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Show that the polynomial does not have any rational zeros. $$ P(x)=x^{3}-x-2 $$

Short Answer

Expert verified
The polynomial \( P(x) = x^3 - x - 2 \) has no rational zeros.

Step by step solution

01

State the Rational Root Theorem

The Rational Root Theorem states that any rational solution, written as a fraction \( \frac{p}{q} \), of the polynomial equation \( a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 = 0 \) must satisfy that \( p \) divides the constant term \( a_0 \) and \( q \) divides the leading coefficient \( a_n \). For the polynomial \( P(x) = x^3 - x - 2 \), the constant term is \(-2\), and the leading coefficient is \(1\). Therefore, potential rational roots can be the factors of \(-2\) divided by the factors of \(1\).
02

Determine possible rational zeros

The factors of the constant term \(-2\) are \( \pm 1 \) and \( \pm 2 \). The factors of the leading coefficient \(1\) are \( \pm 1 \). Thus, the possible rational zeros, according to the Rational Root Theorem, are \( \pm 1 \) and \( \pm 2 \).
03

Test the possible zeros

We will substitute each possible zero into the polynomial \( P(x) = x^3 - x - 2 \) to check if it equals zero.- For \( x = 1 \): \[ P(1) = (1)^3 - 1 - 2 = 1 - 1 - 2 = -2 \] This is not zero.- For \( x = -1 \): \[ P(-1) = (-1)^3 - (-1) - 2 = -1 + 1 - 2 = -2 \] This is not zero.- For \( x = 2 \): \[ P(2) = (2)^3 - 2 - 2 = 8 - 2 - 2 = 4 \] This is not zero.- For \( x = -2 \): \[ P(-2) = (-2)^3 - (-2) - 2 = -8 + 2 - 2 = -8 \] This is not zero.
04

Conclusion

Since none of the potential rational zeros (\( \pm 1 \) and \( \pm 2 \)) result in the polynomial evaluating to zero, we conclude that the polynomial \( P(x) = x^3 - x - 2 \) does not have any rational zeros.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polynomial Zeros
The zeros of a polynomial are the values of its variable that make the entire polynomial equal to zero. In simpler terms, if you have a polynomial function like \( P(x) = x^3 - x - 2 \), the zeros are the values of \( x \) that satisfy the equation \( P(x) = 0 \). These are also called roots or solutions of the equation.

Identifying these zeros is fundamental in algebra as it can help us understand the function's behavior, solve equations, or even derive other mathematical concepts. However, not all polynomials have rational zeros. In our exercise, we tried potential rational zeros and confirmed that the polynomial \( P(x) = x^3 - x - 2 \) doesn't have any. Testing involves substituting possible values derived from the Rational Root Theorem. If none of the trials zero out the polynomial, it indicates a lack of rational roots.
Factors of a Constant Term
In the context of finding rational roots using the Rational Root Theorem, the constant term of a polynomial plays a vital role. The theorem states that for a polynomial written as \( a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 = 0 \), the possible rational roots \( \frac{p}{q} \) must have \( p \),dividing the constant term \( a_0 \).

The constant term is the part of the polynomial without a variable attached. For the polynomial \( P(x) = x^3 - x - 2 \), the constant term is \(-2\). Here, the factors of \(-2\) are \( \pm 1 \) and \( \pm 2 \). These factors are critical because they indicate which values of \( p \) to test when determining potential rational roots.

This step serves as the first filtering process, narrowing down the number of possible rational solutions we need to verify.
Leading Coefficient
The leading coefficient is the coefficient of the term with the highest degree in a polynomial. It serves as an important part of the Rational Root Theorem, which helps us figure out the possible rational zeros of a polynomial.

In our given polynomial, \( P(x) = x^3 - x - 2 \), the highest degree is 3, and the corresponding term is \( x^3 \). For this term, the coefficient is \( 1 \), which we refer to as the leading coefficient. According to the Rational Root Theorem, any possible rational root \( \frac{p}{q} \) of the polynomial must have \( q \) dividing this leading coefficient.

For \( P(x) \), the leading coefficient is \( 1 \), whose factors are \( \pm 1 \). This considerably simplifies the process because it indicates that the denominator in our potential rational roots is either +1 or -1. This aspect of having a leading coefficient of 1 allows us to directly equate potential roots with just the factors of theconstant term.

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Most popular questions from this chapter

\(66-68=\) So far we have worked only with polynomials that have real coefficients. These exercises involve polynomials with real and imaginary coefficients. (a) Show that 2\(i\) and \(1-i\) are both solutions of the equation $$ x^{2}-(1+i) x+(2+2 i)=0 $$ but that their complex conjugates \(-2 i\) and \(1+i\) are not. (b) Explain why the result of part (a) does not violate the Conjugate Zeros Theorem.

Find integers that are upper and lower bounds for the real zeros of the polynomial. $$ P(x)=2 x^{3}-3 x^{2}-8 x+12 $$

Volume of a Box An open box with a volume of 1500 \(\mathrm{cm}^{3}\) is to be constructed by taking a piece of cardboard 20 \(\mathrm{cm}\) by 40 \(\mathrm{cm}\) , cutting squares of side length \(x \mathrm{cm}\) from each corner, and folding up the sides. Show that this can be done in two different ways, and find the exact dimensions of the box in each case.

\(41-58=\) Find all zeros of the polynomial. $$ P(x)=x^{4}-6 x^{3}+13 x^{2}-24 x+36 $$

The Cubic Formula The quadratic formula can be used to solve any quadratic (or second-degree) equation. You may have wondered if similar formulas exist for cubic (third-degree), quartic (fourth-degree), and higher-degree equations. For the depressed cubic \(x^{3}+p x+q=0\) , Cardano (page 344 ) found the following formula for one solution: \(x=\sqrt[3]{\frac{-q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}+\sqrt[3]{\frac{-q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\) A formula for quartic equations was discovered by the Italian mathematician Ferrari in \(1540 .\) In 1824 the Norwegian mathematician Niels Henrik Abel proved that is impossible to write a quartic formula, that is, a formula for fifth-degree equations. Finally, Galois (page 327\()\) gave a criterion for determining which equations can be solved by a formula involving radicals. (a) \(x^{3}-3 x+2=0\) (b) \(x^{3}-27 x-54=0\) (c) \(x^{3}+3 x+4=0\)

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