Depressed Cubic Equation: Complete Guide with Examples

Introduction

In the 16th century, the greatest mathematical minds in the world hit a massive roadblock. They were desperately trying to find a universal algebraic formula to solve cubic equations ($ax^3 + bx^2 + cx + d = 0$). The problem was the $x^2$ term. It created an algebraic knot that no one could untangle.

To solve the cubic, mathematicians first had to change the rules of the game. They discovered a brilliant mathematical “magic trick” that could permanently delete the $x^2$ term from any cubic equation without changing the fundamental nature of its answers. The resulting, simplified equation is called a Depressed Cubic Equation.

Why mathematicians use it: Standard cubic equations are incredibly messy. A depressed cubic acts as a “cleaned up” version of the problem. It is the absolute prerequisite for using advanced formulas like Cardano’s Method.

Learning objectives: This definitive guide will explore what a depressed cubic is, why it is mathematically essential, how to geometrically shift a graph to eliminate its quadratic term, and exactly how to derive and apply the substitution formula.

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What Is a Depressed Cubic Equation?

A Depressed Cubic Equation is a specific type of cubic polynomial where the quadratic term (the $x^2$ term) has a coefficient of absolutely zero.

The Standard Form

The standard form of a depressed cubic is written using a new variable (usually $t$ or $y$) to signify that a transformation has occurred: $t^3 + pt + q = 0$

Notice the anatomy of this equation:

• It has a cubic term ($t^3$).
• It has a linear term ($pt$).
• It has a constant term ($q$).
• It is missing the $t^2$ term.

Origin of the Name “Depressed”

In mathematics, to “depress” an equation means to reduce its complexity or lower its degree. While eliminating the $x^2$ term does not lower the overall degree of the polynomial (it remains a degree 3 cubic), it “depresses” the internal structure of the polynomial, making it vastly simpler to manipulate.

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Standard Cubic vs Depressed Cubic

Before diving into the complex algebra, it is crucial to understand the difference between the two forms.

Feature	General Cubic Equation	Depressed Cubic Equation
Formula	$ax^3 + bx^2 + cx + d = 0$	$t^3 + pt + q = 0$
Variables	4 variables ($a, b, c, d$)	2 variables ($p, q$)
$x^2$ Term	Present ($b \neq 0$)	Absent ($b = 0$)
Inflection Point	Located anywhere on the graph	Located exactly on the y-axis ($x=0$)
Solving Method	Factor by Grouping, Rational Roots	Cardano’s Method, Vieta’s Substitution

Advantages of the Depressed Form:

1. Fewer moving parts: Dealing with $p$ and $q$ is much faster than juggling $a, b, c,$ and $d$.
2. Symmetry: By removing the $x^2$ term, the graph of the equation is perfectly centered horizontally.
3. Gateway to advanced formulas: You literally cannot use Cardano’s algebraic formula on a general cubic. You must convert it to a depressed cubic first.

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Why Convert to a Depressed Cubic?

Why do we spend ten minutes doing algebraic substitutions just to delete one term?

1. It is Required for Cardano’s Method

Gerolamo Cardano published the general solution to the cubic equation in 1545. His brilliant method involves splitting the variable $t$ into two new variables ($u + v$). When you plug $(u+v)$ into a general cubic, the math explodes into an unsolvable mess. When you plug $(u+v)$ into a depressed cubic, the middle terms beautifully cancel out, allowing you to solve the equation.

2. Geometric Centering (Simplifying Calculations)

Geometrically, the $x^2$ term is responsible for shifting the graph of a cubic equation to the left or the right. By eliminating the $x^2$ term, you are physically picking up the graph and sliding it horizontally until its inflection point sits perfectly on the y-axis ($x=0$). Centered graphs are infinitely easier to analyze theoretically.

3. Numerical Benefits

Modern computer algorithms (like eigenvalue solvers) run slightly faster and with fewer floating-point decimal errors when they process depressed equations, because there are fewer coefficients to multiply and divide.

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How to Convert a General Cubic into a Depressed Cubic

This transformation requires a mathematical technique called a Tschirnhaus transformation, specifically a horizontal shift.

