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Chapter 3: Problem 71

Decide whether the statement is true or false. Justify your answer. It is possible for a third-degree polynomial function with integer coefficients to have no real zeros.

Short Answer

Expert verified
The statement is true. A third-degree polynomial function with integer coefficients can have exactly one real root and a pair of complex conjugate roots.

Step by step solution

01

Understanding the polynomial function

A third-degree polynomial function means it can be written in the form \( ax^3 + bx^2 + cx + d = 0 \), in which \( a, b, c, d \) are integers and \( a \neq 0 \). The number of roots this polynomial can have is exactly 3 according to fundamental theorem of algebra.
02

Understanding the nature of polynomial zeros

The roots of a polynomial function can be either real or complex, and complex roots always appear in conjugate pairs. The roots a+bi and a-bi are considered to be a conjugate pair. Since the coefficients are all integers and we know that the equation must has 3 roots, the roots could be all real or one real and other two a complex conjugate pair.
03

Conclusion

It is indeed possible for a third-degree polynomial function with integer coefficients to have no real zeros, except the case where this polynomial has one real root and a pair of complex conjugate roots, it can't have no real roots at all. So the statement is true.

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Most popular questions from this chapter

Use synthetic division to verify the upper and lower bounds of the real zeros of \(f .\) Then find all real zeros of the function. \(f(x)=x^{4}-4 x^{3}+15\) Upper bound: \(x=4\) Lower bound: \(x=-1\)

Find the rational zeros of the polynomial function. $$f(z)=z^{3}+\frac{11}{6} z^{2}-\frac{1}{2} z-\frac{1}{3}=\frac{1}{6}\left(6 z^{3}+11 z^{2}-3 z-2\right)$$

Find all real zeros of the polynomial function. $$f(x)=5 x^{4}+9 x^{3}-19 x^{2}-3 x$$

Divide using long division. $$\left(2 x^{4}+x^{2}-11\right) \div\left(x^{2}+5\right)$$

Simplify the expression. $$\left(\frac{x}{8}\right)^{-3}$$
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