Problem 21 - Precalculus: Functions and Graphs

Question

Find all solutions of the equation. $$6 x^{5}+19 x^{4}+x^{3}-6 x^{2}=0$$

Step-by-Step Solution

Answer

The solutions are \(x = 0, -1, \frac{1}{2}, -\frac{8}{3}\).

Step 1: Factor the Common Term

Notice that every term in the equation has a common factor of \(x^2\). Let's factor \(x^2\) out of each term: \[ 6x^5 + 19x^4 + x^3 - 6x^2 = x^2(6x^3 + 19x^2 + x - 6) \] Thus, the equation can be written as \[ x^2(6x^3 + 19x^2 + x - 6) = 0 \].

Step 2: Solve for the Roots of the Factored Equation

Use the Zero Product Property to solve for \(x\):1. The first factor, \(x^2 = 0\), gives a root: \[ x = 0 \] 2. The second factor is \(6x^3 + 19x^2 + x - 6 = 0\). This cubic equation needs to be solved for its roots.

Step 3: Apply the Rational Root Theorem

To solve \(6x^3 + 19x^2 + x - 6 = 0\), apply the Rational Root Theorem, which suggests possible rational roots are factors of the constant term (\(-6\)) divided by factors of the leading coefficient (\(6\)). The possible rational roots are \( \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2} \).

Step 4: Test Potential Roots

Test the potential roots using substitution. Substitute each into the cubic equation:- Substitute \(x = 1\): \[ 6(1)^3 + 19(1)^2 + 1 - 6 = 20 \quad \text{(not zero)} \]- Substitute \(x = -1\): \[ 6(-1)^3 + 19(-1)^2 + (-1) - 6 = 0 \quad \text{(zero, \(x = -1\) is a root)} \]Since \(x = -1\) is a root, use polynomial division or synthetic division to reduce the cubic polynomial.

Step 5: Polynomial Division

Perform synthetic division of \(6x^3 + 19x^2 + x - 6\) by \(x + 1\) (since \(x = -1\) is a root):The quotient is \(6x^2 + 13x - 6\).So the equation becomes \[ x^2(x + 1)(6x^2 + 13x - 6) = 0 \]

Step 6: Solve the Quadratic Equation

Solve the quadratic \(6x^2 + 13x - 6 = 0\) using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \(a = 6\), \(b = 13\), and \(c = -6\).Calculate the discriminant:\[ b^2 - 4ac = 13^2 - 4 \times 6 \times (-6) = 361 \]Calculate the roots:\[ x = \frac{-13 \pm \sqrt{361}}{12} \]\( \sqrt{361} = 19 \), so\[ x = \frac{-13 + 19}{12} = \frac{1}{2}, \quad x = \frac{-13 - 19}{12} = -\frac{8}{3} \]

Step 7: List All Solutions

Combine all solutions:From \(x^2 = 0\), we have \(x = 0\).From \(x + 1 = 0\), we have \(x = -1\).From the quadratic equation, we have \(x = \frac{1}{2}\) and \(x = -\frac{8}{3}\).Therefore, the complete solutions are: \(x = 0, -1, \frac{1}{2}, -\frac{8}{3}\).

Key Concepts

Zero Product Property

The Zero Product Property is a fundamental concept in algebra that states if a product of multiple factors equals zero, at least one of the factors must be zero. This property is vital for solving polynomial equations.
To understand this concept, think of multiplication resulting in zero. If you multiply any number by zero, the product is naturally zero. Therefore, if you have an equation like \[ ab = 0 \] either \(a\) or \(b\) (or both) must be zero.
This theorem simplifies the solving process for polynomial equations significantly. For instance, consider a polynomial equation already factored in the form \[ (x^2)(6x^3 + 19x^2 + x - 6) = 0 \].With the Zero Product Property, you can split it into simpler equations:

• \( x^2 = 0\) which implies \( x = 0 \)
• \(6x^3 + 19x^2 + x - 6 = 0\), which needs further solving.

This property is a powerful tool because it allows us to break down complex expressions into manageable parts, making it easier to find the solutions.

Rational Root Theorem

The Rational Root Theorem is a useful technique when solving polynomial equations. It helps determine possible rational roots of a polynomial equation.
The theorem states that any rational solution, or root, of the polynomial equation \[ p(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \]accumulated in the form \( \frac{p}{q} \), where \(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading coefficient \(a_n\).
This means if you're dealing with an equation like \[6x^3 + 19x^2 + x - 6 = 0\],you will analyze the factors of

• The constant term, \(-6\): \( \pm 1, \pm 2, \pm 3, \pm 6 \).
• The leading coefficient, \(6\): \( \pm 1, \pm 2, \pm 3, \pm 6 \).

By the Rational Root Theorem, possible rational roots include these divisors, and by systematically testing them, you can identify legitimate roots. This process greatly narrows the search and emphasizes efficiency in polynomial solving.

Synthetic Division

Synthetic Division is a streamlined method of dividing polynomials. Here, it’s used when one root of a polynomial is known, helping in simplifying the equation further.
Unlike long division, synthetic division saves time, especially with simpler polynomials. It works primarily with the coefficients of the polynomial.To perform synthetic division for \[6x^3 + 19x^2 + x - 6\],consider \(x + 1\) since \(x=-1\) is already established as a root. You’ll use

• The root \(-1\) and the coefficients \([6, 19, 1, -6]\).

Execute synthetic division to refine the polynomial to \[6x^2 + 13x - 6\].By splitting the polynomial, you reduce its degree, making it less complex and moving closer to finding all solution roots. Employing synthetic division is beneficial because it efficiently simplifies polynomials with known roots.

Quadratic Formula

The Quadratic Formula is a mathematical formula that provides the solution to quadratic equations of the form \[ ax^2 + bx + c = 0 \].
When used, it involves calculating the roots using the formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]The formula is comprehensive because it applies regardless of whether the quadratic can be factored easily.
Take the quadratic derived from our synthetic division, \[6x^2 + 13x - 6 = 0\],and substitute \(a = 6\), \(b = 13\), and \(c = -6\) into the formula. Calculate the discriminant \[b^2 - 4ac = 361\], which is positive, indicating two distinct real roots.

• The formula then offers the exact solutions \(x = \frac{1}{2} \) and \(x = -\frac{8}{3}\).

This method rounds off the solution process, providing precise values for any quadratic found within a polynomial expression.