Open any algebra textbook, any board paper, any JEE practice set, and you will see the instruction printed in bold above the problems: factor completely. Three problems in, you stop reading the word "completely" — it feels like decoration, like the textbook is just being polite. It is not. That word is a stopping rule, and getting it wrong costs marks on every paper you sit.
"Factor completely" means: keep factoring until every factor in your final answer is irreducible — cannot be broken down any further. The moment one of your factors still has a hidden factorisation inside it, you are not done. The teacher's red pen is waiting.
This article gives you the precise stopping rule, the checks to apply at each step, and the common traps that drain marks from otherwise correct work.
What "irreducible" means
A polynomial is irreducible over the real numbers if it cannot be written as a product of two lower-degree polynomials with real coefficients. That last clause matters — "lower-degree" is what blocks the trivial factorisation p(x) = 1 \cdot p(x).
Three cases tell you everything:
- Linear polynomials (degree 1, like x - 3 or 2x + 7): always irreducible. You cannot break a linear into a product of two lower-degree polynomials, because lower than degree 1 means degree 0, which is just a constant.
- Quadratics (degree 2, like x^2 + 3x + 2): irreducible if and only if the discriminant D = b^2 - 4ac < 0. If D \geq 0, the quadratic splits into two real linear factors (x - r_1)(x - r_2).
- Higher degree (cubic, quartic, etc.): generally reducible. The Fundamental Theorem of Algebra guarantees that every polynomial of degree \geq 2 over the reals factors into linear and quadratic-irreducible pieces. Spotting those pieces is the work.
The stopping rule
At every step of your factoring, look at each factor and ask: is this reducible?
- If the factor is linear: stop this branch. Linear is always atomic.
- If the factor is quadratic with D \geq 0: factor it into (x - r_1)(x - r_2).
- If the factor is quadratic with D < 0: stop this branch. Irreducible over the reals.
- If the factor is higher degree: try difference of squares, sum or difference of cubes, grouping, or the rational root theorem. Then re-check each new factor.
You are done only when every factor in your answer is either linear or an irreducible quadratic.
Worked example 1 — the chain
Factor x^4 - 16 completely.
Step 1: recognise difference of squares. x^4 - 16 = (x^2)^2 - 4^2 = (x^2 - 4)(x^2 + 4).
Now check each factor:
- x^2 - 4: also a difference of squares. Factor again: (x - 2)(x + 2).
- x^2 + 4: a sum of squares. Discriminant is 0 - 16 = -16 < 0. Irreducible over the reals.
Step 2: combine. x^4 - 16 = (x - 2)(x + 2)(x^2 + 4).
Final check: two linears and one irreducible quadratic. Done.
Worked example 2 — partial factoring loses marks
Factor 2x^2 - 8 completely.
A common student answer: 2(x^2 - 4). The GCF was pulled. The student moves on.
This is incomplete. The factor x^2 - 4 is itself a difference of squares — it still factors. Correct answer:
The teacher sees an unfactored quadratic sitting in the answer and deducts marks. The fix took ten seconds. Always re-scan after pulling the GCF.
Worked example 3 — when "complete" looks like more than it is
Factor x^4 + 4x^2 + 4 completely.
Substitute u = x^2: the expression becomes u^2 + 4u + 4 = (u + 2)^2, so x^4 + 4x^2 + 4 = (x^2 + 2)^2.
Is (x^2 + 2)^2 fully factored? Look at x^2 + 2. Discriminant is 0 - 8 = -8 < 0. Irreducible over the reals. So (x^2 + 2)^2 is the complete factorisation. You stop.
A repeated irreducible factor still counts as fully factored — you do not need to "split" the square.
Worked example 4 — keep going until you cannot
Factor x^6 - 1 completely.
Step 1: difference of squares. x^6 - 1 = (x^3)^2 - 1^2 = (x^3 - 1)(x^3 + 1).
Step 2: each cubic uses a cube identity.