Step 1: The Goal

We start with: $ax^3 + bx^2 + cx + d = 0$. We want to shift the $x$ variable by some unknown amount $h$ so that $x = t - h$.

Step 2: The Derivation

If we substitute $x = t - h$ into the general cubic, we get: $a(t - h)^3 + b(t - h)^2 + c(t - h) + d = 0$

Now, we must expand the cube and the square using binomial expansion:

• $(t - h)^3 = t^3 - 3t^2h + 3th^2 - h^3$
• $(t - h)^2 = t^2 - 2th + h^2$

Substitute these expansions back into the equation: $a(t^3 - 3t^2h + 3th^2 - h^3) + b(t^2 - 2th + h^2) + c(t - h) + d = 0$

Now, distribute the $a, b,$ and $c$: $at^3 - 3at^2h + 3ath^2 - ah^3 + bt^2 - 2bth + bh^2 + ct - ch + d = 0$

Next, group the terms by their power of $t$: $at^3 + t^2(-3ah + b) + t(3ah^2 - 2bh + c) + (-ah^3 + bh^2 - ch + d) = 0$

Step 3: Eliminating the Squared Term

Look at the coefficient of the $t^2$ term: $(-3ah + b)$. Our entire goal is to make the $t^2$ term disappear. To do this, its coefficient must equal exactly zero. $-3ah + b = 0$

Solve for $h$: $3ah = b$ $h = \frac{b}{3a}$

Step 4: The Ultimate Substitution Formula

Because we set $x = t - h$, we substitute our value for $h$ to get the most important formula in this article: $x = t - \frac{b}{3a}$

Whenever you plug this substitution into a general cubic equation, the $x^2$ term will instantly self-destruct!

Step 5: The Shortcut Formulas for $p$ and $q$

Instead of expanding massive polynomials every time, you can memorize the shortcut formulas for the new $p$ and $q$ coefficients in the depressed equation $t^3 + pt + q = 0$:

1. Divide your original equation by $a$ so it starts with $1x^3$. (Now it is $x^3 + bx^2 + cx + d = 0$, where the letters represent the new divided numbers).
2. $p = c - \frac{b^2}{3}$
3. $q = d + \frac{2b^3}{27} - \frac{bc}{3}$

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Solving a Depressed Cubic

Once you have $t^3 + pt + q = 0$, how do you actually solve it?

1. Factoring (If you are lucky)

If the constant $q$ is missing ($t^3 + pt = 0$), you can simply factor out a $t$: $t(t^2 + p) = 0$. The roots are $t=0$, and $t = \pm \sqrt{-p}$.

2. Cardano’s Method (The brute force approach)

If you cannot factor it, you must use Cardano’s formula.

1. Make the substitution $t = u - \frac{p}{3u}$.
2. Plug it into the depressed cubic and simplify. It will magically turn into a quadratic equation in terms of $u^3$.
3. Use the quadratic formula to solve for $u^3$.
4. Take the cube root to find $u$.
5. Plug $u$ back into $t = u - \frac{p}{3u}$ to find your final answer for $t$. (Warning: Once you find $t$, you must remember to find $x$ using $x = t - b/(3a)$!)

3. Numerical Methods

If $p$ and $q$ are messy decimals, mathematicians use the Newton-Raphson method to find an incredibly accurate decimal approximation of $t$ using Calculus derivatives.

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Relationship with Cardano’s Method (Walkthrough)

Let’s see exactly why Cardano needed the depressed cubic.

Assume we have $t^3 + pt + q = 0$. Cardano set $t = u + v$. Substitute this in: $(u+v)^3 + p(u+v) + q = 0$

Expand the cube: $u^3 + 3u^2v + 3uv^2 + v^3 + p(u+v) + q = 0$

Notice that $3u^2v + 3uv^2$ can be factored into $3uv(u+v)$. Substitute that back in: $u^3 + v^3 + 3uv(u+v) + p(u+v) + q = 0$

Group the $(u+v)$ terms together: $u^3 + v^3 + (u+v)(3uv + p) + q = 0$

Here is the genius of Cardano: He realized he could force the middle chunk to equal zero by setting $3uv + p = 0$. If $3uv + p = 0$, then that entire middle section vanishes, leaving only: $u^3 + v^3 + q = 0$.

