Day 6 — Systems, Polynomials & Factoring

Day 6 — Systems, Polynomials & Factoring

Day 14 — Systems of linear equations (\(2\times 2\))

Stage II · concept day
Goal: Solve \(2\times 2\) systems by substitution and elimination; classify unique / none / infinitely many solutions geometrically; use determinant \(ad-bc\); write matrix form \(Ax=b\); apply Cramer’s rule; invert a \(2\times 2\) matrix when \(\det\neq 0\).

Why this matters

Two constraints and two unknowns appear in balancing, circuit equations (idealized), and parameter fitting of lines. Matrices are the language of computer graphics transforms, linear models, and later numerical linear algebra. Determinant zero is “singular / non-invertible.”

Theory

Standard form

A general \(2\times 2\) linear system: \[ \begin{cases} a_{11}x+a_{12}y=b_1\\ a_{21}x+a_{22}y=b_2 \end{cases} \]

Geometric picture

Each equation \(ax+by=c\) (with \((a,b)\neq(0,0)\)) is a line in the plane.

Configuration	Solutions
Two lines intersect at one point	Unique solution
Parallel distinct lines	No solution (inconsistent)
Same line (coincident)	Infinitely many solutions

Substitution method

Solve one equation for one variable; substitute into the other; back-substitute.

Elimination method

Multiply equations by constants so that adding/subtracting cancels one variable.

Determinant

For matrix \[A=\begin{pmatrix}a&b\\ c&d\end{pmatrix},\] define \[\det(A)=ad-bc.\]

Theorem (classification via determinant). For the system \(Ax=b\) with \[A=\begin{pmatrix}a&b\\ c&d\end{pmatrix},\quad x=\begin{pmatrix}x\\ y\end{pmatrix},\quad b=\begin{pmatrix}e\\ f\end{pmatrix} \] (using \(e,f\) as RHS):

• If \(\det(A)\neq 0\), unique solution.
  
• If \(\det(A)=0\), either no solution or infinitely many (need further check: whether equations are consistent).

Intuition: \(\det=0\) means rows are scalar multiples (lines parallel or coincident).

Matrix form \(Ax=b\)

Matrix-vector product for \(2\times 2\): \[ \begin{pmatrix}a&b\\ c&d\end{pmatrix} \begin{pmatrix}x\\ y\end{pmatrix} = \begin{pmatrix}ax+by\\ cx+dy\end{pmatrix}. \]

The system is exactly \(Ax=b\).

Cramer’s rule (\(2\times 2\))

If \(\Delta=\det(A)=ad-bc\neq 0\), then \[ x=\frac{\det\begin{pmatrix}e&b\\ f&d\end{pmatrix}}{\Delta}=\frac{ed-bf}{\Delta},\qquad y=\frac{\det\begin{pmatrix}a&e\\ c&f\end{pmatrix}}{\Delta}=\frac{af-ec}{\Delta}. \]

Proof sketch. Solve by elimination algebraically; the formulas match. Or note \(A\,\mathrm{adj}(A)=\det(A)I\) for the adjugate below.

Inverse of a \(2\times 2\) matrix

If \(\Delta=ad-bc\neq 0\), then \[ A^{-1}=\frac{1}{\Delta}\begin{pmatrix}d&-b\\ -c&a\end{pmatrix}. \]

Check: \(A A^{-1}=I=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}\).

Solution via inverse: if \(A\) invertible, \(x=A^{-1}b\).

Theorem. \(A\) is invertible iff \(\det(A)\neq 0\).

Infinite / no solution algebraically

If both equations become \(0=0\) after reduction: infinite (parameterize, e.g. free \(t\), \(x=t\), \(y=\ldots\)).
If \(0=1\) appears: inconsistent, no solution.

Worked examples

Example 1 — Elimination

\(\begin{cases}x+y=5\\ 2x-y=4\end{cases}\). Add: \(3x=9\), \(x=3\), \(y=2\).

Example 2 — Substitution

\(\begin{cases}y=2x+1\\ 3x+2y=20\end{cases}\). \(3x+2(2x+1)=20\). \(7x+2=20\). \(x=\frac{18}{7}\), \(y=\frac{43}{7}\).

Example 3 — Determinant unique

\(A=\begin{pmatrix}2&1\\ 5&3\end{pmatrix}\), \(\det=6-5=1\neq 0\). Unique for any RHS.

Example 4 — Cramer

\(\begin{cases}2x+y=5\\ 5x+3y=13\end{cases}\). \(\Delta=1\), \(x=\det\begin{pmatrix}5&1\\13&3\end{pmatrix}=15-13=2\), \(y=\det\begin{pmatrix}2&5\\5&13\end{pmatrix}=26-25=1\).

Example 5 — Inverse

\(A=\begin{pmatrix}2&1\\5&3\end{pmatrix}\), \(A^{-1}=\begin{pmatrix}3&-1\\ -5&2\end{pmatrix}\).
Check: \(2\cdot 3+1\cdot(-5)=1\), etc.

