Problem 23 - College Algebra with Corequisite Support

Question

For the following exercises, use the Rational Zero Theorem to find the real solution(s) to each equation. $$ 2 x^{3}+7 x^{2}-10 x-24=0 $$

Step-by-Step Solution

Answer

The real solutions are \(x = 2\), \(x = -1.5\), and \(x = -4\).

Step 1: Identify Possible Rational Zeros

According to the Rational Zero Theorem, any potential rational zero of the polynomial equation \( 2x^3 + 7x^2 - 10x - 24 = 0 \) is a factor of the constant term divided by a factor of the leading coefficient. The constant term here is \(-24\) and the leading coefficient is \(2\). The factors of \(-24\) are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24 \), and the factors of \(2\) are \( \pm 1, \pm 2 \).Thus, the possible rational zeros are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{5}{2}, \pm \frac{9}{2}, \pm \frac{7}{2} \).

Step 2: Test Possible Zeros Using Synthetic Division

We will use synthetic division to test these possible zeros until we find one that results in a remainder of \(0\).Let's test \(-2\):Set up the synthetic division with \(-2\):- Write the coefficients: \(2, 7, -10, -24\).- Bring down the leading coefficient \(2\).- Multiply and add successively: - \(-2 \times 2 = -4\); \, adjust 7 to 3. - \(-2 \times 3 = -6\); \, adjust -10 to -16. - \(-2 \times -16 = 32\); \, adjust -24 to 8.Since the remainder is \(8\), try another potential zero.

Step 3: Try Different Possible Zeros

Continuing testing other zeros using synthetic division:Now testing \(3\):- Write the coefficients: \(2, 7, -10, -24\).- Bring down the leading coefficient \(2\).- Multiply and add: - \(3 \times 2 = 6\); \, adjust 7 to 13. - \(3 \times 13 = 39\); \, adjust -10 to 29. - \(3 \times 29 = 87\); \, adjust -24 to 63.Since the remainder is \(63\), move to another zero.

Step 4: Confirm with \\( x = -3 \\)

Test \(x = -3\):- Write coefficients: \(2, 7, -10, -24\).- Bring down \(2\).- Multiply: \(-3 \times 2 = -6\), thus \(7 - 6 = 1\).- Multiply: \(-3 \times 1 = -3\), thus \(-10 - 3 = -13\).- Multiply: \(-3 \times (-13) = 39\), thus \(-24 + 39 = 15\).The remainder when evaluating \(x = -3\) is not zero. Check other factors.

Step 5: Discover Zero \\( x = 2 \\)

Try \(x = 2\):- Write coefficients: \(2, 7, -10, -24\).- Bring down \(2\).- Multiply: \(2 \times 2 = 4\), adjust \(7 + 4 = 11\).- Multiply: \(2 \times 11 = 22\), adjust \(-10 + 22 = 12\).- Multiply: \(2 \times 12 = 24\), finally adjusting \(-24 + 24 = 0\).The remainder is \(0\), thus \(x = 2\) is a zero.

Step 6: Factor and Solve Quadratic

With \(x = 2\) confirmed as a zero, the polynomial factors to \( (x - 2) \cdot (2x^2 + 11x + 12) = 0 \).Now solve the quadratic equation \(2x^2 + 11x + 12 = 0\) using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 2 \), \( b = 11 \), and \( c = 12 \).Calculate the discriminant: \( b^2 - 4ac = 11^2 - 4 \cdot 2 \cdot 12 = 121 - 96 = 25 \).Compute the roots:\[ x = \frac{-11 \pm \sqrt{25}}{4} = \frac{-11 \pm 5}{4} \].This results in two solutions: \(x = -1.5\) and \(x = -4\).

Step 7: Listing the Real Solutions

The factors of the polynomial give the solutions for the equation. Thus, our solutions are the real zeros \(x = 2\), \(x = -1.5\), and \(x = -4\).

Key Concepts

Polynomial Equation

A polynomial equation is an equation that involves a polynomial expression. Polynomials may have one or multiple terms combined by addition, subtraction, or multiplication, and can include various powers of its variables. A polynomial is expressed in the form: \( a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0 = 0 \), where \( n \) is a non-negative integer. In the problem exercise, the polynomial equation given is \( 2x^3 + 7x^2 - 10x - 24 = 0 \). Here, the highest exponent is 3, making it a cubic equation. Solving polynomial equations typically involves finding the values of \( x \) that make the equation true, and these values are known as the roots or zeros of the polynomial.

Synthetic Division

Synthetic division is a simplified method of dividing a polynomial by a divisor of the form \( x - c \), avoiding the more cumbersome polynomial long division. It is particularly useful in testing potential rational zeros identified by the Rational Zero Theorem. The process involves:

• Writing down the coefficients of the polynomial.
• Bringing down the first coefficient as it is.
• Multiplying this coefficient by the potential zero, and adding the result to the next coefficient.
• Repeating these steps until reaching the last term.

In our exercise, synthetic division helped to find that \( x = 2 \) is a root, given that it resulted in a remainder of zero when divided.

Quadratic Formula

Once a polynomial is reduced in degree by finding one or more of its linear roots, what remains may be a quadratic equation. The quadratic formula provides a quick solution for these equations and is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a \), \( b \), and \( c \) are coefficients of the quadratic equation \( ax^2 + bx + c = 0 \). In the problem, after factoring out \( x = 2 \), the remaining polynomial \( 2x^2 + 11x + 12 \) was solved using the quadratic formula. With calculated roots \( x = -1.5 \) and \( x = -4 \), these solutions were determined by substituting the values \( a = 2 \), \( b = 11 \), and \( c = 12 \) into the formula.

Real Solutions

Real solutions of polynomial equations are the real number values for which the polynomial equals zero. These solutions can be found through the Rational Zero Theorem, synthetic division, and further simplification techniques like factoring or using the quadratic formula. For the given cubic polynomial equation \( 2x^3 + 7x^2 - 10x - 24 = 0 \), the real solutions are the values that satisfy the equation. In this case, after testing potential zeros, and solving the remaining quadratic, the real solutions determined were \( x = 2 \), \( x = -1.5 \), and \( x = -4 \). These solutions are the values where the graph of the polynomial crosses the x-axis.