Day 16 — Factoring
Day 16 — Factoring
Stage II · concept day
Goal: Factor by GCF, grouping, trinomial patterns, difference of squares and cubes; use the factor theorem idea; solve quadratic equations by factoring and the zero-product property.
Why this matters
Factoring reverses expansion and reveals roots. Zero-product reasoning is how many discrete equations get solved. Difference of squares appears in algorithmic identities and algebraic rewrites that avoid cancellation (Day 9).
Theory
Goal of factoring
Write a polynomial as a product of lower-degree polynomials (usually with integer or rational coefficients). Completely factored over a ring means factors are irreducible there (e.g. linear factors over \(\mathbb{R}\) when possible).
Zero-product property
For real (or integral domain) numbers: \[ab=0\quad\Longleftrightarrow\quad a=0\text{ or }b=0.\]
Corollary. If \((x-r)(x-s)=0\), then \(x=r\) or \(x=s\).
Greatest common factor (GCF)
Factor largest common monomial: \(6x^3+9x^2=3x^2(2x+3)\).
Factoring by grouping
Four terms: \(ax+ay+bx+by=a(x+y)+b(x+y)=(a+b)(x+y)\).
Trinomials \(x^2+bx+c\)
Find \(p,q\) with \(p+q=b\), \(pq=c\); then \(x^2+bx+c=(x+p)(x+q)\).
Trinomials \(ax^2+bx+c\) (\(a\neq 1\))
AC method: find numbers summing to \(b\) with product \(ac\); split middle; group.
Or trial factors of \(a\) and \(c\).
Difference of squares
\[a^2-b^2=(a-b)(a+b).\]
Proof. Expand right-hand side. \(\square\)
Sum/difference of cubes
\[ \begin{align*} a^3-b^3&=(a-b)(a^2+ab+b^2),\\ a^3+b^3&=(a+b)(a^2-ab+b^2). \end{align*} \]
Perfect square trinomials
\[a^2+2ab+b^2=(a+b)^2,\qquad a^2-2ab+b^2=(a-b)^2.\]
Factor theorem (idea)
If \(p(r)=0\), then \((x-r)\) is a factor of \(p\).
Trial rational roots (rational root theorem lite): any rational root \(\frac{p}{q}\) in lowest terms has \(p\mid\) constant term, \(q\mid\) leading coefficient.
Solving quadratics by factoring
- Write \(ax^2+bx+c=0\).
- Factor left side.
- Apply zero-product.
- Check.
Not all quadratics factor nicely over the integers; Day 19 gives formula and completing the square.
Strategy checklist
- GCF first.
- Count terms: \(2\) → difference of squares/cubes; \(3\) → trinomial/perfect square; \(4\) → grouping.
- Factor further until irreducible.
- Check by expansion.
Worked examples
Example 1 — GCF
\(12x^4-18x^2=6x^2(2x^2-3)\).
Example 2 — Grouping
\(x^3+2x^2+3x+6=x^2(x+2)+3(x+2)=(x^2+3)(x+2)\).
Example 3 — Simple trinomial
\(x^2-5x+6=(x-2)(x-3)\).
Example 4 — AC method
\(6x^2+7x-3\). \(ac=-18\); pair \(9,-2\): \(6x^2+9x-2x-3=3x(2x+3)-1(2x+3)=(3x-1)(2x+3)\).
Example 5 — Difference of squares
\(9x^2-16=(3x-4)(3x+4)\).
\(x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)\).
Example 6 — Cubes
\(x^3-8=(x-2)(x^2+2x+4)\).
\(x^3+27=(x+3)(x^2-3x+9)\).
Example 7 — Perfect square
\(x^2-6x+9=(x-3)^2\).
Example 8 — Factor theorem trial
\(p(x)=x^3-2x^2-x+2\). \(p(1)=0\). Divide / synthetic: \((x-1)(x^2-x-2)=(x-1)(x-2)(x+1)\).
Example 9 — Solve by factoring
\(x^2-x-12=0\Rightarrow(x-4)(x+3)=0\Rightarrow x=4\) or \(x=-3\).
Example 10 — Quadratic no GCF miss
\(2x^2+4x=0\Rightarrow 2x(x+2)=0\Rightarrow x=0\) or \(x=-2\). (Do not divide by \(x\) and lose \(x=0\).)
Example 11 — Higher
\(x^3-x=x(x^2-1)=x(x-1)(x+1)\).
Example 12 — Multivariable
\(a^2-2ab+b^2-c^2=(a-b)^2-c^2=(a-b-c)(a-b+c)\).
Example 13 — Check
\((2x-1)(x+5)=2x^2+10x-x-5=2x^2+9x-5\).
Example 14 — Irreducible over reals?
\(x^2+1\) has no real linear factors; over complexes \((x-i)(x+i)\).
