Problem 42

Question

Solve the polynomial equation. $$ 2 x^{3}+5 x^{2}+x+12=0 $$

Step-by-Step Solution

Verified
Answer
The real root of the polynomial is \(x = -3\).
1Step 1: Identify the Polynomial
The given equation is a cubic polynomial: \(2x^3 + 5x^2 + x + 12 = 0\). It is important to recognize that it is a degree 3 polynomial, which means it can have up to 3 real roots.
2Step 2: Use Rational Root Theorem
According to the rational root theorem, any rational solution \(\frac{p}{q}\), where \(p\) is a factor of the constant term 12 and \(q\) is a factor of the leading coefficient 2, must be checked. The possible rational roots are \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{2}\).
3Step 3: Test Possible Roots
Plug each possible rational root into the polynomial to find a root. \(\frac{1}{2}\), -1, -2, -3, -4, -6, -12, and 1 are tested. On testing, we find that \(-3\) is a root because substituting \(x = -3\) gives zero: \(2(-3)^3 + 5(-3)^2 + (-3) + 12 = 0\).
4Step 4: Perform Synthetic Division
Once \(-3\) is found as a root, perform synthetic division on the polynomial \(2x^3 + 5x^2 + x + 12\) by \(x + 3\) to factor out \((x + 3)\). After division, the quotient polynomial is \(2x^2 - x + 4\).
5Step 5: Solve Quadratic Equation
The quotient polynomial \(2x^2 - x + 4 = 0\) is a quadratic equation. Solve this using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -1\), \(c = 4\). Substitute these values: \(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot 4}}{2 \cdot 2}\).
6Step 6: Calculate the Discriminant
Calculate the discriminant of the quadratic equation: \(b^2 - 4ac = (-1)^2 - 4 \cdot 2 \cdot 4 = 1 - 32 = -31\). Since the discriminant is negative, \(2x^2 - x + 4 = 0\) has no real solutions.
7Step 7: Write the Final Solution
The solution of the polynomial \(2x^3 + 5x^2 + x + 12 = 0\) includes one real root, \(x = -3\), and two complex roots from the quadratic part, which cannot be simplified to real numbers.

Key Concepts

Cubic PolynomialRational Root TheoremSynthetic DivisionQuadratic Formula
Cubic Polynomial
A cubic polynomial is a type of polynomial equation that is defined as having the highest exponent of the variable being three. This can be expressed in the general form:
  • \(ax^3 + bx^2 + cx + d = 0\)
where \(a\), \(b\), \(c\), and \(d\) are constants, with \(a eq 0\).

Cubic polynomials are fascinating because they can have up to three real roots, depending on the values of the coefficients. Each root represents a solution to the equation, meaning that it makes the entire expression equal to zero when substituted in place of \(x\).

These polynomials come in handy in various applications such as physics, engineering, and computer science, where understanding their roots helps interpret systems and models. Understanding the relationship between the coefficients and the roots can provide deeper insights into the behavior of the given equation.
Rational Root Theorem
The Rational Root Theorem is a useful tool in algebra that helps to identify potential rational roots of a polynomial equation. It involves examining the factors of the constant term and the leading coefficient to suggest possible solutions for the equation.
  • Given a polynomial \(ax^n + bx^{n-1} + ... + k = 0\), the Rational Root Theorem states that any rational solution \(\frac{p}{q}\) must have \(p\) as a factor of the constant term \(k\), and \(q\) as a factor of the leading coefficient \(a\).
Using this theorem, we can list all potential rational roots without solving the entire polynomial immediately. Typically, this results in a finite list of potential roots, which can then be tested by substitution until a valid root is found.

This theorem does not guarantee rational roots but provides a systematic way to check for them in polynomial equations. Finding a valid rational root allows us to simplify the equation further by factoring, making it easier to solve.
Synthetic Division
Synthetic division is a simplified method of dividing a polynomial by a linear expression of the form \(x - r\). It is particularly useful because it streamlines the division process, reducing the labor involved in polynomial long division.
  • Synthetic division requires coefficients from the polynomial and the potential root to carry out the division process iteratively.
  • The division results in a quotient polynomial and a remainder, where finding a zero remainder indicates that \(r\) is indeed a root of the polynomial.
Here's how it works:1. Write down the coefficients of the polynomial.2. Use the potential root, \(r\), but change its sign for the synthetic division setup.3. Bring down the leading coefficient to start the process.4. Multiply this coefficient by \(r\) and add the result to the next coefficient. Repeat for all coefficients.5. The remainder indicates whether \(r\) is a root (remainder is zero).

Synthetic division not only helps in verifying roots but also reduces the polynomial's degree, making it easier to solve the remaining expression.
Quadratic Formula
The quadratic formula is a powerful algebraic tool used to find the roots of a quadratic equation of the form \(ax^2 + bx + c = 0\). This equation can be solved by:
  • Using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)\>.
  • Substitute values of \(a\), \(b\), and \(c\) from the quadratic equation into the formula.
The expression under the square root sign, \(b^2 - 4ac\), is known as the discriminant.
  • If the discriminant is greater than zero, there are two distinct real roots.
  • If it equals zero, the equation has one real root (or two identical real roots).
  • If it is less than zero, the equation has no real roots but instead two complex conjugate roots.
In our exercise, the quadratic formula helped identify that the equation resulted in two complex roots because the discriminant was negative. Using the quadratic formula is an established procedure that reliably determines whether or not real solutions exist, and if not, it provides the complex solutions. This is invaluable for solving quadratic equations that arise in both academic and real-world problem-solving scenarios.