Example 6 — Solve via inverse

\(b=\begin{pmatrix}5\\13\end{pmatrix}\), \(x=A^{-1}b=\begin{pmatrix}3\cdot5+(-1)\cdot13\\ -5\cdot5+2\cdot13\end{pmatrix}=\begin{pmatrix}2\\1\end{pmatrix}\).

Example 7 — No solution

\(\begin{cases}x+y=1\\ 2x+2y=3\end{cases}\). \(\det=0\). Second is not multiple of first with same RHS ratio. \(2(x+y)=2\neq 3\). Empty.

Example 8 — Infinite

\(\begin{cases}x+y=1\\ 2x+2y=2\end{cases}\). Second \(=2\times\) first. Solutions: \(y=1-x\), \(x=t\) free.

Example 9 — Geometry

\(x-y=0\) and \(x+y=2\) intersect at \((1,1)\). Parallel: \(x+y=1\), \(x+y=2\).

Example 10 — Word system

\(3\) apples + \(2\) bananas \(= \$7\); \(1\) apple + \(4\) bananas \(=\$6\).
\(\begin{cases}3a+2b=7\\ a+4b=6\end{cases}\). From second \(a=6-4b\). \(3(6-4b)+2b=7\). \(18-12b+2b=7\). \(-10b=-11\). \(b=\frac{11}{10}\), \(a=\frac{8}{5}\)? Check carefully: \(18-10b=7\), \(-10b=-11\), \(b=1.1\), \(a=6-4.4=1.6\).

Example 11 — Matrix multiply

\(\begin{pmatrix}1&2\\ 3&4\end{pmatrix}\begin{pmatrix}5\\6\end{pmatrix}=\begin{pmatrix}17\\39\end{pmatrix}\).

Example 12 — det zero rows proportional

\(\begin{pmatrix}2&4\\ 1&2\end{pmatrix}\), \(\det=0\), row2 \(=\frac{1}{2}\) row1.

Example 13 — Parameter

For \(\begin{cases}x+ky=1\\ kx+y=1\end{cases}\), \(\Delta=1-k^2\). Unique if \(k\neq\pm 1\). If \(k=1\): equations \(x+y=1\) twice → infinite. If \(k=-1\): \(x-y=1\) and \(-x+y=1\) i.e. \(x-y=1\) and \(x-y=-1\) → empty.

Example 14 — Verify inverse formula

\(A A^{-1}=\frac{1}{\Delta}\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}=\frac{1}{\Delta}\begin{pmatrix}ad-bc&0\\0&ad-bc\end{pmatrix}=I\).

Exercises

Easy

1. Solve \(\begin{cases}x+y=4\\ x-y=2\end{cases}\) by elimination.
2. Solve \(\begin{cases}y=3x\\ 2x+y=10\end{cases}\) by substitution.
3. Compute \(\det\begin{pmatrix}3&1\\ 4&2\end{pmatrix}\).
4. Write \(\begin{cases}2x-y=3\\ x+4y=1\end{cases}\) as \(Ax=b\).
5. How many solutions geometrically if two lines are parallel and distinct?

Medium

6. Solve by Cramer: \(\begin{cases}3x+2y=7\\ x-y=1\end{cases}\).
7. Find \(A^{-1}\) for \(A=\begin{pmatrix}1&2\\ 3&5\end{pmatrix}\); solve \(Ax=\begin{pmatrix}1\\1\end{pmatrix}\).
8. Classify and solve: \(\begin{cases}2x+6y=4\\ x+3y=1\end{cases}\).
9. Classify and solve: \(\begin{cases}2x+6y=4\\ x+3y=2\end{cases}\).
10. Word: adult tickets \(\$12\), child \(\$5\); \(10\) tickets total cost \(\$78\). How many of each?
11. Compute \(\begin{pmatrix}0&1\\ 1&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\) and interpret.
12. For which \(h\) does \(\begin{cases}x+2y=3\\ 2x+hy=6\end{cases}\) have infinite solutions? None? Unique?
13. Verify Cramer’s formulas by solving Example 4 with elimination.

Hard / proof

14. Prove that if \(\det(A)\neq 0\), the inverse formula works (\(AA^{-1}=I\)).
15. Prove: if two lines \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\) have \(\det\begin{pmatrix}a_1&b_1\\a_2&b_2\end{pmatrix}\neq 0\), they intersect in exactly one point.
16. Show that row swap multiplies determinant by \(-1\) for \(2\times 2\).
17. Show \(\det(AB)=\det(A)\det(B)\) for \(2\times 2\) by direct expansion (tedious but finite).
18. Parameterize all solutions when the system reduces to a single independent equation.
19. Prove: if \(A\) is invertible and \(Ax=b\), then \(x\) is unique.
20. Explain why division by \(\det=0\) makes Cramer inapplicable; connect to geometry.

Challenge / CS-flavored

21. Graphics: \(2\times 2\) scale matrix \(\begin{pmatrix}s&0\\0&t\end{pmatrix}\) applied to \(\begin{pmatrix}x\\y\end{pmatrix}\). Invert if \(s,t\neq 0\).
22. Linear interpolation constraints: find line through \((0,1)\) and \((2,5)\) by solving system for \(y=mx+c\).
23. Homogeneous \(Ax=0\) with \(\det\neq 0\): only solution \(x=0\). Why?
24. Numerical: if \(\det\) is tiny, small RHS noise can swing solutions—conditioning link to Day 9.
25. Count operations: elimination vs Cramer for \(2\times 2\) (both \(O(1)\); contrast large \(n\) later).

