Problem 14
Question
Show that the equation has no rational root. $$2 x^{5}+3 x^{3}+7=0$$
Step-by-Step Solution
Verified Answer
The equation \(2x^5 + 3x^3 + 7 = 0\) has no rational root because none satisfy the equation when tested.
1Step 1: Understanding Rational Root Theorem
The Rational Root Theorem states that any rational root of a polynomial equation with integer coefficients must be of the form \( p/q \), where \( p \) is a factor of the constant term and \( q \) is a factor of the leading coefficient.
2Step 2: Identify factors of the constant term and leading coefficient
The constant term of the polynomial \( 2x^5 + 3x^3 + 7 = 0 \) is 7, which has factors \( \pm 1, \pm 7 \). The leading coefficient is 2, which has factors \( \pm 1, \pm 2 \).
3Step 3: List all possible rational roots
Based on Step 2, the possible rational roots are \( \pm 1, \pm 7, \pm \frac{1}{2}, \pm \frac{7}{2} \).
4Step 4: Test each possible rational root
Substitute each possible rational root into the polynomial:1. For \( x = 1 \), \( 2(1)^5 + 3(1)^3 + 7 = 2 + 3 + 7 = 12 eq 0 \).2. For \( x = -1 \), \( 2(-1)^5 + 3(-1)^3 + 7 = -2 - 3 + 7 = 2 eq 0 \).3. For \( x = 7 \), \( 2(7)^5 + 3(7)^3 + 7 \) is much larger than zero.4. For \( x = -7 \), \( 2(-7)^5 + 3(-7)^3 + 7 \) is very negative.5. For \( x = \frac{1}{2} \), \( 2(\frac{1}{2})^5 + 3(\frac{1}{2})^3 + 7 eq 0 \).6. For \( x = -\frac{1}{2} \), \( 2(-\frac{1}{2})^5 + 3(-\frac{1}{2})^3 + 7 eq 0 \).7. For \( x = \frac{7}{2} \), \( 2(\frac{7}{2})^5 + 3(\frac{7}{2})^3 + 7 eq 0 \).8. For \( x = -\frac{7}{2} \), \( 2(-\frac{7}{2})^5 + 3(-\frac{7}{2})^3 + 7 eq 0 \).
5Step 5: Conclusion
Since none of the possible rational roots make the polynomial equal to zero, the equation \( 2x^5 + 3x^3 + 7 = 0 \) has no rational roots.
Key Concepts
Polynomial equationsRational rootsLeading coefficientConstant term
Polynomial equations
Polynomial equations are expressions that equate a polynomial to zero. A polynomial itself is a mathematical expression comprising variables, coefficients, and constants, arranged in terms raised to whole number powers. For example, in the polynomial equation \(2x^5 + 3x^3 + 7 = 0\), the highest power of the variable \(x\) is 5, indicating it's a fifth-degree polynomial.
There are several key parts to every polynomial:
There are several key parts to every polynomial:
- Terms: Each product of a coefficient and a variable raised to an exponent like \(2x^5\).
- Degree: The largest exponent, which determines the degree of the polynomial.
- Coefficients: The numbers multiplying the terms, such as 2 in \(2x^5\).
- Constant Term: A term that does not have a variable, like 7 in this case.
Rational roots
Rational roots refer to solutions of a polynomial equation that can be expressed as a fraction. An essential tool for identifying rational roots is the Rational Root Theorem.
The Rational Root Theorem asserts that if a polynomial has a rational root \(\frac{p}{q}\), then:
The Rational Root Theorem asserts that if a polynomial has a rational root \(\frac{p}{q}\), then:
- \(p\) (the numerator) is a factor of the constant term of the polynomial.
- \(q\) (the denominator) is a factor of the leading coefficient of the polynomial.
Leading coefficient
The leading coefficient is the coefficient of the term with the highest degree in a polynomial. It plays a crucial role in the Rational Root Theorem.
In the equation \(2x^5 + 3x^3 + 7 = 0\), the leading coefficient is 2 because it's associated with \(x^5\), the term with the highest power. When using the Rational Root Theorem, the possible values for the denominator \(q\) in the rational root \(\frac{p}{q}\) must be factors of the leading coefficient. In this case, the factors of 2 are \(\pm 1\) and \(\pm 2\).
The leading coefficient also affects the behavior of the graph of the polynomial, influencing its end behavior and how the polynomial's roots impact the x-axis crossings.
In the equation \(2x^5 + 3x^3 + 7 = 0\), the leading coefficient is 2 because it's associated with \(x^5\), the term with the highest power. When using the Rational Root Theorem, the possible values for the denominator \(q\) in the rational root \(\frac{p}{q}\) must be factors of the leading coefficient. In this case, the factors of 2 are \(\pm 1\) and \(\pm 2\).
The leading coefficient also affects the behavior of the graph of the polynomial, influencing its end behavior and how the polynomial's roots impact the x-axis crossings.
Constant term
The constant term is the term in a polynomial that does not contain any variables; it remains constant as the values of variables change. In \(2x^5 + 3x^3 + 7 = 0\), the constant term is 7.
When determining possible rational roots using the Rational Root Theorem, the numerator \(p\) of any potential rational root \(\frac{p}{q}\) must be a factor of this constant term. For the polynomial in question, the factors of 7 are \(\pm 1\) and \(\pm 7\).
This term plays a key role in simplifying and limiting the potential numbers we need to check when identifying possible roots, thereby making the problem more manageable.
When determining possible rational roots using the Rational Root Theorem, the numerator \(p\) of any potential rational root \(\frac{p}{q}\) must be a factor of this constant term. For the polynomial in question, the factors of 7 are \(\pm 1\) and \(\pm 7\).
This term plays a key role in simplifying and limiting the potential numbers we need to check when identifying possible roots, thereby making the problem more manageable.
Other exercises in this chapter
Problem 13
Find all values of \(x\) such that \(f(x)>0\) and all \(x\) such that \(f(x)
View solution Problem 13
Use the factor theorem to show that \(x-c\) is a factor of \(f(x)\). $$f(x)=x^{3}+x^{2}-2 x+12 ; \quad c=-3$$
View solution Problem 14
Hooke's law Hooke's law states that the force \(F\) required to stretch a spring \(x\) units beyond its natural length is directly proportional to \(x\) (a) Exp
View solution Problem 14
Sketch the graph of \(f\) $$f(x)=\frac{(5 x+3)(x+1)}{(3 x-7)(x+1)}$$
View solution