Problem 29 - College Algebra with Corequisite Support

Question

For the following exercises, use the Rational Zero Theorem to find the real solution(s) to each equation. $$ 2 x^{3}-3 x^{2}-x+1=0 $$

Step-by-Step Solution

Answer

The real solutions are \(x = \frac{1}{2}\), \(x = \frac{1 + \sqrt{5}}{2}\), \(x = \frac{1 - \sqrt{5}}{2}\).

Step 1: Identifying Possible Rational Zeros

According to the Rational Zero Theorem, the possible rational solutions of the polynomial equation \(2x^3 - 3x^2 - x + 1 = 0\) can be found by taking the factors of the constant term (1) over the factors of the leading coefficient (2). The factors of 1 are \(\pm 1\), and the factors of 2 are \(\pm 1, \pm 2\). Therefore, the possible rational zeros are \(\pm 1, \pm \frac{1}{2}, \pm 2\).

Step 2: Testing Possible Zeros with Synthetic Division

Test each possible zero by substituting into the polynomial or using synthetic division. Start with \(x = 1\):Using synthetic division with \(x = 1\):```1 | 2 -3 -1 1 | 2 -1 -2 ---------------- 2 -1 -2 -1```The remainder is -1, so \(x = 1\) is not a zero.Repeat the process for \(x = -1\), \(x = \frac{1}{2}\), and so on until a remainder of 0 is found.

Step 3: Finding a Rational Solution

Continue with \(x = -1\):Using synthetic division:```-1 | 2 -3 -1 1 | -2 5 -6 ---------------- 2 -5 4 -5```The remainder is -5, so \(x = -1\) is not a zero.Try \(x = \frac{1}{2}\):```\frac{1}{2} | 2 -3 -1 1 | | 1 -1 -1 ------------------- 2 -2 -2 0```The remainder is 0, therefore \(x = \frac{1}{2}\) is a zero.

Step 4: Reducing the Polynomial

With \(x = \frac{1}{2}\) confirmed as a zero, the polynomial is reduced by synthetic division yielding the quotient polynomial:\(2x^3 - 3x^2 - x + 1 = (x - \frac{1}{2})(2x^2 - 2x - 2)\).

Step 5: Factoring the Quotient Polynomial

Next, factor \(2x^2 - 2x - 2\). Begin by taking out the common factor:\[2(x^2 - x - 1)\]Now, use the quadratic formula to solve \(x^2 - x - 1 = 0\): \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{5}}{2}\]

Step 6: Identifying All Real Solutions

The real solutions of the original equation are \(x = \frac{1}{2}\), \(x = \frac{1 + \sqrt{5}}{2}\), and \(x = \frac{1 - \sqrt{5}}{2}\).

Key Concepts

Rational Zeros

The Rational Zero Theorem is a vital tool in algebra for finding the possible rational zeros of a polynomial. It limits our potential rational solutions to a manageable list. This theorem states that if a polynomial has rational zeros, they can be expressed as the ratio of the factors of the constant term to the factors of the leading coefficient. For example, in the equation \(2x^3 - 3x^2 - x + 1 = 0\), the constant term is 1 and the leading coefficient is 2.

• The factors of 1 are \(\pm 1\).
• The factors of 2 are \(\pm 1\) and \(\pm 2\).

This means the possible rational zeros are \(\pm 1, \pm \frac{1}{2}, \pm 2\). These values represent the starting points to find actual zeros using further testing.

Synthetic Division

Synthetic Division is a streamlined way to divide polynomials, especially useful for testing the potential zeros provided by the Rational Zero Theorem. It simplifies calculations by focusing on the coefficients alone and is less cumbersome than traditional long division.

• Write down the coefficients of the polynomial.
• Use the potential zero on the left to 'divide'.
• The final number in the bottom row is the remainder.

For example, testing \(x = \frac{1}{2}\) on \(2x^3 - 3x^2 - x + 1\) using Synthetic Division gives a remainder of 0. This confirms \(x = \frac{1}{2}\) is indeed a zero. The result also simplifies the polynomial, making it easier to extract additional roots.

Factoring Polynomials

Once a zero is found, the polynomial can be reduced to a simpler form, often revealing more zeros. Factoring involves identifying common factors and using methods like grouping or special formulae to simplify expressions.In our example, discovering \(x = \frac{1}{2}\) led to reducing the polynomial to \(2x^2 - 2x - 2\). Factoring out the greatest common factor yields:\[2(x^2 - x - 1)\]From here, other techniques, like the Quadratic Formula, may be employed to find any remaining roots.

Quadratic Formula

The Quadratic Formula is a powerful tool for finding the zeros of quadratic polynomials of the form \(ax^2 + bx + c = 0\). It provides the zeros directly by plugging coefficients into the formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For the polynomial \(x^2 - x - 1 = 0\) resulting from the process of factoring, the values are calculated as:

• \(a = 1\)
• \(b = -1\)
• \(c = -1\)

Substituting these into the Quadratic Formula, we derive:\[x = \frac{1 \pm \sqrt{5}}{2}\]Thus, the real solutions of the original equation include \(x = \frac{1}{2}\), \(x = \frac{1 + \sqrt{5}}{2}\), and \(x = \frac{1 - \sqrt{5}}{2}\), offering a comprehensive understanding of this process.