Factoring is the habit of rewriting expressions as products to expose roots, divisibility, cancellation, or hidden structure. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.
Factoring rewrites a sum as a product, turning addition into multiplication so roots, divisibility, and cancellation become visible.
Why It Works
1
Expand $(x + a)(x + b)$ using the distributive law: $x\cdot x + x\cdot b + a\cdot x + a\cdot b$.
2
Collect like terms: $x^2 + (a + b)x + ab$ — exactly the trinomial on the left.
3
So to factor $x^2 + px + q$, search for two numbers $a, b$ with sum $p$ and product $q$; then the polynomial is $(x+a)(x+b)$.
4
A product equals $0$ exactly when one factor is $0$, so the roots are $x = -a$ and $x = -b$ — this is why factoring solves equations.
Worked Examples
Example 1
Solve $x^2 - 5x + 6 = 0$ over the integers.
1
Look for two integers with product $6$ and sum $-5$: the pair $-2$ and $-3$ works since $(-2)(-3) = 6$ and $-2 + -3 = -5$.
2
Factor: $x^2 - 5x + 6 = (x - 2)(x - 3)$.
3
Set each factor to zero: $x - 2 = 0$ or $x - 3 = 0$.
Answer:
$x = 2$ or $x = 3$.
Warm-up
Factor $6x^2 + 11x - 10$ over the integers.
1
Use the AC method: $a\cdot c = 6 \cdot (-10) = -60$; find two integers with product $-60$ and sum $11$ — namely $15$ and $-4$.
2
Split the middle term: $6x^2 + 15x - 4x - 10$, then group: $3x(2x + 5) - 2(2x + 5)$.
3
Factor out the common $(2x + 5)$: $(2x + 5)(3x - 2)$.
Answer:
$6x^2 + 11x - 10 = (2x + 5)(3x - 2)$
Contest level
Factor $x^3 - 2x^2 - 9x + 18$ completely over the integers.
1
Group in pairs: $(x^3 - 2x^2) + (-9x + 18) = x^2(x - 2) - 9(x - 2)$.
2
Factor out the common $(x - 2)$: $(x - 2)(x^2 - 9)$.
3
Recognize $x^2 - 9$ as a difference of squares: $(x - 3)(x + 3)$.
Answer:
$x^3 - 2x^2 - 9x + 18 = (x - 2)(x - 3)(x + 3)$
Olympiad / Challenge
Factor $x^4 + x^2 + 1$ completely over the integers.
1
This has no real roots, so look for a product of two quadratics. Add and subtract $x^2$: $x^4 + x^2 + 1 = (x^4 + 2x^2 + 1) - x^2$.
2
The first group is a perfect square: $(x^2 + 1)^2 - x^2$, now a difference of squares.
3
Factor: $(x^2 + 1 - x)(x^2 + 1 + x) = (x^2 - x + 1)(x^2 + x + 1)$. Each quadratic has discriminant $1 - 4 = -3 < 0$, so neither splits further over the reals.
Answer:
$x^4 + x^2 + 1 = (x^2 - x + 1)(x^2 + x + 1)$
Going Deeper
Generalization: every polynomial with integer coefficients factors uniquely into irreducibles over $\mathbb{Z}$ (Gauss's lemma); over $\mathbb{C}$ it splits completely into linear factors (Fundamental Theorem of Algebra). The 'add and subtract a square' trick generalizes to $x^4 + ax^2 + b$ whenever the leftover is itself a perfect square.
Where it appears: factoring is the universal solvent — it solves polynomial equations (set each factor to $0$), proves divisibility/compositeness in number theory, and exposes cancellation that collapses contest expressions.
Pitfall: 'factor completely' means over the stated ring. $x^2 - 2 = (x - \sqrt 2)(x + \sqrt 2)$ is reducible over $\mathbb{R}$ but irreducible over $\mathbb{Q}$ — never stop early, and never invent factors that introduce non-permitted numbers.
Spot the Signal
Look for problems where the key step is rewriting expressions as products to expose roots, divisibility, cancellation, or hidden structure.
You can describe the hard part as rewriting expressions as products to expose roots, divisibility, cancellation, or hidden structure, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.
Learn the Move
Start by identify the expression or equation that calls for factoring, then rewrite the givens around it.
Name the relation that makes Factoring legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.
Avoid These Traps
Do not use Factoring just because the surface looks familiar; verify the required condition first.
Applying Factoring because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.
MathGrit Coach Note
Let factoring reveal the algebraic structure; then compute only what remains.
Try it on:
Practice a contest problem where the key step is rewriting expressions as products to expose roots, divisibility, cancellation, or hidden structure.