MISSION · Algebra
Dividing Polynomials
Long division, with x in place of digits
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The idea in one sentence
Dividing a polynomial by \((x - a)\) works exactly like long division with numbers, and the remainder it leaves is simply \(f(a)\).
The remainder theorem
When \(f(x)\) is divided by \((x - a)\), the remainder is \(f(a)\).
That is a remarkable shortcut. To find the remainder of \(x^3 - 4x + 1\) divided by \((x - 2)\), don’t divide — substitute:
\[f(2) = 8 - 8 + 1 = 1\]The remainder is 1, found in one line.
Why it works: division says \(f(x) = (x - a) \cdot Q(x) + R\), where \(R\) is a constant. Put \(x = a\) into that and the first term vanishes, because \((a - a) = 0\). What’s left is \(f(a) = R\).
The factor theorem
A remainder of zero means the division came out exactly, which means \((x - a)\) is a factor:
\((x - a)\) is a factor of \(f(x)\) if and only if \(f(a) = 0\).
So testing for factors becomes substitution. For \(x^3 - 7x + 6\): try \(f(1) = 1 - 7 + 6 = 0\) — so \((x - 1)\) is a factor, found without dividing anything.
That’s the standard way into a cubic. The quadratic formula handles degree 2; for degree 3 you hunt one root by substitution, divide it out, and the quadratic that’s left is solvable by ordinary means.
Where to look for that first root: any whole-number root must divide the constant term. For \(x^3 - 7x + 6\), the constant is 6, so try \(\pm 1, \pm 2, \pm 3, \pm 6\) and no others. That turns an infinite search into eight substitutions.
Doing the division
Long division with polynomials follows the digit version step for step: divide the leading terms, multiply back, subtract, bring down the next term, repeat.
\[\frac{x^3 - 7x + 6}{x - 1} = x^2 + x - 6\]Two cautions. Write in descending powers, and include a zero for any missing power — \(x^3 - 7x + 6\) has no \(x^2\) term, so it must be treated as \(x^3 + 0x^2 - 7x + 6\). Leaving the gap out misaligns everything after it, the same way misaligned columns wreck numeric long division.
Then finish the factorisation: \(x^2 + x - 6 = (x + 3)(x - 2)\), so the roots are 1, \(-3\), and 2.
Common mix-ups
- Dropping a missing power. Put the zero in.
- Getting the sign of \(a\) backwards. For \((x - 2)\) substitute \(x = 2\); for \((x + 3)\) substitute \(x = -3\).
- Subtracting only the first term of the product at each step.
- Stopping at the first factor rather than factorising what’s left.
Check yourself
Find the remainder when \(x^3 + 2x^2 - 5\) is divided by \((x + 1)\), and say whether it’s a factor.
Substitute \(x = -1\): \(-1 + 2 - 5 = -4\). The remainder is \(-4\), so \((x + 1)\) is not a factor.
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