Vieta's Formulas for Cubic Equations: Complete Guide with Examples

Introduction

In standard algebra, finding the roots of a polynomial is viewed as a one-way street. You start with a big equation, you use tools like factoring or synthetic division to break it apart, and you end up with the roots.

But what if you wanted to walk that street backward? What if you already knew two of the roots, and you wanted to find the third one without doing any division? What if you wanted to instantly verify that the answers on your math test were correct without having to plug all the numbers back into the massive equation?

To do this, you must understand the invisible mathematical strings that tie the final roots directly to the starting equation. You must understand Vieta’s Formulas.

What Vieta’s Formulas are: They are a set of three mathematical rules that prove that the roots of any polynomial are perfectly linked to the coefficients (the numbers attached to the $x$‘s) of that polynomial.

Why they are important: Vieta’s Formulas allow you to manipulate, predict, and verify roots without actually having to solve the equation. They allow you to look at a cubic equation and instantly know what the three answers will add up to.

How they simplify solving cubic equations: If you guess one root using the Rational Root Theorem, you can use Vieta’s formulas to instantly calculate the remaining two roots using basic arithmetic, completely bypassing tedious long division.

Applications in algebra, engineering, and mathematics: In structural engineering and physics, mathematicians frequently use Vieta’s formulas to “reverse-engineer” a system. If an engineer knows the physical stress limits (the roots) a bridge must withstand, they use Vieta’s formulas to calculate exactly what the mathematical properties (the coefficients) of the physical materials must be.

Learning objectives: This massive, exhaustive guide will teach you exactly how to derive and use Vieta’s formulas for cubic equations. We will explore the Sum, Pairwise, and Product formulas, contrast them with the Discriminant, use them to reverse-engineer missing coefficients, and solidify your mastery with 20 complete worked examples and 30 practice problems. Let’s begin.

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Who Was François Viète?

Before diving into the formulas, let’s look at the genius who invented them.

A brief history: François Viète (1540–1603) was a French mathematician and lawyer who served as a privy councillor to Kings Henry III and Henry IV of France. In his spare time, he cracked enemy ciphers and revolutionized mathematics.

His contributions to algebra: Viète is largely considered the “father of modern algebra.” Before him, mathematics was written entirely in words (e.g., “Three things squared minus four things equals five”). Viète was the very first person in history to use letters (like $x, y, a, b$) to represent known and unknown quantities.

Why the formulas bear his name: In 1579, by introducing letters into algebra, Viète was able to prove that there was a universal relationship between the final answers (roots) of an equation and the starting numbers (coefficients). Because he was the first to formalize this using variables, the relationships are globally known as Vieta’s Formulas (using the Latinized version of his name).

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What Are Vieta’s Formulas?

To use Vieta’s formulas, we must look at a cubic equation in standard form.

The standard cubic equation: $ax^3 + bx^2 + cx + d = 0$

In this equation:

• $a$ is the leading coefficient.
• $b$ is the quadratic coefficient.
• $c$ is the linear coefficient.
• $d$ is the constant term.

According to the Fundamental Theorem of Algebra, this cubic equation will have exactly three roots (which we will call $r_1, r_2$, and $r_3$).

The Three Formulas: Vieta’s formulas state that you can combine these three roots in specific ways to perfectly recreate the $a, b, c, d$ coefficients.

1. The Sum of Roots: If you add all three roots together, they will always equal the negative second coefficient divided by the first. $r_1 + r_2 + r_3 = -\frac{b}{a}$

2. The Sum of Pairwise Products: If you multiply every possible pair of roots and add them together, they will equal the third coefficient divided by the first. $r_1r_2 + r_2r_3 + r_1r_3 = \frac{c}{a}$

3. The Product of Roots: If you multiply all three roots together, they will equal the negative last coefficient divided by the first. $r_1r_2r_3 = -\frac{d}{a}$

Intuitive Explanation: Notice the pattern! The formulas cycle through the coefficients ($b, c, d$). They are always divided by the leading coefficient ($a$). And the plus/minus sign always alternates: Negative, Positive, Negative.