- x^3 - 1 = (x - 1)(x^2 + x + 1).
- x^3 + 1 = (x + 1)(x^2 - x + 1).
Combine: x^6 - 1 = (x - 1)(x + 1)(x^2 + x + 1)(x^2 - x + 1).
Are the quadratics irreducible? Both have discriminant 1 - 4 = -3 < 0. Yes — irreducible over the reals.
Fully factored. Two linears, two irreducible quadratics, done.
The level of factorisation depends on the field
"Irreducible" is not absolute — it depends on which numbers you allow as coefficients.
- Over the complex numbers \mathbb{C}: every polynomial splits into linear factors. The Fundamental Theorem of Algebra. So x^2 + 1 = (x - i)(x + i).
- Over the real numbers \mathbb{R}: linear factors and irreducible quadratics (those with D < 0). x^2 + 1 is irreducible.
- Over the rationals \mathbb{Q}: more restrictive. x^2 - 2 is irreducible over \mathbb{Q} (its roots are \pm\sqrt{2}, not rational), but factors over \mathbb{R} as (x - \sqrt{2})(x + \sqrt{2}).
In Indian school and JEE problems, the default is "over the reals" unless stated otherwise.
Checking your answer is fully factored
Run this checklist on every factor in your final answer:
- Is it linear? If yes, atomic. Move on.
- Is it quadratic? Compute D = b^2 - 4ac. If D < 0, irreducible — atomic. If D \geq 0, you missed a step. Factor it.
- Is it higher degree? Try difference of squares (a^2 - b^2), sum or difference of cubes (a^3 \pm b^3), grouping, or the rational root theorem to find a linear factor and divide.
- Repeat until every piece passes step 1 or step 2 with D < 0.
A constant multiplier out front (the GCF) is fine and expected — it is not a "factor" that needs further treatment.
Common student errors
- Stopping at partial GCF pull. Pulling out the 2 in 2x^2 - 8 and forgetting the difference-of-squares inside.
- Treating x^2 + 4 as factorable. It is not — over the reals. Some students chase this for two minutes and write nonsense.
- Confusing x^2 - 4 with x^2 + 4. The minus sign is the difference of squares; the plus sign is irreducible. Read carefully.
- Forgetting to re-check after each step. Every time you produce new factors, check each one.
What if the problem specifies the field?
Sometimes a problem says "factor over the rationals" or "factor over the complex numbers". This changes the stopping rule.
- Over \mathbb{Q}: stop when each factor has rational coefficients and is irreducible over \mathbb{Q}. x^2 - 2 stays as is.
- Over \mathbb{R}: x^2 - 2 = (x - \sqrt{2})(x + \sqrt{2}).
- Over \mathbb{C}: x^2 + 1 = (x - i)(x + i), and every quadratic splits.
When no field is specified in an Indian school or JEE setting, factor over \mathbb{R}.
Recognition drill
For each, decide whether it is fully factored over the reals:
- (x - 1)(x - 2)(x^2 + x + 1) — the quadratic has D = 1 - 4 = -3 < 0. Irreducible. Fully factored.
- (x - 1)(x^2 - 4) — the quadratic factors as (x - 2)(x + 2). Not fully factored. Correct: (x - 1)(x - 2)(x + 2).
- 2(x - 1)(x - 2) — constant out front, two linears. Fully factored.
- (x^2 - 1)(x^2 + 1) — left factor is a difference of squares. Not fully factored. Correct: (x - 1)(x + 1)(x^2 + 1).
- 3(x^2 + 1) — quadratic has D = -4 < 0. Irreducible. Fully factored.
The closing rule
"Factor completely" means every factor in your final answer is irreducible. Discriminant for quadratics, difference-of-squares for even powers, sum and difference of cubes for cubic forms — apply, re-check, apply again, stop only when every piece is atomic.
A partial factorisation looks almost right and earns almost-zero. The full factorisation takes another thirty seconds and saves the marks. Spend the thirty seconds, every time.