A problem asks you to evaluate (x+1)(x+2)(x+3)(x+4) at x = -2.5. Another asks for the coefficient of x² in (x+1)(x+2)(x+3)...(x+10). Your first instinct, if you have been trained only in FOIL, is to multiply out. Don't. At degree 4 the expansion is tedious; at degree 10 it is impossible to do on paper in exam time. Questions that hand you a long chain of factors are almost never asking for the full expansion. They are asking "do you see the shortcut?" Your job is to look for the shortcut first, and pick up the pen only after you have found it.

The habit

The moment you see a product of more than three or four binomial factors, stop. Do not reach for pen and paper. Read the question again and ask what is actually being asked. Long products tend to fall into one of five templates, and each has its own shortcut.

Run through these templates in your head before you expand anything. Nine times out of ten, one of them fits.

Worked example 1 — specific-value substitution

Evaluate (x+1)(x+2)(x+3)(x+4) at x = -5.

The brute path is to expand the full degree-4 polynomial, collect terms, and then substitute. That is about ten minutes of arithmetic and four opportunities to sign-flip wrong.

The smart path: you are not asked for the polynomial. You are asked for its value at one specific x. Plug in directly.

(-5+1)(-5+2)(-5+3)(-5+4) = (-4)(-3)(-2)(-1) = 24.

Ten seconds. Notice how the question quietly chose x = -5, a value that makes every factor a small negative integer. That is the setter telling you to substitute.

Worked example 2 — coefficient of a specific power

Find the coefficient of x⁸ in (x+1)(x+2)(x+3)...(x+10).

This is a degree-10 polynomial. The full expansion has eleven terms and requires tracking every possible product of roots. On paper, intractable.

Use Vieta's. For the polynomial (x + r₁)(x + r₂)...(x + r_n), the coefficient of x^(n-k) equals the sum of all products of the r_i taken k at a time. Here n = 10 and you want the coefficient of x⁸, which sits at position n - 2 = 8, so k = 2. The roots (in the Vieta sense, with the sign convention of the expansion (x+r)) are r_i = i for i = 1, 2, ..., 10.

You need Σ r_i r_j for i < j. The identity you want is

Σ_{i<j} r_i r_j = [ (Σ r_i)² − Σ r_i² ] / 2.

Compute each piece. Σ r_i = 1 + 2 + ... + 10 = 55. Σ r_i² = 1² + 2² + ... + 10² = 10·11·21/6 = 385. So

Σ_{i<j} r_i r_j = (55² − 385) / 2 = (3025 − 385) / 2 = 2640 / 2 = 1320.

The coefficient of x⁸ is 1320. You never expanded a single pair of brackets.

Worked example 3 — telescoping product

Simplify (1 − 1/2)(1 − 1/3)(1 − 1/4)...(1 − 1/n).

Brute: compute each factor as a decimal and multiply. For n = 100, that is 99 multiplications and a rounding error in every one of them.

Smart: rewrite each factor as a single fraction. 1 − 1/k = (k−1)/k. So the product becomes

(1/2)(2/3)(3/4)(4/5)...((n−1)/n).

Now look at the numerators and denominators. The numerator of the second factor (2) cancels the denominator of the first (2). The numerator of the third (3) cancels the denominator of the second (3). And so on, all the way down. Every number from 2 through n−1 appears once on top and once on the bottom. What survives is the first numerator (1) and the last denominator (n).

Product = 1/n.

For n = 100, the answer is 1/100. You got it in thirty seconds.

Worked example 4 — symmetric pairs

Expand (x−1)(x+1)(x−2)(x+2)(x−3)(x+3).

Brute: six factors, each with two terms, so 64 products to sum. Miserable.

Smart: pair the opposites.

[(x−1)(x+1)] · [(x−2)(x+2)] · [(x−3)(x+3)] = (x²−1)(x²−4)(x²−9).

Now it is a product of three factors in the variable x². Let u = x². You are expanding (u−1)(u−4)(u−9), which is a cubic — a standard box-method job. You get u³ − 14u² + 49u − 36, so the answer is

x⁶ − 14x⁴ + 49x² − 36.

The pairing trick cut 64 products to the effort of expanding one cubic.

Recognising when brute-force is actually fine

Shortcuts are not always right. Use brute force when it really is the fastest path.

If you see four or more factors, the default assumption flips: a pattern is probably hiding.

Vieta's identities, restated

For a polynomial a_n x^n + a_{n−1} x^{n−1} + ... + a_0 with roots r₁, r₂, ..., r_n:

If your polynomial is in the factored form (x + c₁)(x + c₂)...(x + c_n), then the "roots" (by Vieta's sign convention) are the c_i. You can read off any symmetric function of the roots without expanding anything. Use this whenever the question asks for a specific coefficient of a long product.

Exam strategy heuristics

The "look before you leap" discipline

There is one habit that separates students who finish the paper from students who run out of time on question four. Before you start any computation, read the question fully. Identify the exact output the question wants — is it a number, a coefficient, a full polynomial, a simplified form? Pick the method that delivers that output in the fewest steps. Only then pick up the pen.

If you start expanding before you have asked "what does the question actually want," you are committing to the longest possible route without checking whether a shortcut exists. On paper, that costs minutes. In an exam, it costs marks.

Recognition drill

For each, identify the shortcut before computing anything.

Notice how every one of these is a ten-second problem if you see the shortcut, and a multi-minute grind if you don't.

Closing

Long products almost never want brute expansion. They want you to notice something — a substitution, a telescoping, a symmetric pair, a coefficient formula, an identity. The setter is testing pattern recognition, and the reward for seeing the pattern is time. Spend the first twenty seconds of any long-product problem scanning for the shortcut. You find it more often than you expect.