Solving Low-Degree Polynomials: Quadratics, Cubics, Quartics, and Symmetry
Purpose, Context, and Learning Outcomes
The quadratic formula is usually the first universal symbolic formula for roots that students meet.
It is tempting to think of it as a trick to memorize. That would miss the real mathematical point.
A formula for roots tells us that the equation has enough internal structure that the roots can be
recovered from the coefficients by a finite sequence of algebraic operations. For degree $2$, that
structure appears through completing the square. For degrees $3$ and $4$, the story becomes richer:
one introduces new variables, builds auxiliary equations, and exploits symmetry among the roots.
- derive the quadratic formula by completing the square;
- state and use Vieta’s formulas for quadratics, cubics, and quartics;
- explain what a symmetric polynomial is and why coefficients are symmetric functions of the roots;
- carry out the reduction of a general cubic to a depressed cubic and explain Cardano’s substitution;
- carry out the reduction of a general quartic to a depressed quartic and explain Ferrari’s method;
- describe, in accessible language, why the search for formulas naturally leads toward groups and Galois theory.
- Confusing the coefficients of a polynomial with the roots themselves. The coefficients are symmetric combinations of the roots.
- Forgetting to normalize first when deriving formulas. For cubics and quartics, the first move is to divide by the leading coefficient.
- Thinking Cardano’s or Ferrari’s method is “magic.” In both cases, an auxiliary equation appears because we intentionally create a perfect square or a quadratic relation.
- Choosing cube roots independently in Cardano’s formula. The two cube roots must be chosen compatibly so that $uv=-p/3$.
Lecture Roadmap and Teaching Plan
Conceptual roadmap.
| Part | Mathematical task | Big idea |
|---|---|---|
| 1 | Review roots and coefficients for quadratics | Coefficients already encode symmetric information about the roots. |
| 2 | Define symmetric polynomials and elementary symmetric polynomials | The coefficients are the elementary symmetric polynomials, up to sign. |
| 3 | Derive the quadratic formula | A universal formula comes from completing the square. |
| 4 | Derive Cardano’s method for cubics | A cubic can be reduced to a quadratic in the new variables $u^3$ and $v^3$. |
| 5 | Derive Ferrari’s method for quartics | A quartic can be reduced to a cubic resolvent and then to two quadratics. |
| 6 | Connect formulas to symmetry and groups | Root permutations control which expressions remain accessible from the coefficients. |
Suggested pacing for a 90-minute class.
- 10 min: motivation, Vieta review, and symmetric polynomials;
- 15 min: complete derivation of the quadratic formula and the discriminant;
- 25 min: reduction of a cubic to depressed form and Cardano’s substitution;
- 20 min: reduction of a quartic to depressed form and Ferrari’s factorization trick;
- 10 min: worked examples and CAS observations;
- 10 min: reflection, symmetry, groups, and homework.
Roots, Coefficients, and the First Appearance of Symmetry
Suppose a polynomial factors as
f(x)=a(x-r_1)(x-r_2)\cdots(x-r_n), \qquad a\neq 0.
\]
When we expand this product, the coefficients are built from sums of products of the roots.
This is already a symmetry statement: the coefficient does not care which root was called
$r_1$, which was called $r_2$, and so on.
Definition (symmetric polynomial). A polynomial $P(r_1,\dots,r_n)$ is symmetric if its value does not change when the variables are permuted. In other words, for every permutation $\sigma$ of $\{1,2,\dots,n\}$,
P(r_1,\dots,r_n)=P(r_{\sigma(1)},\dots,r_{\sigma(n)}).
\]
Elementary symmetric polynomials. For $n$ variables $r_1,\dots,r_n$, define
e_1 = \sum_i r_i,
\qquad
e_2 = \sum_{i<j} r_ir_j,
\qquad
e_3 = \sum_{i<j<k} r_ir_jr_k,
\qquad \dots \qquad
e_n=r_1r_2\cdots r_n.
\]
These are the basic symmetric building blocks.