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Derivation of Vieta’s Formulas

How do we know these formulas are mathematically true? We can prove it algebraically.

A beginner-friendly derivation: Let’s pretend we have a cubic equation where the leading coefficient is 1 ($a = 1$). $x^3 + bx^2 + cx + d = 0$

We know that if an equation has roots $r_1, r_2, r_3$, it can be written in factored form: $(x - r_1)(x - r_2)(x - r_3) = 0$

Let’s expand the first two parentheses using FOIL: $(x^2 - r_1x - r_2x + r_1r_2)(x - r_3) = 0$ $[x^2 - (r_1+r_2)x + r_1r_2] \times (x - r_3) = 0$

Now multiply everything by $(x - r_3)$: $x^3 - r_3x^2 - (r_1+r_2)x^2 + r_3(r_1+r_2)x + r_1r_2x - r_1r_2r_3 = 0$

Combine the like terms ($x^2$ and $x$): $x^3 - (r_1 + r_2 + r_3)x^2 + (r_1r_2 + r_2r_3 + r_1r_3)x - (r_1r_2r_3) = 0$

Compare Coefficients! Look at our final expanded formula, and compare it to the original standard form $x^3 + bx^2 + cx + d = 0$.

• Look at the $x^2$ column: $b$ is exactly equal to $-(r_1 + r_2 + r_3)$.
• Look at the $x$ column: $c$ is exactly equal to $(r_1r_2 + r_2r_3 + r_1r_3)$.
• Look at the constant: $d$ is exactly equal to $-(r_1r_2r_3)$.

(If the leading coefficient $a$ wasn’t 1, we would just divide the whole equation by $a$, giving us the exact $b/a, c/a,$ and $d/a$ formulas we use today).

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Understanding Each Formula

Let’s break down exactly what each of the three formulas does, and how you can use them.

1. The Sum of Roots Formula

$r_1 + r_2 + r_3 = -\frac{b}{a}$ This is the most frequently used formula. It tells you exactly what the roots will add up to.

• Example: In the equation $2x^3 - 6x^2 + 4x - 1 = 0$, $a=2$ and $b=-6$.
• The sum of the roots is $-(-6) / 2 = 3$.
• Even though we have no idea what the roots are, we know with 100% certainty that if we add them together, the answer will be 3.

2. The Sum of Pairwise Products

$r_1r_2 + r_2r_3 + r_1r_3 = \frac{c}{a}$ This is the “symmetric sum.” It takes every unique combination of two roots, multiplies them, and adds the results.

• Example: In $2x^3 - 6x^2 + 8x - 1 = 0$, $a=2$ and $c=8$.
• The pairwise product is $8 / 2 = 4$.

3. The Product of Roots

$r_1r_2r_3 = -\frac{d}{a}$ This tells you what happens when you multiply all three answers together.

• Example: In $x^3 + 5x^2 - 2x + 10 = 0$, $a=1$ and $d=10$.
• The product of the roots is $-10 / 1 = -10$.
• If you know two roots are $x=2$ and $x=5$, you know $(2 \times 5 \times r_3) = -10$. Therefore, $10r_3 = -10$, so the final root MUST be $-1$. (You just found the final root without doing any division!).

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How to Use Vieta’s Formulas

Here is the mechanical, step-by-step process for applying Vieta’s formulas to an algebraic problem.

Step 1: Write the equation in standard form

Ensure it reads $ax^3 + bx^2 + cx + d = 0$. If a term is missing (like $x^2$), you MUST write $+0x^2$.

Step 2: Identify coefficients

Literally write down on your paper: $a = ?$ $b = ?$ $c = ?$ $d = ?$ Pay extreme attention to the negative signs.