Row operations and equivalence

Elementary operations on the augmented matrix \([A|b]\):

1. Swap equations
   
2. Multiply an equation by nonzero constant
   
3. Add multiple of one equation to another

These preserve the solution set. Elimination is systematic row reduction for \(2\times 2\).

Homogeneous systems

\(Ax=0\) always has at least the trivial solution \(x=0\).
If \(\det A\neq 0\), only trivial. If \(\det A=0\), nontrivial solutions exist (free variables).

Geometry of \(\det\)

\(|\det A|\) is the area of the parallelogram spanned by the column vectors of \(A\).
\(\det=0\) means columns are linearly dependent (parallel vectors).

Inverse properties

• \((A^{-1})^{-1}=A\)
  
• \((AB)^{-1}=B^{-1}A^{-1}\) when both invertible
  
• \(\det(A^{-1})=1/\det(A)\)

Extra exercises

26. Solve by row reduction the system \(3x+y=5\), \(2x-y=0\).
    
27. For \(A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\), show \(A\,\mathrm{adj}(A)=(\det A)I\) with \(\mathrm{adj}=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\).
    
28. Find all \(k\) so \(\begin{cases}kx+y=1\\ x+ky=1\end{cases}\) has unique / none / infinite solutions.
    
29. Interpret \(\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}\) as rotation; compute \(\det\).
    
30. Show that if \(Ax=b\) and \(Ax=c\) both have solutions and \(A\) is invertible, then \(b=c\) whenever \(x\) is the same—tautological check of uniqueness.

CS connection

Matrices encode linear transforms; invertibility means reversible transforms. Systems appear in constraint layouts, simple physics engines, and least squares (overdetermined later). Singular matrices cause solver failures—check \(\det\) or rank.

Common pitfalls

Pitfall	What to do instead
Arithmetic sign errors in elimination	Multiply carefully; check in both equations
Declaring no solution when \(\det=0\) always	Distinguish parallel vs coincident
Wrong inverse signs	\(\begin{pmatrix}d&-b\\-c&a\end{pmatrix}/\Delta\)
Cramer when \(\Delta=0\)	Use geometry / reduction cases
Forgetting to verify	Plug \((x,y)\) into both originals
Mixing row vectors / column vectors	Stick to \(Ax=b\) column convention

Fully worked extra examples

E15 — Full Cramer with numbers.
\(\begin{cases}4x+y=6\\ -x+2y=7\end{cases}\). \(\Delta=8-(-1)=9\).
\(x=\frac{\det\begin{pmatrix}6&1\\7&2\end{pmatrix}}{9}=\frac{12-7}{9}=\frac59\).
\(y=\frac{\det\begin{pmatrix}4&6\\ -1&7\end{pmatrix}}{9}=\frac{28-(-6)}{9}=\frac{34}{9}\).

E16 — Inverse multiply.
\(A=\begin{pmatrix}4&1\\ -1&2\end{pmatrix}\), \(\Delta=9\), \(A^{-1}=\frac19\begin{pmatrix}2&-1\\ 1&4\end{pmatrix}\).
\(x=A^{-1}\begin{pmatrix}6\\7\end{pmatrix}=\frac19\begin{pmatrix}12-7\\ 6+28\end{pmatrix}=\begin{pmatrix}5/9\\ 34/9\end{pmatrix}\).

E17 — Infinite family parameter.
\(x+2y=4\) only constraint: \(x=4-2t\), \(y=t\), \(t\in\mathbb{R}\).

E18 — Inconsistent detection.
After elimination: \(0x+0y=5\) appears → empty set immediately.

End-of-day synthesis problems

S1. Solve \(\begin{cases}3x-y=7\\ 2x+3y=1\end{cases}\) by elimination, Cramer, and inverse; compare.

S2. For \(A=\begin{pmatrix}1&2\\ 2&4\end{pmatrix}\) and \(b=\begin{pmatrix}3\\ 6\end{pmatrix}\) vs \(b=\begin{pmatrix}3\\ 5\end{pmatrix}\), classify systems.

S3. Prove \(AA^{-1}=I\) for the general \(2\times 2\) inverse formula when \(\Delta\neq 0\).

S4. Geometry: sketch \(x+y=2\) and \(x+y=3\); conclude solution count.

S5. Word: \(2\) coffees \(+1\) muffin \(=\$7\); \(1\) coffee \(+2\) muffins \(=\$8\). Prices?

S6. Compute \(\det\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}\) and inverse (rotation by \(90^{\circ}\) idea).

S7. Homogeneous \(\begin{cases}x+2y=0\\ 3x+6y=0\end{cases}\): all solutions.

S8. Parameter \(t\): \(\begin{cases}x+ty=1\\ tx+y=1\end{cases}\) solution cases.