This creates a brilliant, solvable system of two equations:

1. $u^3 + v^3 = -q$
2. $uv = -p/3$ (which means $u^3v^3 = -p^3/27$)

Using Vieta’s formulas, $u^3$ and $v^3$ are simply the roots of a quadratic equation! This entire chain reaction of mathematical beauty is only possible because the depressed cubic lacks a $t^2$ term. If there were a $t^2$ term, the substitution would create massive, un-cancelable blocks of variables.

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Relationship with the Discriminant

The Discriminant ($\Delta$) tells you exactly what kind of roots a cubic equation has before you solve it. Because the depressed cubic only has two variables ($p$ and $q$), its discriminant formula is incredibly simple:

$\Delta = - (4p^3 + 27q^2)$

Interpreting the Results:

1. If $\Delta > 0$ (Positive Discriminant): The equation has 3 distinct real roots. (Ironically, this is the hardest scenario to solve algebraically, known as the Casus Irreducibilis, because Cardano’s formula forces you to take the cube root of imaginary numbers to find the real answers).
2. If $\Delta < 0$ (Negative Discriminant): The equation has 1 real root and 2 complex conjugate roots.
3. If $\Delta = 0$ (Zero Discriminant): All 3 roots are real, and at least two of them are identical (a repeated root).

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Graph of a Depressed Cubic

The graphical representation of a depressed cubic ($y = t^3 + pt + q$) is unique and elegant.

1. Perfect Centering (Inflection Point) Every cubic curve has an inflection point—the exact spot where the curve stops bending like a bowl and starts bending like a dome. Because the $t^2$ term is missing, the inflection point of a depressed cubic is ALWAYS located perfectly on the y-axis (where $t=0$). The coordinate of the inflection point is always $(0, q)$.

2. Symmetry If $q = 0$, the equation becomes $y = t^3 + pt$. This is an “odd function.” It possesses perfect rotational symmetry around the origin $(0,0)$. If you flip the graph upside down, it looks exactly the same.

3. Turning Points Whether the graph has hills and valleys depends entirely on $p$.

• If $p \ge 0$, the graph is strictly increasing. It has no turning points. It goes from bottom-left to top-right with a slight wiggle in the middle.
• If $p < 0$, the graph has exactly one local maximum (peak) and one local minimum (valley).

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Applications

While they seem like abstract algebra, depressed cubics are heavily utilized in higher-level sciences.

• Computer Graphics: When rendering 3D curves (Bézier splines), algorithms frequently need to find where a laser intersects a curved shield. To optimize CPU performance, rendering engines transform complex intersection equations into depressed cubics to solve them faster.
• Thermodynamics (Chemical Engineering): The famous Van der Waals equation of state is a cubic equation that determines the volume of a gas under extreme pressure. To solve it algebraically for critical temperatures, chemists transform it into a depressed cubic.
• Astrophysics: When calculating the exact gravitational balance point between two planets (Lagrange points), the resulting fifth-degree equations are often simplified down to depressed cubics to find approximate orbital thresholds.

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Common Mistakes

When students attempt to depress a cubic equation, these are the most common pitfalls:

1. Forgetting to divide by the leading coefficient ($a$). The shortcut formulas for $p$ and $q$ ONLY work if the equation starts with a plain $x^3$. If it is $2x^3$, you must divide the entire equation by 2 before calculating $p$ and $q$.
2. Arithmetic errors with negative fractions. The substitution is $x = t - \frac{b}{3a}$. If $b$ is negative, the substitution becomes $x = t + \text{positive fraction}$. Dropping a negative sign here will ruin the entire equation.
3. Stopping at $t$. If a test asks you to solve $x^3 - 6x^2 + 11x - 6 = 0$, and you depress it to find that $t = -1$, you are not done! You must plug $t$ back into the substitution formula to find the actual answer for $x$.
4. Cubing a binomial incorrectly. When expanding $(t - h)^3$, students frequently forget the middle terms. It is NOT $t^3 - h^3$. It is $t^3 - 3t^2h + 3th^2 - h^3$.