Step 3: Apply each formula

Write down the three mathematical relationships:

• Sum = $-b/a$
• Pairwise = $c/a$
• Product = $-d/a$

Step 4: Interpret the results

If the problem gave you some roots, plug those numbers into $r_1, r_2, r_3$ and solve for the missing root algebraically. If the problem gave you the roots and asked for the equation, plug the roots into the left side of the formulas, calculate the numbers, and you have instantly found the $a,b,c,d$ coefficients!

Step 5: Verify answers

Once you have your proposed roots, multiply them all together. If the answer does not perfectly equal $-d/a$, you made an arithmetic error.

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Vieta’s Formulas and Factoring

Factoring massive polynomials by grouping is notoriously difficult. Vieta simplifies it.

If you are asked to factor $x^3 - 4x^2 - 11x + 30 = 0$:

1. $a=1, b=-4, c=-11, d=30$.
2. The roots must multiply to $-d/a$, which is $-30$.
3. The roots must add to $-b/a$, which is $+4$.
4. What three numbers multiply to $-30$ and add to $4$?
5. Through mental arithmetic: $2, 3$, and $-5$.
   • $2 \times 3 \times -5 = -30$.
   • $2 + 3 + (-5) = 0$. (Wait, this equals 0, not 4. Wrong numbers!)
6. Try $2, -5, 3$? No.
7. Try $2, -3, 5$.
   • $2 \times -3 \times 5 = -30$.
   • $2 + (-3) + 5 = 4$. (Perfect!).
8. The roots are $2, -3, 5$.
9. Factored equation: $(x-2)(x+3)(x-5)=0$.

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Vieta’s Formulas and the Rational Root Theorem

The Rational Root Theorem provides a massive list of candidates ($\pm p/q$) that might be roots. Instead of synthetically dividing the candidates to see if they work, you can use Vieta.

Comparison Example: Equation: $x^3 - 6x^2 + 11x - 6 = 0$. The Rational Root Theorem says the candidates are factors of 6: $\pm 1, \pm 2, \pm 3, \pm 6$. Instead of testing them via long division, apply Vieta:

• Roots must add up to $-(-6)/1 = 6$.
• Roots must multiply to $-(-6)/1 = 6$. Which three candidates add to 6 AND multiply to 6? $1 + 2 + 3 = 6$. $1 \times 2 \times 3 = 6$. The roots are exactly $1, 2, 3$. You solved the cubic equation in 15 seconds.

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Vieta’s Formulas and the Discriminant

The Discriminant ($\Delta$) tells you the nature of the roots (Real, Complex, Repeated). Vieta’s formulas apply to ALL roots, whether they are real or imaginary.

Complex Root Verification: If a cubic equation has $\Delta < 0$, it has one real root and two complex conjugate roots (e.g., $a+bi$ and $a-bi$). If you add them together using Vieta’s Sum Formula: $r_1 + (a+bi) + (a-bi) = -b/a$. The $+bi$ and $-bi$ perfectly cancel each other out! This proves mathematically why adding the roots of an equation with real coefficients will always result in a real number, even if two of the roots are imaginary.

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Vieta’s Formulas and Graphs

How do Vieta’s formulas affect the physical graph of the cubic equation?

Intercepts: The product formula ($r_1r_2r_3 = -d/a$) is directly tied to the y-intercept. The $d$ term in any polynomial is literally where the graph crosses the y-axis. Vieta proves that the physical y-intercept is entirely dictated by multiplying all the x-intercepts together!

Multiplicity: If a graph bounces on the x-axis, it is a double root. (e.g., roots are $2, 2, 5$). Vieta treats the double root as two separate entities. The sum formula will calculate $2 + 2 + 5 = 9$.

Symmetry: When cubic equations lack an $x^2$ term (e.g., $x^3 - 7x + 6 = 0$), they are called “depressed cubics.” Because $b=0$, the Sum Formula states: $r_1 + r_2 + r_3 = 0$. If the sum of the roots is zero, the roots are perfectly balanced around the y-axis!

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Applications

Why do professionals use Vieta’s Formulas?