Expansion principle. If
(x-r_1)(x-r_2)\cdots(x-r_n)
\]
is expanded, then the coefficient of $x^{n-k}$ is $(-1)^k e_k$.
\prod_{i=1}^n (x-r_i)
=x^n-e_1x^{n-1}+e_2x^{n-2}-e_3x^{n-3}+\cdots+(-1)^n e_n.
\]
Vieta’s formulas, general pattern. If
a x^n+b_1x^{n-1}+b_2x^{n-2}+\cdots+b_n = a\prod_{i=1}^n (x-r_i),
\]
then after dividing by $a$ we get
e_1=-\frac{b_1}{a},
\qquad
e_2=\frac{b_2}{a},
\qquad
e_3=-\frac{b_3}{a},
\qquad \dots \qquad
e_n=(-1)^n\frac{b_n}{a}.
\]
Quadratic. If $ax^2+bx+c=0$ has roots $r_1,r_2$, then
r_1+r_2=-\frac{b}{a},
\qquad
r_1r_2=\frac{c}{a}.
\]
Cubic. If $ax^3+bx^2+cx+d=0$ has roots $r_1,r_2,r_3$, then
r_1+r_2+r_3=-\frac{b}{a},
\]
\[
r_1r_2+r_1r_3+r_2r_3=\frac{c}{a},
\]
\[
r_1r_2r_3=-\frac{d}{a}.
\]
Quartic. If $ax^4+bx^3+cx^2+dx+e=0$ has roots $r_1,r_2,r_3,r_4$, then
r_1+r_2+r_3+r_4=-\frac{b}{a},
\]
\[
\sum_{i<j}r_ir_j=\frac{c}{a},
\]
\[
\sum_{i<j<k} r_ir_jr_k=-\frac{d}{a},
\]
\[
r_1r_2r_3r_4=\frac{e}{a}.
\]
Any symmetric polynomial in the roots can be rewritten in terms of $e_1,e_2,\dots,e_n$, and therefore in terms of the coefficients. This is why formulas often search for expressions that are either symmetric, or become symmetric after squaring, cubing, or otherwise combining them.
Quadratic Equations: Methods, Derivation, and the Discriminant
For a quadratic equation $ax^2+bx+c=0$ with $a\neq 0$, there are several useful methods.
Mathematical maturity means learning to choose the method that best matches the structure of the problem.
| Method | When it is natural | What it emphasizes |
|---|---|---|
| Factoring | When the roots are simple rational numbers or visible by inspection | Multiplicative structure and the zero-product principle |
| Graphing | When approximations or geometric interpretation matter | Shape, intercepts, vertex, and number of real roots |
| Completing the square | When vertex form matters or when deriving a universal formula | Structural reorganization of the expression |
| Quadratic formula | When a universal exact method is needed | A closed symbolic formula for the roots |
Derivation of the quadratic formula. Start with
ax^2+bx+c=0, \qquad a\neq 0.
\]
- Divide by $a$:
\[
x^2+\frac{b}{a}x+\frac{c}{a}=0.
\] - Move the constant term to the other side:
\[
x^2+\frac{b}{a}x=-\frac{c}{a}.
\] - Add the square of half the coefficient of $x$ to both sides:
\[
x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2
=-\frac{c}{a}+\left(\frac{b}{2a}\right)^2.
\] - The left-hand side is now a perfect square:
\[
\left(x+\frac{b}{2a}\right)^2
=-\frac{c}{a}+\frac{b^2}{4a^2}.
\] - Write the right-hand side over a common denominator:
\[
\left(x+\frac{b}{2a}\right)^2
=\frac{-4ac+b^2}{4a^2}
=\frac{b^2-4ac}{4a^2}.
\] - Take square roots:
\[
x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}.
\] - Solve for $x$:
\[
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.
\]
\boxed{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}
\]
The discriminant. The quantity
\Delta=b^2-4ac
\]
is called the discriminant. It distinguishes the basic root patterns:
- If $\Delta>0$, there are two distinct real roots.
- If $\Delta=0$, there is one repeated real root.
