The Remainder and Factor Theorems
Divide a polynomial \(p(x)\) by \((x-a)\) and the remainder is always just \(p(a)\) — no division required. When that remainder is 0, \((x-a)\) divides evenly, so it's a factor of \(p(x)\).
By the end you'll be able to find a division remainder by evaluating \(p(a)\) directly, and use \(p(a) = 0\) to decide whether \((x-a)\) is a factor without doing the full division.
Predict: if \(p(3) = 0\), is \((x-3)\) a factor of \(p(x)\)? Set c = 3 on the slider below and check the remainder.
This is \(p(x) = x^3 - 4x^2 + x + 6\). Drag c and watch two independent calculations — plugging c straight into \(p(x)\), and running synthetic division by \((x-c)\) — land on the exact same number. Whenever that number is 0, \((x-c)\) is a factor, and the graph's x-intercept lights up to match.
The Remainder Theorem turns a division problem into a plug-in-and-evaluate problem, and its special case — the Factor Theorem — turns root-finding and factoring into the same question.
Any division of \(p(x)\) by a linear divisor \((x-a)\) can be written \(p(x) = (x-a)\cdot q(x) + r\), where \(r\) is a constant because the divisor has degree 1. Substitute \(x = a\): the \((x-a)\) factor becomes 0, so \(p(a) = 0\cdot q(a) + r = r\). The remainder is \(p(a)\) — that's the side-by-side match you saw in the diagram above. Setting \(r = 0\) gives the Factor Theorem: \((x-a)\) is a factor of \(p(x)\) exactly when \(p(a) = 0\).
Think of \(p(a) = 0\) as saying "x = a makes the polynomial vanish" — a is a root (or zero) of \(p\), and geometrically the graph crosses the x-axis right there, at \((a, 0)\). If a value doesn't zero out the polynomial, dividing by \((x-a)\) still works — you just get whatever nonzero remainder \(p(a)\) happens to equal, and \((x-a)\) is not a factor.
When you're hunting for a polynomial's factors, testing candidate values with \(p(a)\) is far faster than dividing every candidate out by hand. Once you find one \(a\) with \(p(a) = 0\), you divide once (synthetic division is quickest) to knock the degree down by one, then repeat the search on the smaller quotient — the strategy behind rational-root hunting for higher-degree polynomials.
Let \(p(x) = x^3 - 4x^2 + x + 6\). Test \(a = 2\): \(p(2) = 8 - 16 + 2 + 6 = 0\). The remainder is 0, so by the Factor Theorem, \((x-2)\) is a factor. Divide to find the cofactor (synthetic division with \(a = 2\)): quotient \(x^2 - 2x - 3\), so \(p(x) = (x-2)(x^2-2x-3) = (x-2)(x-3)(x+1)\). Now test \(a = 1\): \(p(1) = 1 - 4 + 1 + 6 = 4 \neq 0\), so \((x-1)\) is not a factor — and 4 is exactly the remainder you'd get dividing \(p(x)\) by \((x-1)\).
A different cubic: \(p(x) = x^3 - 2x^2 - 5x + 6\). Test \(a = 1\): \(p(1) = 1 - 2 - 5 + 6 = \) ____. Since the remainder is ____, is \((x-1)\) a factor of \(p(x)\)?
Reveal the answer
\(p(1) = 1 - 2 - 5 + 6 = 0\). The remainder is 0, so \((x-1)\) is a factor — dividing gives quotient \(x^2 - x - 6 = (x-3)(x+2)\), so \(p(x) = (x-1)(x-3)(x+2)\). Set c = 1 on the slider above (using the interactive's own cubic) and confirm the same pattern: remainder 0 lights up the factor banner.
More info — why the remainder has to equal p(a)
Every division of \(p(x)\) by \((x-a)\) fits the same mold as integer division: \(\text{dividend} = \text{divisor}\times\text{quotient} + \text{remainder}\), i.e. \(p(x) = (x-a)q(x) + r\). Because \((x-a)\) has degree 1, whatever is left over after dividing (the remainder \(r\)) can't contain any \(x\) at all — otherwise you could divide further. So \(r\) is just a number, and plugging in \(x=a\) zeroes out the \((x-a)q(x)\) term entirely, leaving \(p(a) = r\). See the Khan Academy link in Dive deeper for a worked walkthrough of this substitution.
Check your understanding
Let \(p(x) = x^3 - 3x^2 - x + 3\). What is the remainder when \(p(x)\) is divided by \((x - 3)\)?
For \(p(x) = x^3 - 3x^2 - x + 3\), you compute \(p(-2) = -15\). What does this tell you about \((x + 2)\)?
Predict, then check: if \(p(3) = 0\) for some polynomial p(x), is \((x - 3)\) a factor of p(x)?
You divide \(p(x) = x^3 + 2x^2 - 5x - 6\) by \((x - 2)\) using synthetic division and get quotient \(x^2 + 4x + 3\) with remainder 0. What must \(p(2)\) equal, and why?
Recap
- Remainder Theorem: dividing \(p(x)\) by \((x-a)\) leaves remainder \(p(a)\) — evaluate instead of dividing.
- Factor Theorem: \((x-a)\) is a factor of \(p(x)\) exactly when \(p(a) = 0\).
- A value \(a\) with \(p(a)=0\) is a root (zero) of \(p\), and \((a,0)\) is an x-intercept of its graph.
- Watch the sign: testing \((x+a)\) means evaluating \(p(-a)\), since \(x+a = x-(-a)\).
Dive deeper
- Khan Academy — Polynomial remainder theorem See why evaluating p(a) yields the division remainder.
- Purplemath — The Remainder Theorem Connect the remainder to the Factor Theorem for root-finding.
Sources
- The Remainder and Factor Theorems