• Engineering: When calibrating a feedback loop in a mechanical system, engineers want the system’s “roots” (frequencies) to add up to a specific stable number. Vieta allows them to instantly adjust the physical materials (coefficients) to hit that exact sum.
• Physics: In quantum mechanics and orbital dynamics, massive polynomials are used to calculate particle states. Vieta’s formulas allow physicists to calculate the “sum of all possible states” without having to solve for each individual state.
• Mathematical Competitions: Exams like the AMC and Math Olympiad frequently feature “impossible” cubic equations. The questions will ask, “What is the sum of the roots of $4x^3 - 299x^2 + 15x - 80 = 0$?” Students who try to solve it fail. Students who use Vieta simply calculate $-(-299)/4$ and get the answer instantly.

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Common Mistakes

Vieta’s formulas are notoriously easy to mess up if you are not paying strict attention.

1. Incorrect coefficient signs: Forgetting that the formulas alternate signs (Negative, Positive, Negative). The sum formula is $-b/a$. The product formula is $-d/a$.
2. Using equations not in standard form: If you apply Vieta to $x^3 = 5x^2 - 4$, you will get the wrong answer. You MUST rewrite it to $x^3 - 5x^2 + 0x + 4 = 0$.
3. Ignoring the leading coefficient: If the equation is $2x^3 - 8x^2 + x - 5 = 0$, students often say the sum of the roots is $+8$. No, you MUST divide by $a$. The sum is $8/2 = 4$.
4. Missing zero placeholders: In $x^3 - 7x + 6 = 0$, students see the second number is -7 and calculate the sum as $+7$. This is a fatal error. The second number is the $x$ term, not the $x^2$ term. $b=0$. The sum is 0.
5. Mixing up the three formulas: Accidentally using $c/a$ for the product, or $d/a$ for the pairwise sum.

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Worked Examples

Let’s execute 20 complete examples using Vieta’s Formulas.

Group 1: Monic Cubic Equations ($a=1$)

Note: When $a=1$, the formulas simplify to Sum = $-b$, Pairwise = $c$, Product = $-d$.

Example 1: Basic Sum and Product Equation: $x^3 - 4x^2 + x + 6 = 0$. Sum of roots: $-(-4)/1 = 4$. Product of roots: $-(6)/1 = -6$.

Example 2: Negative roots Equation: $x^3 + 5x^2 + 2x - 8 = 0$. Sum of roots: $-(5) = -5$. Product of roots: $-(-8) = 8$.

Example 3: Missing $x^2$ term Equation: $x^3 - 7x + 6 = 0$. $b=0$. Sum of roots: $-0 = 0$. $d=6$. Product of roots: $-6$.

Example 4: Missing constant Equation: $x^3 - 6x^2 + 5x = 0$. $d=0$. Product of roots: $-0 = 0$. (This makes sense, as $x$ can be factored out, meaning $x=0$ is a root!).

Example 5: Pairwise sum Equation: $x^3 - 2x^2 - 11x + 12 = 0$. Pairwise sum ($r_1r_2 + r_2r_3 + r_1r_3$): $c/a \rightarrow -11/1 = -11$.

Group 2: General Cubic Equations ($a > 1$)

Example 6: Fractional answers Equation: $2x^3 - 5x^2 + 4x - 1 = 0$. Sum: $-(-5)/2 = 5/2$. Product: $-(-1)/2 = 1/2$. Pairwise: $4/2 = 2$.

Example 7: Complex fractions Equation: $4x^3 + 3x^2 - x + 5 = 0$. Sum: $-3/4$. Product: $-5/4$.

Example 8: Finding a specific root using Product Equation: $3x^3 - 10x^2 + x + 6 = 0$. You are told two of the roots are $1$ and $2$. Find the third root. Product formula: $r_1r_2r_3 = -d/a \rightarrow -(6)/3 = -2$. So, $(1)(2)(r_3) = -2$. $2r_3 = -2 \rightarrow r_3 = -1$. Third root is $-1$.