Checkpoint

• Solve \(2\times 2\) by substitution and elimination
  
• Classify \(1\) / \(0\) / \(\infty\) solutions with geometry and algebra
  
• Compute \(\det\); write \(Ax=b\)
  
• Apply Cramer’s rule when \(\det\neq 0\)
  
• Invert a \(2\times 2\) matrix and use \(x=A^{-1}b\)
  
• One system solved three independent ways (S1)

Write two takeaways in your own words.

Deep dive — \(2\times 2\) systems, det, inverse, Cramer

D1 — One system three ways.
\(\begin{cases}3x-y=7\\ 2x+3y=1\end{cases}\).
Elimination: multiply first by \(3\): \(9x-3y=21\); add to second: \(11x=22\), \(x=2\), then \(y=-1\).
\(\Delta=\det\begin{pmatrix}3&-1\\2&3\end{pmatrix}=9-(-2)=11\).
Cramer: \(x=\frac{\det\begin{pmatrix}7&-1\\1&3\end{pmatrix}}{11}=\frac{21-(-1)}{11}=2\), \(y=\frac{\det\begin{pmatrix}3&7\\2&1\end{pmatrix}}{11}=\frac{3-14}{11}=-1\).
Inverse: \(A^{-1}=\frac{1}{11}\begin{pmatrix}3&1\\ -2&3\end{pmatrix}\), \(A^{-1}\begin{pmatrix}7\\1\end{pmatrix}=\begin{pmatrix}2\\ -1\end{pmatrix}\).

D2 — det zero: infinite vs empty.
\(A=\begin{pmatrix}1&2\\2&4\end{pmatrix}\), \(\det=0\).
With \(b=\begin{pmatrix}3\\6\end{pmatrix}\): second equation is \(2\times\) first → infinite, \(x=3-2t\), \(y=t\).
With \(b=\begin{pmatrix}3\\5\end{pmatrix}\): \(2\times\) left of first is \(6\neq 5\) → inconsistent, empty.

D3 — Parameter family.
\(\begin{cases}x+ty=1\\ tx+y=1\end{cases}\), \(\Delta=1-t^{2}\).
Unique if \(t\neq\pm 1\). If \(t=1\): both \(x+y=1\) → infinite. If \(t=-1\): \(x-y=1\) and \(-x+y=1\) → \(x-y=-1\) contradicts → empty.

D4 — Homogeneous.
\(Ax=0\) always has \(x=0\). If \(\det A\neq 0\), only trivial. If \(\det A=0\), a free variable appears (nontrivial kernel in \(2\)D).

D5 — Geometry of det.
\(|\det A|\) = area of parallelogram from column vectors. \(\det=0\) ⇔ columns parallel ⇔ lines normal vectors proportional ⇔ parallel/coincident lines.

D6 — Word system.
\(2\) coffees + \(1\) muffin \(=\$7\); \(1\) coffee + \(2\) muffins \(=\$8\):
\(\begin{cases}2c+m=7\\ c+2m=8\end{cases}\). From second \(c=8-2m\): \(2(8-2m)+m=7\), \(16-4m+m=7\), \(-3m=-9\), \(m=3\), \(c=2\).

Extra practice set

31. Invert \(A=\begin{pmatrix}4&1\\ -1&2\end{pmatrix}\) and solve \(Ax=\begin{pmatrix}6\\7\end{pmatrix}\).
    
32. Show \(A\,\mathrm{adj}(A)=(\det A)I\) for general \(2\times 2\).
    
33. Classify \(\begin{cases}kx+y=1\\ x+ky=1\end{cases}\) for all real \(k\).
    
34. Rotation matrix \(\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}\): compute \(\det\).
    
35. Prove: if \(A\) is invertible then \(Ax=b\) has exactly one solution.

Tomorrow

Day 15 — Polynomials. Degree, leading coefficient, arithmetic, evaluation, Horner, roots vs factors preview, long division lite.

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Day 15 — Polynomials

Stage II · concept day
Goal: Define polynomials; identify degree and leading coefficient; add and multiply; evaluate efficiently with Horner; preview roots vs factors; perform one polynomial long division.

Why this matters

Polynomials model discrete sums, generating functions (later), Taylor truncations, and simple regressions. Horner’s method is how CPUs and libraries evaluate polynomials with minimal multiplies. Factor/root links unlock equation solving (Day 16).

Theory

Definition

A polynomial in \(x\) over \(\mathbb{R}\) (or \(\mathbb{Q}\), \(\mathbb{Z}\)) is \[p(x)=a_n x^n+a_{n-1}x^{n-1}+\cdots+a_1 x+a_0\] with coefficients \(a_i\) and \(n\in\mathbb{Z}_{\ge 0}\). The zero polynomial is \(p\equiv 0\).

• Degree \(\deg(p)=n\) if \(a_n\neq 0\) (leading coefficient \(a_n\)).
  
• \(\deg(0)\) is defined as \(-\infty\) or left undefined—by convention \(\deg(0)<\) all others.
  
• Monic: leading coefficient \(1\).
  
• Constant polynomials: degree \(0\) if nonzero.

Equality

Two polynomials are equal iff corresponding coefficients are equal (as formal polynomials / as functions on infinite fields like \(\mathbb{R}\)).

