Polynomials: Remainder and Factor Theorems

The Remainder Theorem and Factor Theorem are two of the most useful shortcuts in algebra. They let you find a remainder without long division and factorise cubic polynomials in a few lines, which is why they appear in almost every Matric, FSc and O/A Level paper. This guide explains both theorems and shows you exactly how to use them.

A Quick Polynomial Refresher

A polynomial in \(x\) is an expression such as

$$ p(x) = 2x^3 - 3x^2 + 4x - 5 $$

where the powers of \(x\) are whole numbers. The highest power is the degree (here, 3). Writing \(p(a)\) means substituting \(x = a\).

When you divide a polynomial \(p(x)\) by a divisor \(d(x)\), you get

$$ p(x) = d(x)\,q(x) + r(x) $$

where \(q(x)\) is the quotient and \(r(x)\) is the remainder, with degree less than \(d(x)\). If the divisor is linear, the remainder is just a number.

The Remainder Theorem

If a polynomial \(p(x)\) is divided by \(x - a\), the remainder is \(p(a)\).

Why it works: Write \(p(x) = (x - a)\,q(x) + R\). Put \(x = a\): the first term becomes \(0 \cdot q(a) = 0\), leaving \(p(a) = R\).

For a divisor like \(2x - 1\), set it equal to zero: \(x = \frac{1}{2}\). The remainder is \(p\left(\frac{1}{2}\right)\). For \(x + 3\), use \(x = -3\).

Worked Example 1: Finding a Remainder

Find the remainder when \(p(x) = 2x^3 - 3x^2 + 4x - 5\) is divided by \(x - 2\).

Step 1: The divisor is \(x - 2\), so \(a = 2\).

Step 2: Substitute:

$$ p(2) = 2(8) - 3(4) + 4(2) - 5 = 16 - 12 + 8 - 5 = 7 $$

Answer: the remainder is 7.

Check With Synthetic Division

Write the coefficients 2, −3, 4, −5 and divide by 2:

1. Bring down 2.
2. \(2 \times 2 = 4\); \(-3 + 4 = 1\).
3. \(1 \times 2 = 2\); \(4 + 2 = 6\).
4. \(6 \times 2 = 12\); \(-5 + 12 = 7\).

The last number, 7, is the remainder, and the quotient is \(2x^2 + x + 6\). Both methods agree.

The Factor Theorem

The Factor Theorem is a special case of the Remainder Theorem:

\(x - a\) is a factor of \(p(x)\) if and only if \(p(a) = 0\).

In other words, a remainder of zero means the divisor is a factor. This gives you a fast way to test possible factors.

Which Values Should You Try?

For a polynomial with integer coefficients and leading coefficient 1, any integer root must be a divisor of the constant term. So for a constant of −6, try \(\pm 1, \pm 2, \pm 3, \pm 6\).

Worked Example 2: Factorising a Cubic

Factorise completely \(p(x) = x^3 - 6x^2 + 11x - 6\).

Step 1: List trial values. The constant is −6, so try \(\pm 1, \pm 2, \pm 3, \pm 6\).

Step 2: Test \(x = 1\).

$$ p(1) = 1 - 6 + 11 - 6 = 0 $$

So \(x - 1\) is a factor.

Step 3: Divide to find the quotient. Synthetic division of 1, −6, 11, −6 by 1:

1. Bring down 1.
2. \(1 \times 1 = 1\); \(-6 + 1 = -5\).
3. \(-5 \times 1 = -5\); \(11 - 5 = 6\).
4. \(6 \times 1 = 6\); \(-6 + 6 = 0\).

The quotient is \(x^2 - 5x + 6\).

Step 4: Factorise the quadratic. Find two numbers that multiply to 6 and add to −5: −2 and −3.

$$ x^2 - 5x + 6 = (x - 2)(x - 3) $$

Answer:

$$ x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3) $$

Check: \(p(2) = 8 - 24 + 22 - 6 = 0\) and \(p(3) = 27 - 54 + 33 - 6 = 0\). Correct.

Worked Example 3: Finding an Unknown Constant

Find \(k\) if \(x + 2\) is a factor of \(p(x) = x^3 + kx^2 - 4x + 12\).

Step 1: Since \(x + 2\) is a factor, \(p(-2) = 0\).

Step 2: Substitute:

$$ (-2)^3 + k(-2)^2 - 4(-2) + 12 = -8 + 4k + 8 + 12 = 4k + 12 $$

Step 3: Solve:

$$ 4k + 12 = 0 \quad \Rightarrow \quad k = -3 $$

Check: \(p(x) = x^3 - 3x^2 - 4x + 12\). Then \(p(-2) = -8 - 12 + 8 + 12 = 0\). Correct.

As a bonus, this polynomial factorises by grouping: \(x^2(x - 3) - 4(x - 3) = (x - 3)(x - 2)(x + 2)\).

Worked Example 4: Two Conditions, Two Unknowns

When \(p(x) = x^3 + ax^2 + bx - 4\) is divided by \(x - 1\) the remainder is 2, and \(x + 1\) is a factor. Find \(a\) and \(b\).

Step 1: \(p(1) = 2\):

$$ 1 + a + b - 4 = 2 \quad \Rightarrow \quad a + b = 5 $$

Step 2: \(p(-1) = 0\):

$$ -1 + a - b - 4 = 0 \quad \Rightarrow \quad a - b = 5 $$

Step 3: Add the two equations: \(2a = 10\), so \(a = 5\) and \(b = 0\).

Check: \(p(x) = x^3 + 5x^2 - 4\). Then \(p(1) = 1 + 5 - 4 = 2\) and \(p(-1) = -1 + 5 - 4 = 0\). Correct.

Common Mistakes

• Using the wrong sign. For \(x + 3\), substitute \(x = -3\), not \(+3\).
• Missing terms in synthetic division. For \(x^3 - 2x + 1\), write coefficients as 1, 0, −2, 1, including a 0 for the missing \(x^2\) term.
• Stopping after one factor. "Factorise completely" means breaking the quadratic quotient down too.
• Errors with powers of negatives. \((-2)^2 = 4\) but \((-2)^3 = -8\).

Tips

• Always try \(x = 1\) first: \(p(1)\) is simply the sum of the coefficients.
• \(x = -1\) is a root if the sum of coefficients of even powers equals the sum of coefficients of odd powers.
• Verify each factor by substituting back; it takes seconds and saves marks.

Key Takeaways

• Remainder Theorem: dividing \(p(x)\) by \(x - a\) leaves remainder \(p(a)\).
• Factor Theorem: \(x - a\) is a factor exactly when \(p(a) = 0\).
• To factorise a cubic, find one root by trial, divide, then factorise the quadratic.
• Unknown constants can be found by turning each given condition into an equation.

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