Exercises
Easy
- Factor \(15x^3-10x\).
- Factor \(x^2+7x+12\).
- Factor \(x^2-9\).
- Solve \((x-2)(x+5)=0\).
- Factor \(x^2-10x+25\).
Medium
- Factor \(6x^2-x-2\).
- Factor \(x^3+3x^2-x-3\) by grouping.
- Factor \(27x^3-1\).
- Factor \(x^4-16\) completely over the reals.
- Solve \(2x^2-5x-3=0\) by factoring.
- Solve \(x^2=5x\) without losing a root.
- Factor \(2x^3+4x^2-6x\) completely.
- Use factor theorem: show \(x+2\) divides \(x^3+6x^2+11x+6\); factor fully.
Hard / proof
- Prove \(a^2-b^2=(a-b)(a+b)\) and \(a^3-b^3=(a-b)(a^2+ab+b^2)\) by expansion.
- Prove zero-product property from field axioms (no zero divisors).
- Explain why dividing an equation by \(x\) can lose solutions; give example.
- Factor \(x^6-1\) completely over the reals (use difference of squares/cubes as needed).
- For which \(c\) does \(x^2+cx+1\) factor with real linear factors?
- Show that if \(p\) has roots \(r_1,\ldots,r_k\) distinct and \(\deg p=k\), then \(p(x)=a(x-r_1)\cdots(x-r_k)\).
- Disprove: every quadratic factors over the integers.
Challenge / CS-flavored
- Boolean algebra analogy: factoring vs expanding—when do you want CNF-like product form?
- Complexity identity: \(n^2-1=(n-1)(n+1)\) for analyzing neighboring sizes.
- Solve \(n(n-1)=20\) for integer \(n\) (combinatorial \(P(n,2)\) feel).
- Polynomials over \(\mathbb{Z}/2\mathbb{Z}\): factor \(x^2+x\) — note \(x^2+x=x(x+1)\) and every element is root (Freshman’s dream awareness).
- Rewrite \(x^2-y^2\) to factor a difference of measurements in code-free algebra.
Rational root theorem (stated)
If \(p(x)=a_n x^n+\cdots+a_0\) has integer coefficients and rational root \(\frac{u}{v}\) in lowest terms, then \(u\mid a_0\) and \(v\mid a_n\).
Trial list is finite—pair with synthetic division.
Irreducibles over \(\mathbb{R}\)
Over \(\mathbb{R}\), irreducible polynomials are degree \(1\), or degree \(2\) with negative discriminant. Higher degrees always factor into those (fundamental theorem of algebra + real coefficients force conjugate pairs)—stated for orientation, not proved.
Completing the square vs factoring
Not every monic quadratic with real roots factors “nicely” over \(\mathbb{Z}\), but it factors over \(\mathbb{R}\) as \(a(x-r)(x-s)\). Completing the square / formula always works over \(\mathbb{R}\) when \(\Delta\ge 0\).
Zero-product extended
\(a_1 a_2\cdots a_k=0\) iff some \(a_i=0\) (in a field / integral domain). Use after complete factorization.
Extra exercises
- Factor \(x^{3}+1\) completely over the reals.
- Solve \(x(x-3)(x+2)=0\).
- Factor \(a^{4}-b^{4}\) completely over the reals.
- Use rational root theorem candidates for \(2x^{3}+3x^{2}-1=0\); find one rational root if any.
- Explain why \(x^{2}-2\) is irreducible over \(\mathbb{Q}\) but reducible over \(\mathbb{R}\).
CS connection
Factoring reveals structure for partial fractions (Day 17) and closed forms of recurrences. Zero-product mirrors “if product of flags is zero…” less than boolean OR, but the logic of cases is the same. Irreducibility over a field matters in coding theory and crypto polynomials.
Common pitfalls
| Pitfall | What to do instead |
|---|---|
| Forgetting GCF | Always pull GCF first |
| Sign errors in trinomials | Check by FOIL |
| Dividing variables out of equations | Factor and use zero-product |
| Stopping at \(x^2-1\) | Continue to \((x-1)(x+1)\) if complete |
| Assuming all quadratics factor over \(\mathbb{Z}\) | Discriminant / formula later |
| Dropping \((x^2+1)\) factors over reals | Know when irreducible |
Fully worked extra examples
E15 — Grouping four terms carefully.
\(xy+5x+3y+15=x(y+5)+3(y+5)=(x+3)(y+5)\).
E16 — Difference of squares twice.
\(x^{4}-16=(x^{2}-4)(x^{2}+4)=(x-2)(x+2)(x^{2}+4)\). Stop over reals; over complexes continue \(x^{2}+4=(x-2i)(x+2i)\).
E17 — Solve higher degree.
