Vieta's Formulas — Mastering Polynomial Relationships

What Are Vieta’s Formulas?

Vieta’s Formulas are a set of elegant relationships that connect the roots of a polynomial to its coefficients. Here’s the magic: you can solve many polynomial problems without ever calculating the actual roots. This is incredibly useful when roots are messy, irrational, or complex.

The formulas were discovered by François Viète in the 16th century and have been a cornerstone of algebra ever since.

Why Vieta's Formulas Matter

Solve polynomial problems without finding roots explicitly
Connect coefficients and roots through elegant relationships
Work with quadratic, cubic, and higher-degree polynomials
Essential for competition mathematics and advanced algebra

Vieta’s Formulas for Quadratic Polynomials

For a quadratic polynomial with roots and , Vieta’s Formulas tell us:

These two relationships are incredibly powerful. Once you know them, you can find things like , , and even higher powers without solving for and individually.

Question

What does Vieta’s Formula say about the sum of roots in

Answer

The sum of roots equals

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Question

What does Vieta’s Formula say about the product of roots in

Answer

The product of roots equals

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Proof of Vieta’s Formulas (Quadratic)

Let’s prove this using the quadratic formula. For with :

Sum of roots:

Product of roots:

Find Sum and Product Using Vieta’s

Problem: Let and be the two roots of the quadratic . Without using the quadratic formula, find:

Solution:

For , we have , ,

(a) Sum:

(b) Product:

Going Beyond: Expressions with Root Powers

The real power of Vieta’s Formulas shows up when you need to find expressions like , , or . Here’s how you build these from the basic relationships.

Finding

We want to express this in terms of what we know: and .

This identity is worth memorizing! It converts a sum of squares into expressions we can compute directly from Vieta’s Formulas.

Finding

Again, we’re expressing everything in terms of and .

Finding

There’s a useful factorization:

Now substitute :

Worked Examples: Quadratic Case

Sum of Squares

Problem: Let and be the two roots of . Find

Solution:

From : and

First, find

Then,

Reciprocals and Products

Problem: Let and be the two roots of . Find

Solution:

From : and

(We found in the previous problem)

Transformed Roots

Problem: Let and be the two roots of . Find

Solution:

From : and

Higher Powers

Problem: Let and be the two roots of . Find:

Solution:

From : and

(a)

(b)

(c)

Vieta’s Formulas for Cubic Polynomials

For a cubic polynomial with roots , , and , the relationships extend beautifully:

These three equations connect all three roots to the coefficients. Let’s prove the general pattern.

Proof of Vieta’s Formulas (Cubic)

If , , and are the roots of , then we can write:

Expanding the right side:

Comparing coefficients with :

Coefficient of : →
Coefficient of : →
Constant term: →

Question

For a cubic

with roots

, what is

Answer

(sum of products of roots taken two at a time)

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Worked Examples: Cubic Case

Cubic Root Relationships

Problem: Let , , and be the three roots of the cubic . Find:

Solution:

For : , , ,

(a)

(b)

(c)

Cubic Reciprocal Sum

Problem: Let , , and be the three roots of . Find

Solution:

Cubic Sum Expression

Problem: Let , , and be the three roots of . Find:

Solution:

Expand each part:

Sum:

Cubic Product Transformation

Problem: Let , , and be the three roots of . Find

Solution:

Substituting our values:

Cubic Power Sums

Problem: Let , , and be the three roots of . Find:

Solution:

From : , ,

(a) Using :

(b) First find :

Then:

Key Patterns and Strategies

Problem-Solving Toolkit

Start with Vieta’s: Always identify and (or their cubic equivalents)
Use algebraic identities: Know that
Build step-by-step: Complex expressions are built from simpler pieces
Convert to standard form: Reciprocals, products, and powers all use these building blocks
Check dimensions: Make sure your answer matches what was asked

Challenge Problems

Once you’ve mastered the basics, try these deeper problems that combine multiple concepts:

Complex Root Expression

Problem: Let , , and be the three roots of . Find:

Solution:

From : , ,

(a)

(b) Note that isn’t quite right. Instead, think of it as evaluating a polynomial:

If we substitute into the condition , we get , so .

The polynomial with roots , , can be found by the product formula:

Summary and Learning Checklist

Vieta's Formulas Mastery

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Key Takeaways

Vieta’s Formulas connect polynomial roots to coefficients without solving
For quadratics: Use and
For cubics: Add and
Build expressions using algebraic identities and the basic relationships
Practice systematically to internalize these powerful techniques