Problem 39 - Precalculus : Building Concepts and Connections

Question

Find all real solutions of the polynomial equation. $$2 x^{3}-3 x^{2}=11 x-6$$

Step-by-Step Solution

Answer

The real solutions of the given polynomial equation are \(x = 1, -2, 3\).

Step 1: Simplify the Equation

First, rewrite the polynomial equation to be equal to zero. Begin by subtracting \(11x\) and adding \(6\) to both sides to obtain: \(2x^3 - 3x^2 - 11x + 6 = 0\)

Step 2: Factor the Polynomial

Try to factor the polynomial. The standard form of factoring a cubic polynomial is \((x - a)(x - b)(x - c) = 0\). Unfortunately, in this case, factoring is complicated. We need to check for rational roots using the rational root theorem and synthetic division.

Step 3: Apply the Rational Root Theorem and Synthetic Division

The Rational Root Theorem says that any rational root, written in lowest terms, will have a numerator that is a factor of the constant term, and a denominator that is a factor of the leading coefficient. In this case, the potential rational roots are \(\pm1, \pm2, \pm3, \pm6\). Upon trying these out with synthetic division, it is found that \(x = 1\) and \(x = -2\) are roots of the polynomial.

Step 4: Find the Remaining Roots

After finding out that \(x = 1\) and \(x = -2\) are roots, use these to factor the polynomial into \(2(x - 1)(x + 2)(x - 3) = 0\). Here, we can easily identify the remaining real root as \(x = 3\).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a powerful tool for finding potential rational roots of a polynomial equation. It states that if a polynomial has a rational solution, it can be expressed as a fraction \(\frac{p}{q}\), where:

• The numerator \(p\) is a factor of the constant term.
• The denominator \(q\) is a factor of the leading coefficient.

In our example, the leading coefficient is 2, and the constant term is 6. The possible numerators \(p\) could be \(\pm 1, \pm 2, \pm 3, \pm 6\), and the possible denominators \(q\) could be \(\pm 1, \pm 2\). Consequently, the potential rational roots are \(\pm 1, \pm 2, \pm 3, \pm 6\).
Once you have this list, you can test each one within the polynomial to see if it makes the equation equal to zero. This method effectively narrows down possible solutions, allowing us to use synthetic division on the viable candidates.

Synthetic Division

Synthetic division is a simplified form of polynomial division. It's particularly useful when dividing by a linear factor. In our scenario, it's deployed to check the viability of potential rational roots. This method involves fewer steps and less writing than long division, making it faster and more efficient.
To use synthetic division:

• Write down the root you are testing on the left.
• Write the coefficients of the polynomial on the right, in descending order.
• Start by bringing down the leading coefficient.
• Multiply it by the root, and add it to the next coefficient.
• Repeat the multiply-and-add process for all coefficients.

If the result at the end is zero, your test root is indeed a root of the polynomial. This was the case when testing roots \(x = 1\) and \(x = -2\) in the example. They made the remainder zero, confirming them as roots.

Cubic Polynomials

Cubic polynomials are polynomial expressions of degree three. They take the general form \(ax^3 + bx^2 + cx + d = 0\). Solving cubic polynomials requires finding all possible roots, which could be real or complex.
For solving these, techniques include factoring, synthetic division, and the rational root theorem, as shown in our example. Here, initially factoring directly was complicated, so other methods were employed. We discovered that two of the roots were \(x=1\) and \(x=-2\) through the rational root theorem and synthetic division. Once some roots are found, the polynomial can be factorized further into linear factors to find other roots, if any, using basic arithmetic.

• The sum of the roots taken one at a time is \(-b/a\).
• The sum of the product of the roots taken two at a time is \(c/a\).
• The product of the roots is \(-d/a\).

This understanding helps tackle polynomial problems systematically and comprehensively.