Example 9: Finding a specific root using Sum Equation: $2x^3 - 9x^2 + 7x + 6 = 0$. You are told two roots are $2$ and $-1/2$. Find the third root. Sum formula: $r_1 + r_2 + r_3 = -b/a \rightarrow -(-9)/2 = 4.5$. So, $2 + (-0.5) + r_3 = 4.5$. $1.5 + r_3 = 4.5 \rightarrow r_3 = 3$. Third root is $3$.

Example 10: All zero coefficients Equation: $5x^3 - 40 = 0$. $b=0, c=0, d=-40$. Sum: $0/5 = 0$. Pairwise: $0$. Product: $-(-40)/5 = 8$.

Group 3: Finding Missing Coefficients

Example 11: Reverse-engineering the equation You are told a cubic equation with $a=1$ has roots $1, 2, 3$. Find the equation. Sum ($r_1+r_2+r_3$): $1+2+3 = 6$. So $-b=6 \rightarrow b=-6$. Pairwise ($r_1r_2 + r_2r_3 + r_1r_3$): $(1)(2) + (2)(3) + (1)(3) = 2 + 6 + 3 = 11$. So $c=11$. Product ($r_1r_2r_3$): $(1)(2)(3) = 6$. So $-d=6 \rightarrow d=-6$. Equation: $x^3 - 6x^2 + 11x - 6 = 0$.

Example 12: Equation with a double root Roots are $2, 2, -1$. Find the equation ($a=1$). Sum: $2+2+(-1) = 3 \rightarrow b=-3$. Pairwise: $(2)(2) + (2)(-1) + (2)(-1) = 4 - 2 - 2 = 0 \rightarrow c=0$. Product: $(2)(2)(-1) = -4 \rightarrow d=4$. Equation: $x^3 - 3x^2 + 4 = 0$.

Example 13: Equation with complex roots Roots are $1, i, -i$. Find the equation ($a=1$). Sum: $1 + i - i = 1 \rightarrow b=-1$. Pairwise: $(1)(i) + (i)(-i) + (1)(-i) = i - i^2 - i = -(-1) = 1 \rightarrow c=1$. Product: $(1)(i)(-i) = -i^2 = -(-1) = 1 \rightarrow d=-1$. Equation: $x^3 - x^2 + x - 1 = 0$.

Example 14: Finding $k$ Equation: $x^3 - 5x^2 + kx - 12 = 0$. Roots are $2, 3, r_3$. Find $k$. Sum: $2 + 3 + r_3 = 5 \rightarrow 5 + r_3 = 5 \rightarrow r_3 = 0$. Wait! If $r_3 = 0$, then Product = 0. But $d = -12$. This equation is mathematically impossible!

Example 15: Valid finding $k$ Equation: $x^3 - 6x^2 + kx - 6 = 0$. Roots are $1, 2, r_3$. Find $k$. Sum: $1 + 2 + r_3 = 6 \rightarrow r_3 = 3$. Now use Pairwise to find $k$ (since $c=k$): $k = (1)(2) + (2)(3) + (1)(3) = 2 + 6 + 3 = 11$. $k = 11$.

Group 4: Competition Style Questions

Example 16: Sum of squares of roots For $x^3 - 4x^2 + x + 6 = 0$, find $(r_1)^2 + (r_2)^2 + (r_3)^2$. Algebraic identity: $r_1^2 + r_2^2 + r_3^2 = (r_1+r_2+r_3)^2 - 2(r_1r_2 + r_2r_3 + r_1r_3)$. Vieta sum: $4$. Vieta pairwise: $1$. Calculation: $(4)^2 - 2(1) = 16 - 2 = 14$. Answer: 14. (We didn’t even need to find the roots!).

Example 17: Sum of reciprocals For $x^3 - 5x^2 + 2x - 8 = 0$, find $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}$. Common denominator: $\frac{r_2r_3 + r_1r_3 + r_1r_2}{r_1r_2r_3}$. This is literally (Pairwise Sum) / (Product)! Pairwise: $c/a = 2$. Product: $-d/a = 8$. Calculation: $2 / 8 = 1/4$. Answer: 1/4.