- If $\Delta<0$, there are two nonreal complex conjugate roots.
Why the discriminant is a symmetry object. Let the roots be $r_1$ and $r_2$. Then
r_1+r_2=-\frac{b}{a},
\qquad
r_1r_2=\frac{c}{a}.
\]
Therefore
\Delta=b^2-4ac
=a^2(r_1+r_2)^2-4a^2r_1r_2
=a^2\big((r_1+r_2)^2-4r_1r_2\big)
=a^2(r_1-r_2)^2.
\]
So the discriminant is, up to the factor $a^2$, the square of the difference of the two roots.
The sign of $r_1-r_2$ changes when the roots are swapped, but its square does not. That is exactly the pattern of a symmetric quantity.
Example. Solve $x^2+6x+1=0$ by completing the square.
x^2+6x=-1,
\qquad
x^2+6x+9=8,
\qquad
(x+3)^2=8.
\]
\[
x+3=\pm 2\sqrt{2},
\qquad
x=-3\pm 2\sqrt{2}.
\]
This example reminds us that completing the square is not only a derivation tool; it is also a practical solving method.
Symmetric Polynomials More Explicitly: Why They Organize the Formulas
The quadratic discriminant gave our first serious example of an expression that is not obviously written in terms of roots but is still governed by symmetry.
Let us make this more explicit.
Example 1. If $r_1,r_2$ are roots of a quadratic, then
(r_1-r_2)^2=(r_1+r_2)^2-4r_1r_2.
\]
Although $r_1-r_2$ is not symmetric, its square is symmetric and can be rewritten using the coefficients.
Example 2. If $r_1,r_2,r_3$ are roots of a cubic, then
r_1^2+r_2^2+r_3^2=(r_1+r_2+r_3)^2-2(r_1r_2+r_1r_3+r_2r_3).
\]
This also becomes a coefficient expression by Vieta’s formulas.
Fundamental theorem of symmetric polynomials (working statement).
Every symmetric polynomial in $r_1,\dots,r_n$ can be written as a polynomial in the elementary symmetric polynomials $e_1,\dots,e_n$.
Since the $e_i$ are the coefficients, up to sign and scaling, every symmetric root expression can be converted into a coefficient expression.
We will not prove the full theorem here, but we will use its message constantly:
if an expression is symmetric enough, then the coefficients know about it.
The roots themselves are not symmetric quantities, so the coefficients do not directly determine each root separately. The formulas for cubics and quartics therefore search for intermediate expressions that are not fully symmetric, but become symmetric after squaring or cubing. Once that happens, they satisfy an auxiliary equation whose coefficients can be written in terms of the original coefficients.
Cubics, Part I: From the General Cubic to the Depressed Cubic
Start with a general cubic equation
ax^3+bx^2+cx+d=0, \qquad a\neq 0.
\]
Divide through by $a$ and write
x^3+Bx^2+Cx+D=0,
\qquad
B=\frac{b}{a},\ C=\frac{c}{a},\ D=\frac{d}{a}.
\]
The first goal is to remove the $x^2$ term. We do this by shifting the variable:
x=y-\frac{B}{3}.
\]
Computation.
x^3=\left(y-\frac{B}{3}\right)^3
=y^3-By^2+\frac{B^2}{3}y-\frac{B^3}{27},
\]
\[
Bx^2
=B\left(y-\frac{B}{3}\right)^2
=By^2-\frac{2B^2}{3}y+\frac{B^3}{9},
\]
\[
Cx=C\left(y-\frac{B}{3}\right)=Cy-\frac{BC}{3}.
\]
Substituting into $x^3+Bx^2+Cx+D=0$ gives
y^3+\left(C-\frac{B^2}{3}\right)y
+\left(\frac{2B^3}{27}-\frac{BC}{3}+D\right)=0.
\]
Depressed cubic. After the substitution $x=y-\dfrac{B}{3}$, the cubic becomes
y^3+py+q=0,
\]
where
p=C-\frac{B^2}{3},
\qquad
q=\frac{2B^3}{27}-\frac{BC}{3}+D.