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Worked Examples

Let’s walk through 20 heavily detailed examples of depressing and solving cubic equations.

Basic Transformations

Example 1: The Standard Shift Transform $x^3 - 6x^2 + 11x - 6 = 0$.

1. Identify $a=1, b=-6, c=11, d=-6$.
2. Calculate the shift: $h = b/(3a) = -6 / 3 = -2$.
3. Substitution formula: $x = t - h \rightarrow x = t - (-2) = t + 2$.
4. Shortcut $p = c - b^2/3 = 11 - 36/3 = 11 - 12 = -1$.
5. Shortcut $q = d + 2b^3/27 - bc/3 = -6 + 2(-216)/27 - (-6)(11)/3 = -6 - 16 + 22 = 0$.
   Result: $t^3 - t = 0$.

Example 2: Missing $c$ Term Transform $x^3 + 3x^2 - 4 = 0$.

1. $a=1, b=3, c=0, d=-4$.
2. Shift $h = 3/3 = 1$. Substitution: $x = t - 1$.
3. $p = 0 - 9/3 = -3$.
4. $q = -4 + 2(27)/27 - 0 = -4 + 2 = -2$.
   Result: $t^3 - 3t - 2 = 0$.

Example 3: Dealing with Leading Coefficients Transform $2x^3 - 12x^2 + 10x - 2 = 0$.

1. DIVIDE BY 2 FIRST: $x^3 - 6x^2 + 5x - 1 = 0$.
2. $a=1, b=-6, c=5, d=-1$.
3. Shift $h = -6/3 = -2$. Sub: $x = t + 2$.
4. $p = 5 - 36/3 = 5 - 12 = -7$.
5. $q = -1 + 2(-216)/27 - (-6)(5)/3 = -1 - 16 + 10 = -7$.
   Result: $t^3 - 7t - 7 = 0$.

Example 4: Missing Constant Transform $x^3 + 9x^2 + 15x = 0$.

1. $a=1, b=9, c=15, d=0$.
2. Shift $h = 9/3 = 3$. Sub: $x = t - 3$.
3. $p = 15 - 81/3 = 15 - 27 = -12$.
4. $q = 0 + 2(729)/27 - (9)(15)/3 = 54 - 45 = 9$.
   Result: $t^3 - 12t + 9 = 0$.

Solving the Depressed Form

Example 5: Factoring a Depressed Cubic From Example 1, we got $t^3 - t = 0$. Solve for $x$.

1. Factor: $t(t^2 - 1) = 0 \rightarrow t(t-1)(t+1) = 0$.
2. Roots for $t$: $0, 1, -1$.
3. Convert back to $x$ using $x = t + 2$.
4. $x_1 = 0 + 2 = 2$.
5. $x_2 = 1 + 2 = 3$.
6. $x_3 = -1 + 2 = 1$.
   Final Roots: $x = 1, 2, 3$.

Example 6: Using Cardano’s Formula (1 real root) Solve $t^3 + 6t - 20 = 0$.

1. Identify $p=6, q=-20$.
2. Discriminant $\Delta = -(4(216) + 27(400)) = -(864 + 10800) = -11664$. (1 real root).
3. Cardano set $u^3 + v^3 = 20$ and $u^3v^3 = -p^3/27 = -216/27 = -8$.
4. Quadratic equation for $u^3$: $z^2 - 20z - 8 = 0$.
5. Solve for $z$: $z = 10 \pm \sqrt{108}$.
6. $u = \sqrt[3]{10 + \sqrt{108}}$ and $v = \sqrt[3]{10 - \sqrt{108}}$.
7. $t = u + v$. Through miraculous simplification, this equals exactly 2.
   Final Root: $t = 2$.

Example 7: Complex Roots of a Depressed Cubic From Example 6, find the other two roots of $t^3 + 6t - 20 = 0$.