Addition and scalar multiplication

Add coefficients of like powers.
\(\deg(p+q)\le \max(\deg p,\deg q)\), with equality unless leading terms cancel.

Multiplication

\[ \biggl(\sum_{i=0}^{n}a_i x^i\biggr)\biggl(\sum_{j=0}^{m}b_j x^j\biggr)=\sum_{k=0}^{n+m}\biggl(\sum_{i+j=k}a_i b_j\biggr)x^k. \]

Theorem. If \(p,q\neq 0\), then \(\deg(pq)=\deg p+\deg q\), over integral domains like \(\mathbb{R}[x]\) (no zero divisors).

Proof idea. Leading term \(a_n b_m x^{n+m}\) with \(a_n b_m\neq 0\). \(\square\)

Evaluation

\(p(c)\) is the number obtained by substituting \(x=c\).
Remainder theorem preview: dividing by \(x-c\) leaves remainder \(p(c)\) (see division).

Horner’s method

\[p(x)=a_n x^n+\cdots+a_0=(\cdots((a_n x+a_{n-1})x+a_{n-2})\cdots)x+a_0.\]

Same structure as Day 6 base conversion. About \(n\) multiplies and \(n\) adds for degree \(n\).

Roots and factors (preview)

\(c\) is a root (zero) of \(p\) if \(p(c)=0\).
Factor theorem (statement): \(p(c)=0\) iff \((x-c)\) divides \(p(x)\) (proved via division algorithm for polynomials).

Polynomial division (lite)

Divide \(p\) by nonzero \(d\): unique \(q,r\) with \[p=d\cdot q+r,\qquad r=0\text{ or }\deg r<\deg d.\]

Long division: mimic integer division—leading term of current dividend over leading term of divisor; multiply; subtract; repeat.

Example setup: divide \(x^3+0x^2-x-1?\) wait standard: \(x^2+3x+2\) by \(x+1\), etc.

Special products (recall)

\((x+a)(x+b)=x^2+(a+b)x+ab\).
\((x+a)^2=x^2+2ax+a^2\).
\((x+a)^3=x^3+3ax^2+3a^2x+a^3\).
\(x^3\pm y^3=(x\pm y)(x^2\mp xy+y^2)\).
\(x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots+y^{n-1})\).

Remainder and factor theorems (formal)

Division algorithm. For \(p,d\in\mathbb{R}[x]\) with \(d\neq 0\), there exist unique \(q,r\) with \(p=dq+r\) and \(r=0\) or \(\deg r<\deg d\).

Remainder theorem. Dividing by \(x-c\) gives remainder \(p(c)\): \(p(x)=(x-c)q(x)+p(c)\).
Proof. Apply division: remainder is constant \(r\); set \(x=c\): \(p(c)=r\). \(\square\)

Factor theorem. \(p(c)=0\) iff \((x-c)\) divides \(p\) in \(\mathbb{R}[x]\).
Proof. Remainder theorem: \(p(c)=0\) iff remainder \(0\) iff \((x-c)\) divides \(p\). \(\square\)

At most \(n\) roots

Theorem. A nonzero polynomial of degree \(n\) over \(\mathbb{R}\) (or any field) has at most \(n\) roots.
Proof. If \(p(c_1)=0\), write \(p=(x-c_1)q_1\) with \(\deg q_1=n-1\). Induct: each new distinct root factors out another linear factor; after \(n+1\) distinct roots, \(p\) would be zero polynomial. \(\square\)

Leading-term asymptotics (lite)

For large \(|x|\), \(p(x)\sim a_n x^n\): the sign of \(p(x)\) for large positive \(x\) matches \(\operatorname{sign}(a_n)\); for large negative \(x\) it matches \(\operatorname{sign}(a_n(-1)^n)\). End behavior of graphs follows.

Worked examples

Example 1 — Degree

\(p(x)=4x^3-x+7\): degree \(3\), leading coefficient \(4\), constant term \(7\).

Example 2 — Add

\((2x^2+3x-1)+(-2x^2+x+5)=4x+4\). Degree drops due to cancellation.

Example 3 — Multiply

\((x+2)(x^2-x+3)=x^3-x^2+3x+2x^2-2x+6=x^3+x^2+x+6\).

Example 4 — Evaluate naive

\(p(x)=2x^3-5x+1\) at \(x=3\): \(2\cdot 27-15+1=54-15+1=40\).

Example 5 — Horner

Coefficients \(2,0,-5,1\) for \(2x^3+0x^2-5x+1\) at \(x=3\):
\(2\); \(2\cdot3+0=6\); \(6\cdot3+(-5)=13\); \(13\cdot3+1=40\).

Example 6 — Root check

\(p(x)=x^2-5x+6\), \(p(2)=4-10+6=0\). Root \(x=2\). Factors \((x-2)(x-3)\).

Example 7 — Long division

Divide \(p(x)=x^3-2x^2-x+2\) by \(d(x)=x-1\).

• \(x^3/x=x^2\); multiply \(x^2(x-1)=x^3-x^2\); subtract: \(-x^2-x+2\) wait: \((-2x^2)-(-x^2)=-x^2\).
  