\]
A polynomial is called depressed when the next-to-leading term has been removed.
For quadratics this is automatic after dividing by $a$; for cubics and quartics we must shift the variable.
Cubics, Part II: Cardano’s Method Carefully Derived
Now solve the depressed cubic
y^3+py+q=0.
\]
Cardano’s idea is to write the unknown root as a sum
y=u+v.
\]
Expand and organize.
(u+v)^3=u^3+v^3+3uv(u+v).
\]
Therefore
y^3+py+q
=(u+v)^3+p(u+v)+q
=u^3+v^3+(3uv+p)(u+v)+q.
\]
If we choose $u$ and $v$ so that
3uv+p=0,
\qquad
u^3+v^3=-q,
\]
then $y=u+v$ will indeed satisfy the cubic.
We replaced the original cubic equation by two conditions on $u$ and $v$.
Those conditions become manageable after cubing, because $u^3+v^3$ and $u^3v^3$ are symmetric in $u^3$ and $v^3$.
Set
U=u^3,
\qquad
V=v^3.
\]
Then
U+V=-q,
\qquad
UV=(uv)^3=\left(-\frac{p}{3}\right)^3=-\frac{p^3}{27}.
\]
So $U$ and $V$ are the two roots of the quadratic equation
z^2-(U+V)z+UV=0,
\]
\[
z^2+qz-\frac{p^3}{27}=0.
\]
Applying the quadratic formula gives
U=-\frac{q}{2}+\sqrt{\left(\frac{q}{2}\right)^2+\left(\frac{p}{3}\right)^3},
\]
\[
V=-\frac{q}{2}-\sqrt{\left(\frac{q}{2}\right)^2+\left(\frac{p}{3}\right)^3}.
\]
Hence one solution is
y=\sqrt[3]{-\frac{q}{2}+\sqrt{\left(\frac{q}{2}\right)^2+\left(\frac{p}{3}\right)^3}}
+
\sqrt[3]{-\frac{q}{2}-\sqrt{\left(\frac{q}{2}\right)^2+\left(\frac{p}{3}\right)^3}}.
\]
Returning to the original variable, a root of
x^3+Bx^2+Cx+D=0
\]
is
x=y-\frac{B}{3}.
\]
Cardano discriminant indicator. Let
\Delta_C=\left(\frac{q}{2}\right)^2+\left(\frac{p}{3}\right)^3.
\]
- If $\Delta_C>0$, the cubic has one real root and two nonreal complex conjugate roots.
- If $\Delta_C=0$, the cubic has a repeated root.
- If $\Delta_C<0$, the cubic has three distinct real roots, but Cardano’s radicals pass through complex numbers. This phenomenon is called casus irreducibilis.
All three roots. Let $\omega=\dfrac{-1+i\sqrt{3}}{2}$, a primitive cube root of unity. If $u$ and $v$ are chosen so that $u^3=U$, $v^3=V$, and $uv=-p/3$, then the three roots of the depressed cubic are
y_0=u+v,
\qquad
y_1=\omega u+\omega^2 v,
\qquad
y_2=\omega^2 u+\omega v.
\]
This is where the cube roots of unity naturally enter the problem.
Cubic Worked Example and Interpretation
Example. Solve the depressed cubic
y^3-6y-9=0.
\]
Here $p=-6$ and $q=-9$. Compute
\Delta_C=\left(-\frac{9}{2}\right)^2+\left(-2\right)^3
=\frac{81}{4}-8
=\frac{49}{4}.
\]
Thus
U=-\frac{q}{2}+\sqrt{\Delta_C}=\frac{9}{2}+\frac{7}{2}=8,
\qquad
V=-\frac{q}{2}-\sqrt{\Delta_C}=\frac{9}{2}-\frac{7}{2}=1.
\]
So we may choose $u=2$ and $v=1$, giving
y=u+v=3.
\]
Indeed, $3^3-6(3)-9=27-18-9=0$.
