Here is one of those classroom reflexes that eats up exam minutes. A question appears: "Does x = 3 satisfy p(x) = x^2 - 5x + 6?" The student leans in, pulls out the factoring toolkit, and carefully works out p(x) = (x - 2)(x - 3). Then reads off: "yes, x = 3 is a root." Correct answer, wrong tool.

You could have just plugged in. p(3) = 9 - 15 + 6 = 0. Same conclusion, faster. No factoring anxiety, no wondering if you split the middle term correctly, no sign slip in the brackets.

The reflex to factor every polynomial you see is the first thing to unlearn. Factoring is a sledgehammer. For "what is p at this specific number?", you need arithmetic, not a sledgehammer.

Two different questions, two different tools

Polynomial problems come in three flavours, and each flavour has its own tool.

"What is p(5)?" — Pure evaluation. Substitute 5, compute. No factoring required.

"What are the roots of p?" — Structural question. You need all x with p(x) = 0. Plugging in won't work unless you already know the answers. Tool: factor.

"Does 3 satisfy p(x) = 0?" — Verification. Substitute 3, check if the result is 0.

Two of these three are solved by plugging in. Only one requires factoring. Default instinct: plug in first, reach for factoring only when the question demands it.

When to plug in (evaluate)

You plug in whenever a specific number is named in the question. Watch for these phrasings:

In every one of these, a specific number is provided or implied (0 for the y-intercept). You do not need the factored form to get a numerical answer. You just need to substitute.

When to factor

Factoring is the right tool when the question is about structure — about the set of values that do something, not about one specific value.

The test: does the question mention a specific number, or is it asking for the general behaviour? If specific, plug in. If general, factor.

Worked example — evaluation, don't factor

Let p(x) = x^3 - 4x^2 + 3x + 8. What is p(2)?

Plug in: p(2) = 8 - 16 + 6 + 8 = 6. Done in fifteen seconds.

Now imagine the student who decided to "first factor the cubic." They would try the rational roots theorem, test \pm 1, \pm 2, \pm 4, \pm 8, and after two minutes of effort realise this cubic has no rational roots. All that work produced no useful information about p(2).

Worked example — factoring, don't just evaluate

Let p(x) = x^2 - 5x + 6. Question: find all the roots.

Can you plug in to solve this? Only by trying every number in the universe, one at a time. That is not a strategy.

Factor: p(x) = (x - 2)(x - 3). Set each factor to zero. Roots: x = 2 and x = 3.

The structure of "find all roots" demands factoring. No shortcut exists for a general polynomial.

Worked example — plug in as a factor-verification

Suppose someone claims: "p(x) = (x - 3)(x + 2) is the correct factoring of x^2 - x - 6."

You can verify without re-doing the factoring. Just check the claimed roots by plugging into the original.

Plug x = 3: original gives 9 - 3 - 6 = 0. Factored gives 0 \cdot 5 = 0. Matches.

Plug x = -2: original gives 4 + 2 - 6 = 0. Factored gives -5 \cdot 0 = 0. Matches.

Both claimed roots are genuine roots of the original. The factoring is verified without you lifting a pencil to expand brackets. This is plug-in as a quick audit of someone else's (or your own) factored answer.

The "Horner's method" speed trick

Evaluating high-degree polynomials looks tedious, but there is a way to cut the number of multiplications. Take p(x) = 3x^3 - 2x^2 + x - 5 at x = 4.

Naive: 3 \cdot 64 - 2 \cdot 16 + 4 - 5 = 192 - 32 + 4 - 5 = 159. You computed x^2 and x^3 separately, which is extra work.

Horner: rewrite as ((3x - 2)x + 1)x - 5.

Step by step at x = 4: 3 \cdot 4 - 2 = 10. Then 10 \cdot 4 + 1 = 41. Then 41 \cdot 4 - 5 = 159.

Same answer, but you never computed 4^2 or 4^3 as separate quantities. Three multiplications and three additions, total. For a degree-n polynomial, Horner uses n multiplications; the naive approach uses up to 2n - 1.

Horner in one line

The general pattern for p(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0:

p(x) = (\ldots(((a_n \, x + a_{n-1}) \, x + a_{n-2}) \, x + \ldots) \, x + a_1) \, x + a_0

Write the coefficients in descending order of degree. Start with a_n. Multiply by x, add the next coefficient. Multiply by x again, add the next one. Keep going until you hit the constant term. That is your answer.

If a coefficient is zero (say the polynomial skips a degree), write it as 0 in the list. Missing terms are not the same as absent terms.

Recognition drill

For each of these, say the right approach before you compute:

Notice how quickly the right tool becomes obvious once you ask: is a specific number named?

Common confusions

"Factoring is always useful." True in principle, expensive in practice. Reaching for it when you only need to evaluate is like firing up an oven to melt a cube of butter.

"Plug in is only for simple polynomials." Not with Horner's method. A degree-7 polynomial evaluates in seven multiplications. That is faster than factoring it, which for degree 7 might not even be possible in closed form.

"Always verify by plugging in." Good habit after you factor — plug each claimed root into the original and check you get 0. But for a pure evaluation problem, do not manufacture a factoring step just so you can verify it.

Why this matters

Factoring a cubic carefully takes two minutes. Evaluating it at a specific point takes fifteen seconds with Horner's method. If the problem only wants the value, the student who factored has spent 1 minute 45 seconds on nothing. Multiply that across five problems in a paper and the marks vanish.

Closing

Match the tool to the question. Plug in for specific values. Factor for general structure. The first sentence tells you which — if a number is named, substitute; if the set of all roots is named, factor.