• \(-x^2/x=-x\); \((-x)(x-1)=-x^2+x\); subtract: \((-x)-(+x)=-2x\), bring \(+2\): \(-2x+2\).
  
• \(-2x/x=-2\); \((-2)(x-1)=-2x+2\); subtract \(0\).

Quotient \(x^2-x-2\), remainder \(0\). So \(p(x)=(x-1)(x^2-x-2)\).

Example 8 — Remainder

\(p(x)=x^2+x+1\) divided by \(x-2\): \(p(2)=7\), so \(p(x)=(x-2)(x+3)+7\) check: \((x-2)(x+3)=x^2+x-6\), plus \(7=x^2+x+1\).

Example 9 — Degree of product

\(\deg((x^2+1)(x^3+2x))=5\).

Example 10 — Zero divisors?

Over \(\mathbb{R}[x]\), \(pq=0\Rightarrow p=0\) or \(q=0\). (Unlike modular coefficient rings sometimes.)

Example 11 — Even/odd functions preview

\(p(x)=x^3+x\) is odd as a function: \(p(-x)=-p(x)\). \(q(x)=x^2+1\) even.

Example 12 — Nested Horner for multipoint

Same scheme for each evaluation point; reuse coefficients.

Example 13 — Expand

\((2x-1)^3=8x^3-3\cdot4x^2\cdot1+3\cdot2x\cdot1-1=8x^3-12x^2+6x-1\).

Example 14 — Missing terms

Write \(x^3+1\) as \(x^3+0x^2+0x+1\) for Horner/synthetic comfort.

Example 15 — Synthetic division

Divide \(x^3-2x^2-x+2\) by \(x-1\). Coefficients \(1,-2,-1,2\); synthetic with \(c=1\):
bring \(1\); \(1\cdot1-2=-1\); \((-1)\cdot1-1=-2\); \((-2)\cdot1+2=0\).
Quotient \(x^2-x-2\), remainder \(0\) (matches long division earlier).

Example 16 — Sum and difference of cubes

\(x^3-8=(x-2)(x^2+2x+4)\). Check: \(x^3+2x^2+4x-2x^2-4x-8=x^3-8\).
\(x^3+8=(x+2)(x^2-2x+4)\).

Example 17 — At most \(n\) roots application

\(p(x)=x^2-1\) has roots \(\pm 1\) only. Cannot have a third root without being identically zero. Identity theorem lite: if two degree-\(\le n\) polynomials agree at \(n+1\) points, they are equal as polynomials.

Example 18 — End behavior

\(p(x)=-2x^3+x\): as \(x\to+\infty\), \(p\to-\infty\); as \(x\to-\infty\), \(p\to+\infty\) (odd degree, negative lead).

Example 19 — Nested product expand carefully

\((x-1)(x+1)(x^2+1)=((x^2-1)(x^2+1))=x^4-1\). Degree \(4=1+1+2\).

Example 20 — Polynomial identity check

Is \(x^2+2x+1=(x+1)^2\) as polynomials? Expand RHS: yes, coefficients match. Checking three points would also suffice for degree \(\le 2\), but coefficient comparison is definitive.

Exercises

Easy

1. State degree and leading coefficient of \(7-3x+x^4\).
2. Add \((x^2+2x)+(3x^2-5)\).
3. Multiply \((x+3)(x-3)\).
4. Evaluate \(x^2-4x+1\) at \(x=-1\).
5. Is \(x^{-1}+x\) a polynomial? Why or why not?

Medium

6. Multiply \((2x^2-x+1)(x+4)\); give degree.
7. Use Horner to evaluate \(3x^4-2x^2+x-5\) at \(x=2\).
8. Show \(x= -2\) is a root of \(x^3+3x^2-4\); factor out \((x+2)\) via division or inspection.
9. Divide \(x^3+1\) by \(x+1\); find \(q,r\).
10. Prove \(\deg(pq)=\deg p+\deg q\) for nonzero \(p,q\) over \(\mathbb{R}\).
11. Find all roots of \((x-1)(x+2)(x-3)=0\).
12. Expand \((x+y)(x^2-xy+y^2)\) (sum of cubes pattern).
13. For \(p(x)=x^2+bx+c\), express \(p(0)\), \(p(1)\) in coefficients.

Hard

14. Perform long division: \((2x^3-3x^2+x-1)\div(x^2-1)\); state \(q,r\) with \(\deg r<\deg(x^2-1)\).
15. Prove remainder theorem: \(p(x)=(x-c)q(x)+p(c)\) using division algorithm uniqueness.
16. Show that if \(\deg p=n\), then \(p\) has at most \(n\) roots unless \(p\equiv 0\) (use factor theorem repeatedly).
17. Give polynomials \(p,q\) with \(\deg(p+q)<\max(\deg p,\deg q)\).
18. Compare operation counts: naive eval of degree \(n\) vs Horner.
19. Define addition and multiplication of polynomials formally; verify distributive law sketch.
20. Is every function \(\mathbb{R}\to\mathbb{R}\) a polynomial? Justify.

