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Chapter 2: Problem 69

Why must every polynomial equation with real coefficients of degree 3 have at least one real root?

Short Answer

Expert verified
Every cubic polynomial equation with real coefficients has at least one real root because the Fundamental Theorem of Algebra dictates that a polynomial of degree n will have exactly n roots. As cubic polynomials are of degree 3, they have 3 roots. Additionally, non-real roots of a polynomial with real coefficients come in conjugate pairs. Given that complex roots exist in pairs and cubic polynomials have 3 roots, it is impossible for all 3 roots to be non-real. Therefore, at least one real root must exist.

Step by step solution

01

Define Real Roots

In the context of polynomials, a 'real root' refers to the solutions of the polynomial that are real numbers. These are the values of x which make the polynomial equal to zero.
02

Discuss the Fundamental Theorem of Algebra

The fundamental theorem of algebra stipulates that a polynomial of degree n will have exactly n roots. These roots may be real or complex, but they exist within the field of complex numbers.
03

Relate the Degree of Equation to its Roots

As the exercise discusses a polynomial of degree 3, the fundamental theorem of algebra tells us it will have 3 roots. These roots may be real or complex.
04

Explain the Role of Conjugate Pairs

Now important to note is that non-real roots of a polynomial with real coefficients come in conjugate pairs, i.e. if \(a + i*b\) is a root then so is \(a - i*b\). This means that complex roots exist in pairs.
05

Realize Complex Pair Existence and Draw Conclusion

Given that a cubic polynomial has 3 roots (as per the Fundamental Theorem of Algebra) and complex roots exist in pairs, it is impossible for all three roots to be complex. Therefore, in a polynomial equation with real coefficients of degree 3, at least one real root must exist.

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