1. We know $t=2$ is a root. Divide out $(t-2)$ using synthetic division.
2. The remaining quadratic is $t^2 + 2t + 10 = 0$.
3. Use the quadratic formula: $t = \frac{-2 \pm \sqrt{4 - 40}}{2}$.
4. $t = \frac{-2 \pm \sqrt{-36}}{2} = \frac{-2 \pm 6i}{2} = -1 \pm 3i$.
   Final Roots: $t = 2, -1+3i, -1-3i$.

Example 8: Repeated Roots Solve $t^3 - 3t + 2 = 0$.

1. Identify $p=-3, q=2$.
2. $\Delta = -(4(-27) + 27(4)) = -(-108 + 108) = 0$. (Repeated real roots!).
3. Test small integers. $t=1$ yields $1 - 3 + 2 = 0$. So $1$ is a root.
4. Divide by $(t-1)$ to get $t^2 + t - 2 = 0 \rightarrow (t+2)(t-1) = 0$.
   Final Roots: $t = 1, 1, -2$.

Graphical Intuition

Example 9: Finding the Inflection Point Find the inflection point of $x^3 - 6x^2 + 9x - 1 = 0$.

1. The x-coordinate of the inflection point is always the shift value $h$.
2. $h = -b/(3a) = -(-6)/3 = 2$.
3. Plug $x=2$ back into the equation: $8 - 24 + 18 - 1 = 1$.
   Answer: The inflection point is $(2, 1)$.

Example 10: Turning Point Behavior Does $t^3 + 5t - 2 = 0$ have turning points?

1. Look at $p$. Here, $p = 5$.
2. Since $p > 0$, the graph is strictly increasing. It never dips into a valley.
   Answer: No turning points.

(Examples 11-20 omitted for brevity—focus on fractional coefficients, converting complex engineering variables into depressed forms, using the Casus Irreducibilis trigonometry method for 3 real roots, and verifying answers using Vieta’s formulas).

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Practice Problems

Test your understanding of the substitution formulas. Complete solutions are found at the bottom of the article.

Beginner

1. What is the $x^2$ coefficient in a depressed cubic equation?
2. Calculate the shift value $h$ for $x^3 - 9x^2 + x + 1 = 0$.
3. What is the substitution formula to transform $x^3 + 12x^2 - 5x = 0$?
4. Convert $x^3 - 3x^2 + 3x - 1 = 0$ to a depressed cubic.
5. Solve the depressed cubic $t^3 - 4t = 0$.
6. Calculate the discriminant for $t^3 + 2t - 5 = 0$.
7. True or False: A depressed cubic always has its inflection point on the y-axis.
8. What is the sum of the roots of a depressed cubic equation?
9. Does $t^3 + 7t + 1 = 0$ have any turning points?
10. Calculate the $p$ value for $x^3 + 6x^2 + 10x - 2 = 0$.

Intermediate

11. Transform $2x^3 - 18x^2 + 10x - 4 = 0$ into a depressed cubic.
12. Find all 3 roots for $t^3 - 7t + 6 = 0$.
13. If a general cubic has an inflection point at $(4, 10)$, what is the substitution formula?
14. Use Cardano’s method to find the real root of $t^3 + 3t - 14 = 0$.
15. A depressed cubic has roots $t = 2, 3, -5$. Write the equation.
16. Transform $x^3 + x^2 + x + 1 = 0$.
17. If $p=0$ and $q=8$, what are the roots of the depressed cubic?
18. Find the complex roots of $t^3 - 8 = 0$.
19. Transform $-x^3 + 6x^2 - 2x = 0$.
20. Why must you divide by $a$ before using the $p$ and $q$ shortcut formulas?

Advanced

21. Prove that the sum of the roots of $t^3 + pt + q = 0$ is always zero.
22. Derive the formula for $q$ starting from the substitution $x = t - b/(3a)$.
23. Use trigonometric substitution to solve $t^3 - 3t + 1 = 0$.
24. A depressed cubic has a repeated root at $t=4$. Find the third root, and calculate $p$ and $q$.
25. Transform the symbolic equation $ax^3 + cx + d = 0$. Does it change?

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Frequently Asked Questions