Challenge / CS-flavored

21. Horner is the same as parsing base-\(b\) integers—explain with \(p(b)\) for digit polynomial.
22. Complexity: multiplying two degree-\(n\) polynomials naive \(O(n^2)\) coefficient ops.
23. Generating function lite: \(1+x+x^2+\cdots+x^n=\frac{1-x^{n+1}}{1-x}\) for \(x\neq 1\)—polynomial identity for finite sum.
24. Why store coefficients in an array of length \(\deg+1\)? Index meaning?
25. Numeric stability: evaluating \(x^n\) via recurrence \(p\leftarrow p\cdot x\) (Horner-like) vs pow—qualitative.

Polynomial ring intuition

\(\mathbb{R}[x]\) denotes polynomials with real coefficients. Addition and multiplication make it a ring. Units (invertible elements) are nonzero constants. Irreducible polynomials play a prime-like role (Stage later if revisited).

Synthetic division (companion to long division)

For monic linear divisors \(x-c\), synthetic division packs coefficients efficiently—same results as long division. Example: coefficients of \(x^{3}-2x^{2}-x+2\) divided by \(x-1\): bring down, multiply-add cascade → quotient coeffs and remainder \(p(1)\).

Evaluation as nested form

Horner is optimal for multiplication count among algorithms that only use multiplies by \(x\) and adds (in a precise model)—culture fact; naive powering is worse.

Roots bound (optional)

For monic \(x^n+a_{n-1}x^{n-1}+\cdots+a_0\), any real root \(r\) satisfies \(|r|\le 1+\max |a_i|\) (crude bound)—useful for search intervals.

Extra exercises

26. Prove that the sum of two degree-\(n\) polynomials can have degree \(<n\).
    
27. Synthetic or long divide \(x^{3}+x^{2}-4x-4\) by \(x+2\).
    
28. Show \(x^{2}+1\) has no real roots; conclude it does not factor into real linears.
    
29. Count multiplications: Horner for degree \(5\) vs expanding powers \(x,x^{2},\ldots\) naively.
    
30. Write the general product coefficient formula for \((\sum a_i x^{i})(\sum b_j x^{j})\).

CS connection

Polynomials appear in hash polynomials, checksums, graphics curves (Bézier are polynomial), and coding theory. Horner minimizes multiplications—classic algorithm analysis example. Coefficient arrays are the data structure; sparse polynomials need maps.

Common pitfalls

Pitfall	What to do instead
Forgetting missing powers	Write \(0\) coefficients
\(\deg(p+q)=\max\) always	Leading terms may cancel
Calling rational functions polynomials	Denominator must be \(1\)
Horner wrong direction	Highest degree first
Root implies factor without proof	Factor theorem (use division)
Division degree condition on \(r\)	\(\deg r<\deg d\) or \(r=0\)

Fully worked extra examples

E15 — Full long division tableau.
Divide \(2x^{3}+3x^{2}-x+5\) by \(x^{2}+1\):
\(\frac{2x^{3}}{x^{2}}=2x\); \(2x(x^{2}+1)=2x^{3}+2x\); subtract: \(3x^{2}-3x+5\).
\(\frac{3x^{2}}{x^{2}}=3\); \(3(x^{2}+1)=3x^{2}+3\); subtract: \(-3x+2\).
Quotient \(2x+3\), remainder \(-3x+2\). Check: \((x^{2}+1)(2x+3)+(-3x+2)=2x^{3}+3x^{2}+2x+3-3x+2=2x^{3}+3x^{2}-x+5\).

E16 — Horner multiprecision feel.
\(p(x)=x^{5}+x^{4}+x^{3}+x^{2}+x+1\) at \(x=2\): nested \(((((((1)\cdot2+1)2+1)2+1)2+1)2+1)=63\). Indeed \(\frac{2^{6}-1}{2-1}=63\).

E17 — Root multiplicity preview.
\((x-1)^{2}=x^{2}-2x+1\) has double root \(1\): graph touches axis. Factor theorem still applies: \((x-1)\) divides, and again divides the quotient.

E18 — Degree of sum cancellation.
\((x^{3}+x)+( -x^{3}+2)=x+2\) degree \(1\), not \(3\). Leading coefficients summed to zero.

End-of-day synthesis problems

S1. Multiply \((x^{2}-x+1)(x^{2}+x+1)\) and state degree.

S2. Horner-evaluate \(x^{4}-2x^{3}+x-7\) at \(x=-1\).

S3. Long divide \(x^{3}-3x+2\) by \(x-1\); relate remainder to \(p(1)\).

S4. Prove \(\deg(pq)=\deg p+\deg q\) for nonzero \(p,q\) over \(\mathbb{R}\).

S5. Show \(x=2\) is a root of \(x^{3}-6x^{2}+11x-6\); factor completely over \(\mathbb{R}\).

S6. Add \((3x^{3}+x)-(3x^{3}-2x^{2}+4)\) and explain degree drop.

S7. Write \(x^{3}+1\) with all intermediate zero coefficients for Horner.

S8. If \(p\) has degree \(4\) and \(q\) degree \(3\), what are possible degrees of \(p+q\) and \(pq\)?