Factoring confirms this:
y^3-6y-9=(y-3)(y^2+3y+3).
\]
Cardano’s method does not guess the factor $(y-3)$. Instead, it manufactures a quadratic equation for the quantities $u^3$ and $v^3$. The cubic becomes solvable because the new quantities are governed by symmetric relations.
Quartics, Part I: Depressing the Quartic and the Special Biquadratic Case
Start with the general quartic
ax^4+bx^3+cx^2+dx+e=0, \qquad a\neq 0.
\]
After dividing by $a$, write
x^4+Bx^3+Cx^2+Dx+E=0.
\]
As with the cubic, we first remove the next-to-leading term. Set
x=y-\frac{B}{4}.
\]
The result of the substitution. After expansion and collection of like terms, one obtains
y^4+py^2+qy+r=0,
\]
where
p=C-\frac{3B^2}{8},
\]
\[
q=D-\frac{BC}{2}+\frac{B^3}{8},
\]
\[
r=E-\frac{BD}{4}+\frac{B^2C}{16}-\frac{3B^4}{256}.
\]
Depressed quartic. A quartic of the form
y^4+py^2+qy+r=0
\]
is called a depressed quartic because the cubic term is gone.
Special case: biquadratic equations. If $q=0$, the quartic becomes
y^4+py^2+r=0.
\]
Set $z=y^2$. Then we only need to solve the quadratic
z^2+pz+r=0.
\]
After finding $z$, take square roots to recover $y$.
Biquadratic example. Solve
y^4-5y^2+4=0.
\]
Set $z=y^2$. Then
z^2-5z+4=0=(z-1)(z-4).
\]
So $z=1$ or $z=4$, hence
y=\pm 1, \qquad y=\pm 2.
\]
Quartics, Part II: Ferrari’s Method Carefully Derived
Now solve the depressed quartic
y^4+py^2+qy+r=0.
\]
Ferrari’s key move is to force the quartic into a difference of two squares.
Introduce a new parameter $z$ and write
y^4+py^2+qy+r
=\left(y^2+\frac{z}{2}\right)^2
-\left((z-p)y^2-qy+\left(\frac{z^2}{4}-r\right)\right).
\]
The expression inside the second parentheses is a quadratic in $y$.
If we can choose $z$ so that this quadratic is actually a perfect square,
then the entire quartic factors as a difference of squares.
Condition for a quadratic to be a square. A quadratic
Ay^2+By+C
\]
is a perfect square if and only if its discriminant is zero:
B^2-4AC=0.
\]
Here
A=z-p,
\qquad B=-q,
\qquad C=\frac{z^2}{4}-r.
\]
So we require
(-q)^2-4(z-p)\left(\frac{z^2}{4}-r\right)=0.
\]
Equivalently,
q^2=(z-p)(z^2-4r).
\]
Expanding gives the resolvent cubic:
z^3-pz^2-4rz+4pr-q^2=0.
\]
Resolvent cubic for Ferrari’s method. Choose any root $z$ of
z^3-pz^2-4rz+4pr-q^2=0.
\]
Now define
\alpha=\sqrt{z-p},
\qquad
\beta=\frac{q}{2\alpha}.
\]
The resolvent equation guarantees that
\beta^2=\frac{q^2}{4(z-p)}=\frac{z^2}{4}-r.
\]
Hence
(z-p)y^2-qy+\left(\frac{z^2}{4}-r\right)=(\alpha y-\beta)^2.
\]
Therefore
y^4+py^2+qy+r
=\left(y^2+\frac{z}{2}\right)^2-(\alpha y-\beta)^2.
\]
Now factor as a difference of squares:
y^4+py^2+qy+r
=\left(y^2-\alpha y+\frac{z}{2}+\beta\right)
\left(y^2+\alpha y+\frac{z}{2}-\beta\right).
\]
Conclusion. Ferrari’s method reduces the quartic to:
- a cubic equation for $z$ (the resolvent cubic);
- then two quadratic equations in $y$.