\(x^{3}-x^{2}-4x+4=0\). Group: \(x^{2}(x-1)-4(x-1)=(x^{2}-4)(x-1)=(x-2)(x+2)(x-1)=0\). Roots \(2,-2,1\).
E18 — Trinomial with leading coefficient.
\(8x^{2}-10x-3\): \(ac=-24\); pairs $ -12,2$: \(8x^{2}-12x+2x-3=4x(2x-3)+1(2x-3)=(4x+1)(2x-3)\).
End-of-day synthesis problems
S1. Factor completely \(2x^{3}-2x^{2}-12x\).
S2. Factor \(x^{3}+2x^{2}-x-2\) by grouping; solve \(=0\).
S3. Factor \(x^{4}-81\) completely over the reals.
S4. Solve \(x^{2}=7x\) without losing the root \(x=0\).
S5. Prove \(a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})\) by expansion.
S6. AC-factor \(12x^{2}+5x-2\); solve \(=0\).
S7. Use factor theorem: test \(x=1,-1,2\) on \(x^{3}-2x^{2}-x+2\); factor fully.
S8. Explain why you cannot cancel \(x\) in \(x^{2}+x=0\) by dividing by \(x\) alone.
Checkpoint
- Factor with GCF, grouping, trinomials
- Apply difference of squares/cubes
- Use factor theorem trials on cubics
- Solve factorable quadratic equations without losing roots
- Expand to verify factoring
- S4 and S8 as verbal mastery checks
Write two takeaways in your own words.
Quick reference card
| Pattern | Factorization |
|---|---|
| GCF | pull largest common monomial |
| \(x^{2}+bx+c\) | \((x+p)(x+q)\), \(p+q=b\), \(pq=c\) |
| \(a^{2}-b^{2}\) | \((a-b)(a+b)\) |
| \(a^{3}-b^{3}\) | \((a-b)(a^{2}+ab+b^{2})\) |
| \(a^{3}+b^{3}\) | \((a+b)(a^{2}-ab+b^{2})\) |
| Zero product | \(ab=0\Rightarrow a=0\) or \(b=0\) |
Always GCF first; always expand-check when unsure.
Deep dive — factoring patterns and zero-product
D1 — Complete factoring pipeline.
\(2x^{3}-2x^{2}-12x=2x(x^{2}-x-6)=2x(x-3)(x+2)\).
Order: GCF → trinomial → linear factors. Solve \(=0\): \(x=0,3,-2\).
D2 — Grouping cubic.
\(x^{3}+2x^{2}-x-2=x^{2}(x+2)-1(x+2)=(x^{2}-1)(x+2)=(x-1)(x+1)(x+2)\).
Roots \(1,-1,-2\). Always expand-check one intermediate if signs feel slippery.
D3 — Difference of squares twice.
\(x^{4}-81=(x^{2}-9)(x^{2}+9)=(x-3)(x+3)(x^{2}+9)\).
Over \(\mathbb{R}\) stop at irreducible \(x^{2}+9\); over \(\mathbb{C}\) continue if asked.
D4 — AC method with leading coefficient.
\(12x^{2}+5x-2\): \(ac=-24\); pair \(8,-3\):
\(12x^{2}+8x-3x-2=4x(3x+2)-1(3x+2)=(4x-1)(3x+2)\).
Solve \((4x-1)(3x+2)=0\): \(x=\frac14\) or \(x=-\frac23\).
D5 — Never divide away a root.
\(x^{2}=7x\Rightarrow x^{2}-7x=0\Rightarrow x(x-7)=0\Rightarrow x=0\) or \(7\).
Dividing by \(x\) first loses \(x=0\). Zero-product after factoring is the safe habit.
D6 — Sum/difference of cubes expand-check.
\(a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})\): right side \(a^{3}-a^{2}b+ab^{2}+a^{2}b-ab^{2}+b^{3}=a^{3}+b^{3}\).
Same verification for \(a^{3}-b^{3}\).
D7 — Factor theorem trial on a cubic.
\(p(x)=x^{3}-2x^{2}-x+2\). \(p(1)=0\), so \((x-1)\) divides.
Synthetic/division: \(x^{2}-x-2=(x-2)(x+1)\). Full factor \((x-1)(x-2)(x+1)\).
Extra practice set
- Factor completely \(x^{4}-16\) over the reals.
- Factor \(8x^{2}-10x-3\); solve \(=0\).
- Prove \(a^{2}-b^{2}=(a-b)(a+b)\) by expansion.
- Rational root candidates for \(2x^{3}+3x^{2}-1\); test until one works or none.
- Explain why \(x^{2}-2\) is irreducible over \(\mathbb{Q}\) but factors over \(\mathbb{R}\).
Tomorrow
Day 17 — Rational expressions. Domain, simplify, arithmetic, complex fractions, partial fractions setup for distinct linear factors.