Example 18: Arithmetic Progression The roots of $x^3 - 12x^2 + 39x - 28 = 0$ are in arithmetic progression ($r-d, r, r+d$). Find them. Sum: $(r-d) + r + (r+d) = 3r$. Vieta Sum: $-(-12)/1 = 12$. So, $3r = 12 \rightarrow r = 4$. One of the roots is 4. Use synthetic division to find the rest: 1, 4, 7.

Example 19: Geometric Progression The roots of $x^3 - 14x^2 + 56x - 64 = 0$ are in geometric progression ($r/k, r, rk$). Find them. Product: $(r/k) \times (r) \times (rk) = r^3$. Vieta Product: $-(-64)/1 = 64$. So, $r^3 = 64 \rightarrow r = 4$. One root is 4. Synthetically divide and solve: 2, 4, 8.

Example 20: Complex coefficients Even if a polynomial has imaginary coefficients, Vieta still holds true. If $x^3 - ix^2 + 2x + 3i = 0$, the sum of the roots is absolutely $i$.

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Practice Problems

Test your mastery of Vieta’s formulas. Solutions are below.

Beginner Level

1. What is the sum of the roots of $x^3 - 7x^2 + 2x - 5 = 0$?
2. What is the product of the roots of $x^3 + 4x^2 - x - 9 = 0$?
3. In $2x^3 - 8x^2 + 5x - 1 = 0$, what is $a, b, c, d$?
4. What is the sum of the roots for the equation in problem 3?
5. Find the product of the roots for $3x^3 - 12 = 0$.
6. If the roots are $1, -1, 4$, what is their sum?
7. Using the roots from problem 6, what is $-b/a$?
8. If the roots are $2, 3, 5$, what is $d$ (assuming $a=1$)?
9. What is the pairwise sum for $x^3 - 4x^2 + 6x - 2 = 0$?
10. True or false: Vieta’s formulas can only be used if all roots are real.

Intermediate Level

11. Two roots of $x^3 - 6x^2 + 11x - 6 = 0$ are $1$ and $2$. Use Vieta’s sum formula to find the third root.
12. Use Vieta’s product formula to verify your answer to problem 11.
13. Find the sum of the roots for $x^3 - 5x + 2 = 0$.
14. A cubic equation ($a=1$) has roots $2, 2, -3$. Find $b$ and $d$.
15. If the product of the roots is $10$, and two of the roots are $-2$ and $1$, find the third root.
16. Find the sum of the reciprocals of the roots of $x^3 - 4x^2 + 3x - 12 = 0$. (Use the formula $\frac{c}{-d}$).
17. What is the pairwise sum of roots for $4x^3 - 2x^2 - 8x + 1 = 0$?
18. The roots of $x^3 - 9x^2 + 23x - 15 = 0$ are in arithmetic progression. What is the sum of the roots, and what is the middle root?
19. Recreate the cubic equation ($a=1$) with roots $0, 4, -4$.
20. Why must you flip the sign for the sum formula, but not the pairwise sum?

Advanced Level

21. Find $(r_1)^2 + (r_2)^2 + (r_3)^2$ for $x^3 - 2x^2 - 5x + 6 = 0$.
22. A cubic equation has roots $1+i, 1-i, 3$. Write the equation in standard form.
23. The roots of $x^3 - 7x^2 + kx - 8 = 0$ are in geometric progression. Find $k$.
24. Explain why depressed cubics ($x^3 + cx + d = 0$) always have roots that sum to exactly 0.
25. If $r_1+r_2+r_3 = 0$ and $r_1r_2r_3 = 0$, what does the graph of this cubic look like?
26. Solve for the third root if $x^3 - 5x^2 + (k)x - 24 = 0$ and two roots are complex conjugates $a+bi$ and $a-bi$, and the sum of the complex conjugates is $2$.
27. The roots of $2x^3 - 4x^2 + 8x - 10 = 0$ are $r_1, r_2, r_3$. Find the product of $(2r_1)(2r_2)(2r_3)$.
28. Find a cubic equation whose roots are exactly double the roots of $x^3 - 3x^2 + 2x - 1 = 0$.
29. If $r_1 r_2 = 5$ in the equation $x^3 - 4x^2 - 11x + 30 = 0$, find $r_3$.
30. Derive the product of roots formula for a quartic ($x^4$) equation using Vieta’s logic.

