Chapter 5: Problem 61
Solve each equation in the real number system. $$ 3 x^{3}-x^{2}-15 x+5=0 $$
Short Answer
Expert verified
\(x = \frac{1}{3}, x = \pm \sqrt{5} \)
Step by step solution
01
Identify the Polynomial
The given polynomial equation is \(3x^3 - x^2 - 15x + 5 = 0\).
02
Apply Rational Root Theorem
According to the Rational Root Theorem, the possible rational roots of the polynomial \(3x^3 - x^2 - 15x + 5 = 0\) are the factors of the constant term (5) divided by the factors of the leading coefficient (3). Therefore, possible rational roots are \( \pm 1, \pm 5, \pm \frac{1}{3}, \pm \frac{5}{3} \). Test these possible roots to see if they satisfy the equation.
03
Test Possible Rational Roots
Test \(x = 1\): \(3(1)^3 - (1)^2 - 15(1) + 5 = 3 - 1 - 15 + 5 = -8 \), which is not zero. Test \(x = -1\): \(3(-1)^3 - (-1)^2 - 15(-1) + 5 = -3 - 1 + 15 + 5 = 16 \), which is not zero. Test \(x = \frac{1}{3}\): \(3(\frac{1}{3})^3 - (\frac{1}{3})^2 - 15(\frac{1}{3}) + 5 = 3(\frac{1}{27}) - \frac{1}{9} - 5 + 5 = \frac{1}{9} - \frac{1}{9} = 0 \). Thus, \(x = \frac{1}{3}\) is a root.
04
Factor the Polynomial
Since \(x = \frac{1}{3}\) is a root, we can factor one term out: \(3x^3 - x^2 - 15x + 5 = 0\) as \( (x - \frac{1}{3})(3x^2 - fx + a) = 0\). Simplify \(by Polynomial Division\) to get: \((3x - 1)(x^2 + 0x - 5) = 0\).
05
Solve the Quadratic Equation
\(x^2 - 5 = 0 \) gives \(x = \pm \sqrt{5} \). These are the values of x.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rational Root Theorem
The Rational Root Theorem is a useful tool for finding possible rational roots of a polynomial equation. It states that any potential rational root, expressed as a fraction \(\frac{p}{q}\), must be such that \(p\) is a factor of the constant term and \(q\) is a factor of the leading coefficient.
In the given equation, \(3x^3 - x^2 - 15x + 5 = 0\), the constant term is 5 and the leading coefficient is 3.
Therefore, possible rational roots are \( \pm 1, \pm 5, \pm \frac{1}{3}, \pm \frac{5}{3} \). To find out if they are actual roots, test each one by substituting them into the equation and checking if the result is zero.
In the given equation, \(3x^3 - x^2 - 15x + 5 = 0\), the constant term is 5 and the leading coefficient is 3.
Therefore, possible rational roots are \( \pm 1, \pm 5, \pm \frac{1}{3}, \pm \frac{5}{3} \). To find out if they are actual roots, test each one by substituting them into the equation and checking if the result is zero.
Polynomial Division
Once a root is identified, the next step is to factorize the polynomial using polynomial division. This divison allows for reducing the polynomial into simpler components.
In this example, the root \(x = \frac{1}{3} \) allows us to factorize the polynomial \((3x^3 - x^2 - 15x + 5)\) as \((x - \frac{1}{3})(3x^2 + 0x - 5) \).
To do this, rewrite the polynomial in terms of the identified root, then divide the original polynomial by the root factor.
The result is \( (3x - 1)(x^2 - 5) \).
In this example, the root \(x = \frac{1}{3} \) allows us to factorize the polynomial \((3x^3 - x^2 - 15x + 5)\) as \((x - \frac{1}{3})(3x^2 + 0x - 5) \).
To do this, rewrite the polynomial in terms of the identified root, then divide the original polynomial by the root factor.
The result is \( (3x - 1)(x^2 - 5) \).
Quadratic Equation
Finally, solving the remaining polynomial, which is often in the form of a quadratic equation, is the last step. A quadratic equation is typically expressed as \(ax^2 + bx + c = 0\).
In our factored form, we are left with \(x^2 - 5 = 0 \).
Solving this equation involves isolating \(x\) by taking the square root of both sides: \ x^2 - 5 = 0 \ gives us \(x^2 = 5 \), and thus \( x = \pm \sqrt{5} \).
So the complete set of solutions for the given polynomial equation is \( \frac{1}{3}, \sqrt{5}, - \sqrt{5} \).
In our factored form, we are left with \(x^2 - 5 = 0 \).
Solving this equation involves isolating \(x\) by taking the square root of both sides: \ x^2 - 5 = 0 \ gives us \(x^2 = 5 \), and thus \( x = \pm \sqrt{5} \).
So the complete set of solutions for the given polynomial equation is \( \frac{1}{3}, \sqrt{5}, - \sqrt{5} \).