NapkinCalc

Algebra 1B — Systems, Exponents & Factoring

Factoring & the zero-product rule

continues from lesson 2 — values defined earlier in the course stay live here

ELI5: factoring rewrites a sum as a product — and that unlocks the zero-product rule: if A·B = 0, then A = 0 or B = 0. So a quadratic set to zero splits into two easy linear equations. For example x² − 5x + 6 = (x − 2)(x − 3), whose roots are simply 2 and 3.

factor(x2−5⋅x+6)\mathrm{factor}\left(x^{2} - 5 \cdot x + 6\right) = (x−3)⋅(x−2)\left(x-3\right) \cdot \left(x-2\right) the engine factors it: (x − 2)(x − 3) — click Steps to watch the hunt
rootsf=solve(x2−5⋅x+6,x)roots_{f} = \mathrm{solve}\left(x^{2} - 5 \cdot x + 6, x\right) = [2,3][2, 3] and the roots fall out: {2, 3}
✓ pass max(rootsf)==3andmin(rootsf)==2max(roots_{f}) == 3 and min(roots_{f}) == 2 x = 2 and x = 3

Real-world hook: a thrown ball's height is a quadratic in time; factoring tells you exactly when it hits the ground (height = 0). Same trick sizes break-even points and rectangle dimensions.

Try it yourself: x² − x − 12 factors as (x + 3)(x − 4). Enter its larger root.

rootyou=4root_{you} = 4 ✏️ Your turn: find the larger root of x² − x − 12. The check confirms it makes (x + 3)(x − 4) = 0 — without telling you which root.
✓ pass abs((rootyou+3)∗(rootyou−4))<1e−9androotyou>0abs((root_{you} + 3) * (root_{you} - 4)) < 1e-9 and root_{you} > 0 green when your value is the larger root

Where next: Algebra 2A treats x² seriously — functions, parabolas, and the discriminant.