Chapter 3: Problem 90
Show that the polynomial does not have any rational zeros. $$P(x)=x^{50}-5 x^{25}+x^{2}-1$$
Short Answer
Expert verified
The polynomial \(P(x) = x^{50} - 5x^{25} + x^2 - 1\) has no rational zeros.
Step by step solution
01
- Understand the problem
We need to prove that the polynomial \(P(x) = x^{50} - 5x^{25} + x^2 - 1\) does not have any rational zeros. Rational zeros are zeros of the form \( \frac{p}{q} \), where \(p\) and \(q\) are integers and \(q eq 0\).
02
- Use the Rational Root Theorem
The Rational Root Theorem states that any rational solution \( \frac{p}{q} \) of a polynomial equation with integer coefficients is such that \(p\) is a factor of the constant term and \(q\) is a factor of the leading coefficient. Here, the constant term is \(-1\) and the leading coefficient is \(1\).
03
- Determine possible factors from the Rational Root Theorem
For \(P(x)\), the possible rational roots are factors of the constant term \(-1\), which are \(\pm 1\). Thus, the only possible rational zeros are \(1\) and \(-1\).
04
- Test the possible rational roots
Substitute \(x = 1\) into \(P(x)\):\[ P(1) = 1^{50} - 5 \cdot 1^{25} + 1^2 - 1 = 1 - 5 + 1 - 1 = -4 \]Substitute \(x = -1\) into \(P(x)\):\[ P(-1) = (-1)^{50} - 5(-1)^{25} + (-1)^2 - 1 = 1 + 5 + 1 - 1 = 6 \]Neither test results in zero, so neither \(1\) nor \(-1\) is a root.
05
- Conclude based on Rational Root Theorem
Since neither \(1\) nor \(-1\) is a root of \(P(x)\), and these are the only possibilities for rational roots, \(P(x)\) has no rational zeros.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomials
A polynomial is a special kind of mathematical expression. It consists of terms which are built from variables, coefficients (usually integers), and the operations of addition, subtraction, and multiplication. In simpler terms, a polynomial is like a string of math numbers and letters put together in a specific order. For example, the polynomial function \( P(x) = x^{50} - 5x^{25} + x^2 - 1 \) consists of four terms.
- \( x^{50} \) is a term with a degree of 50.
- \( -5x^{25} \) has a degree of 25. Here, -5 is the coefficient.
- \( x^2 \) is a term with a degree of 2.
- \( -1 \) is a constant term with a degree of 0.
Rational Zeros
Rational zeros are the solutions to a polynomial equation that can be expressed as a fraction \( \frac{p}{q} \), where \( p \) and \( q \) are integers, and \( q eq 0 \). Finding rational zeros means we're looking for roots that simplify into those neat little fractions everyone loves or learns to love in math class.
The Rational Root Theorem is a helpful tool for predicting these zeros. It tells us that given a polynomial equation with integer coefficients:
The Rational Root Theorem is a helpful tool for predicting these zeros. It tells us that given a polynomial equation with integer coefficients:
- Possible values for \( p \) are the factors of the constant term.
- Possible values for \( q \) are the factors of the leading coefficient.
Integer Coefficients
Integer coefficients are simply those numbers in a polynomial's terms which are whole numbers, like 1, -5, or 0. They are the multiplier of a variable raised to a power, like in the quadratic term \( -5x^{25} \). Working with integer coefficients can be easier than dealing with decimals or fractions since integers are nice, whole, and quite predictable.
For \( P(x) = x^{50} - 5x^{25} + x^2 - 1 \), all coefficients (1, -5, 1 and -1) are integers. This characteristic makes the polynomial applicable for the Rational Root Theorem.
Polynomials with integer coefficients allow the Rational Root Theorem to tell us something special: Every potential rational root must be a number \( \frac{p}{q} \) where \( p \) and \( q \) are integers that divide the constant term and leading coefficient, respectively. This simplifies our job greatly in analyzing the polynomial.
For \( P(x) = x^{50} - 5x^{25} + x^2 - 1 \), all coefficients (1, -5, 1 and -1) are integers. This characteristic makes the polynomial applicable for the Rational Root Theorem.
Polynomials with integer coefficients allow the Rational Root Theorem to tell us something special: Every potential rational root must be a number \( \frac{p}{q} \) where \( p \) and \( q \) are integers that divide the constant term and leading coefficient, respectively. This simplifies our job greatly in analyzing the polynomial.