← All solutions · E.T.I. Physics · Algebra toolkit
Factoring is not a hoop you jump through before the physics starts. It is how you pull an unknown out of a system of equations when it is stuck in two places at once. Four markers, seven methods, five things you can push on.
What gets pulled out. The piece every term shares.
What is left of the first term once green leaves.
What is left of the second term. Same colour on both sides.
The result. Boxed, because you are going to use it again.
There is only one rule on this page. Everything else is that rule wearing a different hat.
Left of the equals sign: a sum. Right of it: a product. That swap is the entire point, because a product tells you something a sum never can — if a product is zero, one of its factors is zero. A sum equal to zero tells you nothing whatsoever about its parts.
A rectangle (x+a) wide and (x+b) tall. Its area is the product — and also the sum of the four pieces inside it. Same area, two ways of writing it. That is factoring.
Your unknown turns up in two different terms and you cannot isolate it. Factor it out and it appears exactly once, in front of a bracket. Divide by the bracket and you are done. That is literally step 3 of the two-cable derivation — it is on this site, and we rebuild it live further down.
Read the grey line first: it tells you what to look for. Recognising the pattern is the hard part. Executing it is mechanical.
Look for: something present in every term — a number, a variable, or both. Try this first. Always first.
The product is zero, so a factor is zero: t = 0 (launch) or t = 2v0 sin θ/g (landing). Both roots, no quadratic formula. Simulation 04 below runs this one.
Look for: four terms with nothing shared by all four — but shared in pairs. Split into two pairs, factor each, and a common bracket appears.
The bracket (x+y) was the shared piece all along — it only became visible after the first move.
The grouping says something physical: only the total mass matters, and only the sum of the speeds. Four symbols became two meaningful quantities.
Look for: two perfect squares with a minus between them. A plus does not factor over the reals — that rule does not bend.
Take a square of side a, cut a square of side b out of the corner. Slice what is left into two pieces and slide them into a single rectangle. Its sides are (a+b) and (a−b). Nothing was added or removed — so the two areas are equal, which is the identity.
Two factorings in one line: m is a common factor, then the bracket is a difference of squares. Pair it with v² − v0² = 2aΔx and the work–energy theorem drops out in one step.
Look for: first and last terms are perfect squares, and the middle is exactly twice the product of their roots. Check the middle — that is where it usually fails.
A squared bracket is never negative, and it carries a minus. So y is biggest exactly when the bracket is zero. You get the time and the height of the peak without differentiating anything.
Look for: a leading coefficient of 1. Find two numbers that multiply to c and add to b. Start from the product — it has fewer options.
Look for: a leading coefficient that is not 1. Split the middle term: two numbers multiplying to ac and adding to b. Then group, method 02.
ac = 6, and 6×1 = 6 with 6+1 = 7. Split, then group.
Look for: two perfect cubes. Unlike squares, the sum does factor. For the signs: Same, Opposite, Always Positive.
Turns up in scaling: a sphere's volume goes as r³, so the volume between two radii is a difference of cubes.
Hunting for integer pairs that do not exist is the most common way to lose half a class. The discriminant answers the question in one line, before you start.
Height against time is a quadratic. Factor it instead of attacking it with the formula and both answers appear for free — one of them being the launch itself.
Two cables hold a weight. Equilibrium gives two equations, and the unknown is stuck in two terms of a sum — you cannot divide your way out. Step through it.
Solve one equation for one unknown. Substitute into the other. The surviving unknown will now sit in several terms — factor it out, divide by the bracket, done. Factoring is the step that turns "it is in two places" into "it is in one place".
Same move, now on a vector. Every component carries the same magnitude, so it is a common factor.
What stays in the bracket is a unit vector — pure direction, length one. The sentence "a vector is a magnitude times a direction" is not a definition you memorise. It is a factoring.
The Pythagorean identity is sitting inside the bracket and collapses it to 1. Keep the magnitude buried in each component and you never see it.
Two forces along the same line: F1(cos θ, sin θ) + F2(cos θ, sin θ) = (F1+F2)(cos θ, sin θ). The direction factors out, so only magnitudes add. Obvious written this way, and a reliable source of sign errors when it is not.
You may only cancel something that multiplies the whole numerator and the whole denominator. This is the single most common error here.
On the left, a is a term in a sum. On the right it is a factor of a product. Only the second one cancels.
Given t² = 3t, dividing both sides by t gives t = 3 and silently destroys t = 0. Factor instead: t(t − 3) = 0 keeps both. In simulation 04 the root at zero is the launch — a real instant, not a stray solution.
a² + b² does not factor over the reals. Only the difference does. If you catch yourself writing (a+b)², expand it: there is an extra 2ab that was never in the original.
2x² + 8x + 6 looks general and people start hunting for factors of 12. Pull the 2 out first — 2(x²+4x+3) = 2(x+1)(x+3) — and it is a simple trinomial. Method 01 before anything else. Press "common factor hiding" in simulation 03.
Work down the rows in order. The first one that matches is the method to use.
| If you see | Use | Result | Physics it unlocks |
|---|---|---|---|
| Something in every term | Common factor | k(a+b) | Isolate an unknown; flight time |
| Four terms, shared in pairs | Grouping | (x+y)(a+b) | Collapse to meaningful quantities |
| Square − square | Difference of squares | (a−b)(a+b) | Energy change, v²−v0² |
| Middle = 2× product of roots | Perfect square | (a±b)² | Peak height, no calculus |
| x²+bx+c | Two numbers: ×=c, +=b | (x+p)(x+q) | Roots of quadratic motion |
| ax²+bx+c | Split middle by ac, group | (mx+p)(nx+q) | Collisions, motion with drag |
| Cube ± cube | Sum/difference of cubes | (a±b)(a²∓ab+b²) | Volume and scaling |
E.T.I. Physics · Algebra toolkit · Anuar Kafuri Zarzosa · Tecnológico de Monterrey