This is the quartic analogue of Cardano’s insight for cubics: solve the original equation by moving to a carefully designed auxiliary equation.
Ferrari Worked Example
Example. Solve
y^4-6y^2+8y-3=0.
\]
This is already a depressed quartic with
p=-6, \qquad q=8, \qquad r=-3.
\]
The resolvent cubic is
z^3-pz^2-4rz+4pr-q^2=0,
\]
\[
z^3+6z^2+12z+8=0.
\]
This factors as
(z+2)^3=0,
\]
so we may take $z=-2$.
Now compute
\alpha=\sqrt{z-p}=\sqrt{-2-(-6)}=\sqrt{4}=2,
\]
\[
\beta=\frac{q}{2\alpha}=\frac{8}{4}=2.
\]
Ferrari’s factorization becomes
y^4-6y^2+8y-3
=\left(y^2-2y+\frac{-2}{2}+2\right)
\left(y^2+2y+\frac{-2}{2}-2\right),
\]
\[
y^4-6y^2+8y-3=(y^2-2y+1)(y^2+2y-3).
\]
So
y^2-2y+1=(y-1)^2,
\qquad
y^2+2y-3=(y-1)(y+3).
\]
Hence the roots are
y=1 \text{ (with multiplicity three) },
\qquad
y=-3.
\]
The arithmetic is not the point. The method succeeds because we create an auxiliary cubic whose solution makes a stubborn quadratic expression become a perfect square.
How the Cubic and Quartic Methods Reflect Symmetry
The formulas for quadratics, cubics, and quartics are not random collections of algebraic manipulations.
Each method isolates a new quantity that has more symmetry than the roots themselves.
Quadratic.
The individual roots are not symmetric, but the sum $r_1+r_2$ and product $r_1r_2$ are.
The expression $r_1-r_2$ changes sign under a swap, yet $(r_1-r_2)^2$ is symmetric.
That symmetric square appears as the discriminant.
Cubic.
Cardano introduces $u$ and $v$ with $y=u+v$. The quantities $u^3$ and $v^3$ satisfy a quadratic because their sum and product are symmetric in $u$ and $v$.
In other words, we move to a setting where a cubic problem becomes a quadratic one.
Quartic.
Ferrari introduces a parameter $z$ so that the quartic factors into two quadratics. The right value of $z$ is determined by a cubic resolvent.
Again, we do not attack the roots one by one. We first solve for a more symmetric auxiliary quantity.
A resolvent equation is an auxiliary polynomial whose roots encode strategically chosen combinations of the original roots.
The resolvent is often of lower degree than the original problem, and it packages the relevant symmetry in a simpler way.
Why This Leads Toward Groups and Why Degree 5 Is Different
If a polynomial has roots $r_1,\dots,r_n$, then we can reorder them. The set of all reorderings forms the symmetric group $S_n$.
Some root expressions stay unchanged under every permutation; others do not.
The coefficients know only the fully symmetric information.
Invariant example.
r_1+r_2+r_3+r_4
\]
is unchanged under every permutation of the roots.
Non-invariant example.
r_1-r_2
\]
changes sign when $r_1$ and $r_2$ are swapped.
For quadratics, cubics, and quartics, one can still find enough partially symmetric quantities to recover the roots by radicals.
Historically, this led mathematicians to ask whether the same must be true for degree $5$ and beyond.
The answer is no in general: the generic quintic has no formula by radicals.
Important nuance.
We are not proving the impossibility of a general quintic formula here.
The point is conceptual: the success of the formulas for degrees $2$, $3$, and $4$ is controlled by symmetry, and the obstruction that appears for degree $5$ is also a symmetry obstruction.
This realization is one of the roads leading to Galois theory.
The search for formulas began as a computational problem and ended by creating one of the deepest structural subjects in mathematics: modern algebra.
CAS Exploration and What Technology Can Clarify
A computer algebra system is useful here not because it replaces theory, but because it lets us inspect structure quickly.
Quadratic tasks.
- Expand and factor $(x-h)^2-k$ for several choices of $h$ and $k$.
