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Show that the polynomial does not have any rational zeros. $$P(x)=3 x^{3}-x^{2}-6 x+12$$

Short Answer

Expert verified
The polynomial \(P(x) = 3x^3 - x^2 - 6x + 12\) has no rational zeros.

Step by step solution

01

Identify Potential Rational Zeros

According to the Rational Root Theorem, any rational root of the polynomial \(P(x) = 3x^3 - x^2 - 6x + 12\) is a factor of the constant term (12) divided by a factor of the leading coefficient (3). The potential rational zeros can be \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\). Dividing each of these by the factors of 3 (which are \(\pm 1\) and \(\pm 3\)), the possible rational roots are \(\pm 1, \pm \frac{1}{3}, \pm 2, \pm \frac{2}{3}, \pm 3, \pm 4, \pm \frac{4}{3}, \pm 6, \pm \frac{6}{3} = \pm 2, \pm 12, \pm \frac{12}{3} = \pm 4\).
02

Test Each Potential Rational Zero

Evaluate the polynomial \(P(x) = 3x^3 - x^2 - 6x + 12\) at each potential rational zero identified in Step 1. If \(P(r) = 0\), then \(r\) is a root. Plugging each potential root into the polynomial yields non-zero results for all attempts, indicating no rational zeroes.
03

Evaluate Key Calculations

For further illustration, test a few values: \(P(1) = 3(1)^3 - 1^2 - 6(1) + 12 = 8\), \(P(-1) = 3(-1)^3 - (-1)^2 - 6(-1) + 12 = 8\), \(P(2) = 3(2)^3 - 2^2 - 6(2) + 12 = 10\), \(P\left(-\frac{1}{3}\right) = 3(-\frac{1}{3})^3 - (-\frac{1}{3})^2 - 6(-\frac{1}{3}) + 12 = \frac{160}{27}\). None of these calculations yield zero, affirming that the polynomial lacks rational zeros.

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Key Concepts

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

Polynomial roots
Polynomials often have roots, which are values of the variable that make the polynomial equal to zero. Finding these roots, especially rational ones, is an important task in algebra. The roots of a polynomial are also called its solutions or zeros. When we say a polynomial has no roots, we mean it never touches the x-axis when graphed.
To solve a polynomial is to determine its roots. These could be rational numbers, irrational numbers, or complex numbers. Rational roots are typically solutions that can be expressed as a fraction, like \(\frac{1}{2}\) or -3.
It's essential to understand that polynomials can have multiple roots or even none. A cubic polynomial like \(P(x) = 3x^3 - x^2 - 6x + 12\) might have up to three roots, but they aren’t necessarily all rational. In real-world applications, finding these roots helps us analyze equations representing physical phenomena, like motion paths or growth models.
Rational zeros
When trying to find zeros of a polynomial that are rational, we employ the Rational Root Theorem. This powerful theorem gives us a way to list potential rational roots of a polynomial. These potential roots are derived from the factors of the constant term and the factors of the leading coefficient.
The theorem asserts that if a polynomial with integer coefficients has a rational zero \(\frac{p}{q}\), \(p\) is a factor of the constant term, and \(q\) is a factor of the leading coefficient. By testing these possibilities in our equation, we can check if any meet the condition of making the polynomial zero.
Considering the polynomial \(P(x) = 3x^3 - x^2 - 6x + 12\), if we test all prospective rational zeros (like in the solution steps), none satisfy the equation \(P(x) = 0\). Consequently, the polynomial does not have rational zeros based on the theorem's predictions.
Factorization
Factorization of polynomials involves breaking down a polynomial into simpler components, or factors, that when multiplied together give the original polynomial. This method can sometimes reveal roots that aren’t immediately visible, helping to simplify complex problems.
For the polynomial \(P(x) = 3x^3 - x^2 - 6x + 12\), trying to factor it traditionally means searching for a product of polynomials of lesser degrees. Unfortunately, if none of those factors have rational coefficients, factorization won’t reveal rational roots.
Many higher degree polynomials do not factor neatly into smaller ones with rational roots. Instead, they might factor into irreducible polynomials with multiple complex or irrational roots. In such scenarios, while factorization is fruitful for simplifying or solving equations, it may not yield rational numbers, as is true in this case.

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Most popular questions from this chapter

The Conjugate Zeros Theorem says that the complex zeros of a polynomial with real coefficients occur in complex conjugate pairs. Explain how this fact proves that a polynomial with real coefficients and odd degree has at least one real zero.

Find all solutions of the equation and express them in the form \(a+b i\) $$x^{2}+2 x+2=0$$

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 the intercepts and asymptotes, and then sketch a graph of the rational function and state the domain and range. Use a graphing device to confirm your answer. $$r(x)=\frac{x^{2}+3 x}{x^{2}-x-6}$$

Snow began falling at noon on Sunday. The amount of snow on the ground at a certain location at time \(t\) was given by the function $$ \begin{aligned} h(t)=11.60 t &-12.41 t^{2}+6.20 t^{3} \\ &-1.58 t^{4}+0.20 t^{5}-0.01 t^{6}\end{aligned}$$ where \(t\) is measured in days from the start of the snowfall and \(h(t)\) is the depth of snow in inches. Draw a graph of this function, and use your graph to answer the following questions. (a) What happened shortly after noon on Tuesday? (b) Was there ever more than 5 in. of snow on the ground? If so, on what day(s)? (c) On what day and at what time (to the nearest hour) did the snow disappear completely?

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