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Solutions to Practice Problems

Beginner Solutions:

1. $7$.
2. $9$.
3. $a=2, b=-8, c=5, d=-1$.
4. $-(-8)/2 = 4$.
5. $-(-12)/3 = 4$.
6. $1 + (-1) + 4 = 4$.
7. $-b/a = 4$.
8. Product is 30. $-d=30 \rightarrow d=-30$.
9. $c/a = 6/1 = 6$.
10. False. They work perfectly with complex and irrational roots.

Intermediate Solutions: 11. Sum = $6$. $1 + 2 + r_3 = 6 \rightarrow r_3 = 3$. 12. Product = $6$. $(1)(2)(3) = 6$. Correct. 13. $0$ (Because $b=0$). 14. Sum = $1 \rightarrow b=-1$. Product = $-12 \rightarrow d=12$. 15. $(-2)(1)(r_3) = 10 \rightarrow -2r_3 = 10 \rightarrow r_3 = -5$. 16. Pairwise / Product = $c / -d = 3 / 12 = 1/4$. 17. $c/a = -8/4 = -2$. 18. Sum is $9$. $3r = 9 \rightarrow r=3$. The middle root is 3. 19. Sum is 0, Product is 0, Pairwise is -16. $x^3 - 16x = 0$. 20. Because algebraic expansion creates alternating signs: $(x-r)$ multiplies $x$ to create positive components, and $-r$ to create negative components.

Advanced Solutions: 21. $(Sum)^2 - 2(Pairwise) = (2)^2 - 2(-5) = 4 + 10 = 14$. 22. Sum: $(1+i)+(1-i)+3 = 5$. Product: $(1+i)(1-i)(3) = (2)(3) = 6$. Pairwise: $(1+i)(1-i) + 3(1+i) + 3(1-i) = 2 + 3+3i + 3-3i = 8$. Equation: $x^3 - 5x^2 + 8x - 6 = 0$. 23. Geom progression product: $r^3 = -(-8) = 8 \rightarrow r=2$. One root is 2. Plug $x=2$ into equation: $8 - 28 + 2k - 8 = 0 \rightarrow 2k = 28 \rightarrow k=14$. 24. Because $b=0$, the sum formula $-b/a$ equals exactly 0. The positive roots perfectly cancel out the negative roots. 25. It passes perfectly through the origin $(0,0)$ because $d=0$, and its outer roots are perfectly symmetric around the y-axis because $b=0$. 26. Sum of roots = $5$. $(a+bi) + (a-bi) + r_3 = 5$. Since sum of conjugates is $2$, then $2 + r_3 = 5 \rightarrow r_3 = 3$. 27. $8(r_1r_2r_3) = 8(10/2) = 8(5) = 40$. 28. Let $y = 2x$, so $x = y/2$. Substitute: $(y/2)^3 - 3(y/2)^2 + 2(y/2) - 1 = 0 \rightarrow y^3/8 - 3y^2/4 + y - 1 = 0 \rightarrow y^3 - 6y^2 + 8y - 8 = 0$. 29. Product = $-30$. So $(r_1 r_2) \times r_3 = -30 \rightarrow 5r_3 = -30 \rightarrow r_3 = -6$. 30. The signs alternate: $x^4$ is pos, $x^3$ is neg, $x^2$ is pos, $x^1$ is neg, constant is pos. Therefore, product of roots = $+e/a$.

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Frequently Asked Questions