- Verify symbolically that the quadratic formula solves $ax^2+bx+c=0$.
- Compute $\Delta$ and compare it with the graph of the parabola.
Cubic and quartic tasks.
- Shift a general cubic by $x=y-B/3$ and ask the CAS to simplify.
- Solve a depressed cubic symbolically and compare with Cardano’s formula.
- Factor a quartic obtained from Ferrari’s method and inspect the two quadratic factors.
Why do exact formulas become more complicated as the degree increases, even though the conceptual strategy still says “reduce to something simpler”? The short answer is that the symmetry becomes richer, so the auxiliary quantities must encode more information.
Formative Checks and In-Class Questions
FC1. Explain why the coefficient of $x^{n-2}$ in $\prod_{i=1}^n(x-r_i)$ is $\sum_{i<j}r_ir_j$.
FC2. Use Vieta’s formulas to rewrite $(r_1-r_2)^2$ for a quadratic in terms of the coefficients.
FC3. Starting with $x^3+Bx^2+Cx+D=0$, verify directly that the substitution $x=y-B/3$ removes the $y^2$ term.
FC4. In Cardano’s method, why do $U=u^3$ and $V=v^3$ satisfy a quadratic equation rather than another cubic equation?
FC5. In Ferrari’s method, what condition must be imposed so that a certain quadratic expression becomes a perfect square?
Write one sentence explaining the difference between a root and a symmetric function of the roots.
Exercises, Homework, and Next Steps
Recommended exercises.
- Derive the quadratic formula from scratch without looking back at the notes.
- For $2x^2-4x+7=0$, compute the discriminant and classify the roots before solving.
- If a monic quadratic has roots $r_1,r_2$, express $r_1^2+r_2^2$ in terms of the coefficients.
- Expand $(x-r_1)(x-r_2)(x-r_3)$ and identify the three elementary symmetric polynomials.
- Reduce the cubic $x^3+6x^2+9x+1=0$ to depressed form.
- Use Cardano’s method to solve $y^3-6y-9=0$ and compare with direct factoring.
- Reduce the quartic $x^4+4x^3+2x^2-4x-3=0$ to depressed form.
- Solve the biquadratic equation $y^4-13y^2+36=0$.
- Verify that $z=-2$ solves the Ferrari resolvent cubic for $y^4-6y^2+8y-3=0$.
- Explain, in your own words, why cubic and quartic formulas are examples of “solve by introducing a smarter variable.”
- Reflective question: why does the study of equations naturally lead to the study of permutations?
Next lecture preview.
The next conceptual step is to study symmetry more deliberately: permutations of roots, invariant quantities, and the beginnings of group-theoretic thinking. At that stage, formulas stop looking like isolated tricks and start looking like consequences of algebraic structure.
Summary of the Main Mathematical Ideas
- The coefficients of a polynomial are elementary symmetric polynomials in the roots.
- The quadratic formula comes from completing the square.
- The discriminant is a symmetric root expression: $\Delta=a^2(r_1-r_2)^2$.
- A general cubic is reduced to a depressed cubic by shifting the variable.
- Cardano’s method turns the cubic into a quadratic for $u^3$ and $v^3$.
- A general quartic is reduced to a depressed quartic by shifting the variable.
- Ferrari’s method turns the quartic into a cubic resolvent and then two quadratics.
- The existence and shape of these formulas are governed by symmetry.
- This symmetry viewpoint leads naturally to groups and, ultimately, to Galois theory.
References and Attribution
- Carl B. Boyer and Uta C. Merzbach, A History of Mathematics, 3rd ed., Wiley.
- David S. Dummit and Richard M. Foote, Abstract Algebra, 3rd ed., Wiley.
- Israel M. Gelfand and Alexander Shen, Algebra, Birkhäuser.
- Victor J. Katz, A History of Mathematics: An Introduction, 3rd ed.
- R. Bruce King, Beyond the Quartic Equation, Birkhäuser.
- Ian Stewart, Galois Theory, 4th ed., Chapman and Hall/CRC.