Degree and terms
A polynomial is a sum of terms, each a number multiplied by a whole-number power of \(x\), for example \(3x^3 - 2x^2 + 5x - 7\). No negative powers, no roots of \(x\), no \(x\) in a denominator: just non-negative integer powers.
- Term
- One piece of the polynomial, such as \(-2x^2\). It carries a coefficient (the number) and a power of \(x\).
- Degree
- The highest power of \(x\) present. \(3x^3 - 2x^2 + 5x - 7\) has degree \(3\), so it is a cubic.
- Leading coefficient
- The coefficient of the highest-power term. Above it is \(3\). The lone number \(-7\) is the constant term.
| Degree | Name | Example |
|---|---|---|
| 1 | Linear | \(2x - 5\) |
| 2 | Quadratic | \(x^2 - 3x + 1\) |
| 3 | Cubic | \(x^3 - 2x^2 - 5x + 6\) |
| 4 | Quartic | \(x^4 + x - 9\) |
Write in descending order
Before you divide or evaluate, arrange terms from the highest power down and fill any gap with a zero term (for instance \(x^3 + 0x^2 - 5x + 6\)). It keeps columns lined up and stops missed terms.
Algebraic long division
Dividing one polynomial by another works just like the long division of numbers you learned years ago: divide, multiply, subtract, bring down, repeat. The aim is to write
Divide \(x^3 - 2x^2 - 5x + 6\) by \((x - 1)\).
The Remainder Theorem
Long division is thorough but slow if you only want the remainder. The Remainder Theorem gives it instantly:
In words: substitute the value of \(x\) that makes the divisor zero, and the number you get is the remainder. For \((x - a)\) that value is \(x = a\); for \((x + 2)\) it is \(x = -2\).
Find the remainder when \(f(x) = 2x^3 - 3x^2 + x - 5\) is divided by \((x - 2)\).
The Factor Theorem
The Factor Theorem is the special case where the remainder is zero:
So to test whether \((x - a)\) divides a polynomial exactly, just check whether \(f(a) = 0\). If it does, you have found a factor without any division at all. This is the usual first step in factorising a cubic: try small values like \(1, -1, 2, -2\) until one gives zero.
Show that \((x + 2)\) is a factor of \(f(x) = x^3 - 2x^2 - 5x + 6\).
Where this is assessed
Naming degree and applying the theorems is Criterion A. Laying out long division neatly, column by column, is Criterion C. A full "test with the Factor Theorem, then divide, then factorise the quotient" solution shows the joined-up reasoning examiners reward, and links to Criterion B in an investigation.
Mind the sign of \(a\)
For \((x + 2)\) you substitute \(x = -2\), not \(x = 2\). The value of \(a\) is whatever makes the bracket zero. Getting this sign wrong is the most common slip with both theorems.
Check yourself
Work each through before revealing the answer.
1. What is the degree and leading coefficient of \(4 - 7x + 5x^4 - x^2\)? +
Rewrite in descending order: \(5x^4 - x^2 - 7x + 4\). The highest power is \(4\), so the degree is \(4\) (a quartic) and the leading coefficient is \(5\).
2. Find the remainder when \(f(x) = x^3 + 2x^2 - x + 4\) is divided by \((x + 1)\). +
\((x + 1)\) is zero at \(x = -1\), so evaluate \(f(-1) = (-1)^3 + 2(-1)^2 - (-1) + 4 = -1 + 2 + 1 + 4 = 6\). The remainder is \(6\).
3. Is \((x - 3)\) a factor of \(g(x) = x^3 - 4x^2 + x + 6\)? +
Test \(g(3) = 3^3 - 4(3)^2 + 3 + 6 = 27 - 36 + 3 + 6 = 0\). Since \(g(3) = 0\), by the Factor Theorem yes, \((x - 3)\) is a factor.
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