(a\): a constant (in \(ax - b\), \(a \neq 0\))
- \(b\): a constant
“If and only if” means the rule works in both directions:
- if \(x - a\) is a factor, then \(f(a) = 0\) (so \(a\) is a zero of \(f\), and \(x = a\) is a root of \(f(x) = 0\));
- if \(f(a) = 0\), then \(x - a\) is a factor, and \(f(x)\) is exactly divisible by \(x - a\).
Example 4 Determine whether \(x + 2\) is a factor of \(f(x)\).
(a) \(f(x) = x^3 + 3x^2 - 4\) \(\qquad\) (b) \(f(x) = x^2 + 3x(x + 1) - 2\)
Substitute \(x = -2\).
(a) \(f(-2) = -8 + 12 - 4 = 0\). So \(x + 2\) is a factor.
(b) \(f(-2) = 4 + 3(-2)(-1) - 2 = 4 + 6 - 2 = 8 \neq 0\). So \(x + 2\) is not a factor (the remainder is \(8\)).
To use the factor theorem with a quadratic divisor, first factorise the quadratic into two linear factors. Each linear factor must be a factor of the polynomial.
Example 5 The expression \(ax^3 + bx^2 - 13x + 6\) is exactly divisible by \(x^2 + x - 6\). Find the values of \(a\) and \(b\).
\(x^2 + x - 6 = (x + 3)(x - 2)\). So both \(x - 2\) and \(x + 3\) are factors of \(f(x) = ax^3 + bx^2 - 13x + 6\).
\[ f(2) = 8a + 4b - 26 + 6 = 0 \;\Longrightarrow\; 8a + 4b = 20 \;\Longrightarrow\; 2a + b = 5 \quad (1) \]
\[ f(-3) = -27a + 9b + 39 + 6 = 0 \;\Longrightarrow\; -27a + 9b = -45 \;\Longrightarrow\; -3a + b = -5 \quad (2) \]
\((1) - (2)\): \(\;5a = 10\), so \(a = 2\), and then \(b = 1\).
(Check: \(2x^3 + x^2 - 13x + 6 = (x^2 + x - 6)(2x - 1)\) ✓.)
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| Remainder on dividing by \(x + 2\) is \(f(2)\) | It is \(f(-2)\) | Substitute the value that makes the divisor zero. |
| Remainder on dividing by \(2x - 1\) is \(f(1)\) or \(f(2)\) | It is \(f\!\left(\frac{1}{2}\right)\) | Solve \(2x - 1 = 0\). |
| \(f(3) = 0\), so \(x + 3\) is a factor | \(x - 3\) is a factor | \(f(a) = 0 \Leftrightarrow (x - a)\) is a factor. |
| Using \(f(x) = 0\) for “leaves remainder \(9\)” | Use \(f(a) = 9\) | Only a factor gives remainder \(0\). |
| Dividing by \(x^2 + x - 6\) and substituting one value only | Factorise it, use both roots | Two unknowns need two equations. |
Exercises
1 Find the remainder when \(x^3 - 4x^2 + 2x + 7\) is divided by \(x - 3\).
\(f(3) = 27 - 36 + 6 + 7 = 4\).
2 Find the remainder when \(4x^3 + 2x^2 - x + 1\) is divided by \(2x - 1\).
\[ f\!\left(\frac{1}{2}\right) = 4 \times \frac{1}{8} + 2 \times \frac{1}{4} - \frac{1}{2} + 1 = \frac{1}{2} + \frac{1}{2} - \frac{1}{2} + 1 = \frac{3}{2} \]
3 Find the remainder when \(x^4 - 2x + 3\) is divided by \(x + 1\).
\(f(-1) = 1 + 2 + 3 = 6\).
4 When \(x^3 + kx^2 - 4x + 1\) is divided by \(x - 2\), the remainder is \(7\). Find \(k\).
\(f(2) = 8 + 4k - 8 + 1 = 4k + 1 = 7\), so \(k = \dfrac{3}{2}\).
5 (a) Show that \(x - 3\) is a factor of \(f(x) = x^3 - 2x^2 - 5x + 6\).
(b) Hence factorise \(f(x)\) completely.
(a) \(f(3) = 27 - 18 - 15 + 6 = 0\), so \(x - 3\) is a factor by the factor theorem.
(b) Synthetic division by \(x - 3\) gives the coefficients \(1, 1, -2\) and remainder \(0\), so
\[ f(x) = (x - 3)(x^2 + x - 2) = (x - 3)(x + 2)(x - 1) . \]
6 The polynomial \(2x^3 + ax^2 + bx - 6\) has a factor \(x + 2\), and leaves a remainder of \(-6\) when divided by \(x - 1\). Find \(a\) and \(b\).
\(f(-2) = -16 + 4a - 2b - 6 = 0 \Rightarrow 4a - 2b = 22 \Rightarrow 2a - b = 11\).
\(f(1) = 2 + a + b - 6 = -6 \Rightarrow a + b = -2\).
Adding: \(3a = 9\), so \(a = 3\) and \(b = -5\).
7 The expression \(x^3 - 2x^2 + kx + 5\) leaves the same remainder when divided by \(x - 1\) and by \(x + 2\). Find \(k\).
\(f(1) = 1 - 2 + k + 5 = k + 4\) and \(f(-2) = -8 - 8 - 2k + 5 = -2k - 11\).
\(k + 4 = -2k - 11 \Rightarrow 3k = -15 \Rightarrow k = -5\).
8 The expression \(x^3 + ax^2 + bx - 12\) is exactly divisible by \(x^2 - x - 6\). Find \(a\) and \(b\).
\(x^2 - x - 6 = (x - 3)(x + 2)\).
\(f(3) = 27 + 9a + 3b - 12 = 0 \Rightarrow 3a + b = -5\).
\(f(-2) = -8 + 4a - 2b - 12 = 0 \Rightarrow 2a - b = 10\).
Adding: \(5a = 5\), so \(a = 1\) and \(b = -8\).
9 A polynomial \(f(x)\) leaves a remainder of \(3\) when divided by \(x - 1\), and a remainder of \(-6\) when divided by \(x + 2\). Find the remainder when \(f(x)\) is divided by \((x - 1)(x + 2)\).
The divisor is quadratic, so the remainder has the form \(px + q\):
\[ f(x) \equiv (x - 1)(x + 2)\,Q(x) + px + q . \]
\(x = 1\): \(\;p + q = 3\). \(\qquad x = -2\): \(\;-2p + q = -6\).
Subtracting: \(3p = 9\), so \(p = 3\) and \(q = 0\). The remainder is \(3x\).
10 Let \(f(x) = x^3 + ax + b\). Given that \(x + 1\) is a factor of \(f(x)\) and that the remainder is \(4\) when \(f(x)\) is divided by \(x - 1\),
(a) find \(a\) and \(b\);
(b) show that \(f(x) = 0\) has only one real root.
(a) \(f(-1) = -1 - a + b = 0\) and \(f(1) = 1 + a + b = 4\).
Adding: \(2b = 4\), so \(b = 2\) and \(a = 1\).
(b) \(f(x) = x^3 + x + 2 = (x + 1)(x^2 - x + 2)\). For \(x^2 - x + 2\) the discriminant is \(1 - 8 = -7 < 0\), so it has no real roots. The only real root is \(x = -1\).