Deep dive — Why Horner is standard

Naive evaluation of \(a_n x^n+\cdots+a_0\) computing each \(x^k\) separately costs about \(n+(n-1)+\cdots+1=\Theta(n^2)\) multiplies if each power is built from scratch, or \(\Theta(n)\) multiplies if you reuse \(x^{k}=x\cdot x^{k-1}\), plus \(n\) multiplies by coefficients—still more multiplies than Horner’s \(n\). Horner fuses powering and coefficient combination into \(n\) multiplies and \(n\) adds. In code, it is a tight loop with excellent locality. The same nested multiply-add is how base-\(b\) digit strings evaluate to integers (Day 6–7 link).

Synthesis

Polynomials form a ring closed under \(+,\cdot\) with degree additivity for products. Evaluation, Horner, division, and the factor/remainder theorems are one package: division by \(x-c\) is evaluation in disguise. Roots are not free—degree \(n\) allows at most \(n\) of them unless the polynomial is zero.

S9. Prove the remainder theorem in two sentences from the division algorithm.

S10. Use synthetic division to divide \(2x^3-3x+1\) by \(x+1\); report \(q\) and \(r\).

S11. Expand \((x+2)^3\) two ways (binomial and stepwise) and match coefficients.

S12. Show that if \(p\) and \(q\) agree on infinitely many reals, then \(p-q\equiv 0\) as a polynomial (use the root bound).

Checkpoint

• Define polynomial, degree, leading coefficient
  
• Add and multiply polynomials; track degree
  
• Evaluate with Horner
  
• State root \(\leftrightarrow\) factor idea
  
• Carry out one long division with remainder
  
• Finish S3 with full long-division tableau

Write two takeaways in your own words.

Quick reference card

Object	Symbol / formula
Degree	highest power with nonzero coeff
Leading coeff	coeff of \(x^{\deg p}\)
Horner	nested multiply-add
Remainder thm	\(p(x)=(x-c)q(x)+p(c)\)
Product degree	\(\deg p+\deg q\) if \(p,q\neq 0\)

Memorize the card; rebuild one long division from blank paper.

Tomorrow

Day 16 — Factoring. GCF, grouping, trinomials, difference of squares/cubes, factor theorem idea, solve quadratics by factoring.

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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

1. Write \(ax^2+bx+c=0\).
   
2. Factor left side.
   
3. Apply zero-product.
   
4. Check.

Not all quadratics factor nicely over the integers; Day 19 gives formula and completing the square.

Strategy checklist

1. GCF first.
   
2. Count terms: \(2\) → difference of squares/cubes; \(3\) → trinomial/perfect square; \(4\) → grouping.
   
3. Factor further until irreducible.
   
4. 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

1. Factor \(15x^3-10x\).
2. Factor \(x^2+7x+12\).
3. Factor \(x^2-9\).
4. Solve \((x-2)(x+5)=0\).
5. Factor \(x^2-10x+25\).

Medium

6. Factor \(6x^2-x-2\).
7. Factor \(x^3+3x^2-x-3\) by grouping.
8. Factor \(27x^3-1\).
9. Factor \(x^4-16\) completely over the reals.
10. Solve \(2x^2-5x-3=0\) by factoring.
11. Solve \(x^2=5x\) without losing a root.
12. Factor \(2x^3+4x^2-6x\) completely.
13. Use factor theorem: show \(x+2\) divides \(x^3+6x^2+11x+6\); factor fully.

Hard / proof

14. Prove \(a^2-b^2=(a-b)(a+b)\) and \(a^3-b^3=(a-b)(a^2+ab+b^2)\) by expansion.
15. Prove zero-product property from field axioms (no zero divisors).
16. Explain why dividing an equation by \(x\) can lose solutions; give example.
17. Factor \(x^6-1\) completely over the reals (use difference of squares/cubes as needed).
18. For which \(c\) does \(x^2+cx+1\) factor with real linear factors?
19. Show that if \(p\) has roots \(r_1,\ldots,r_k\) distinct and \(\deg p=k\), then \(p(x)=a(x-r_1)\cdots(x-r_k)\).
20. Disprove: every quadratic factors over the integers.

Challenge / CS-flavored

21. Boolean algebra analogy: factoring vs expanding—when do you want CNF-like product form?
22. Complexity identity: \(n^2-1=(n-1)(n+1)\) for analyzing neighboring sizes.
23. Solve \(n(n-1)=20\) for integer \(n\) (combinatorial \(P(n,2)\) feel).
24. 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).
25. 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

26. Factor \(x^{3}+1\) completely over the reals.
    
27. Solve \(x(x-3)(x+2)=0\).
    
28. Factor \(a^{4}-b^{4}\) completely over the reals.
    
29. Use rational root theorem candidates for \(2x^{3}+3x^{2}-1=0\); find one rational root if any.
    
30. 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

31. Factor completely \(x^{4}-16\) over the reals.
    
32. Factor \(8x^{2}-10x-3\); solve \(=0\).
    
33. Prove \(a^{2}-b^{2}=(a-b)(a+b)\) by expansion.
    
34. Rational root candidates for \(2x^{3}+3x^{2}-1\); test until one works or none.
    
35. 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.