Problem 32 - College Algebra with Corequisite Support

Question

For the following exercises, use the Rational Zero Theorem to find the real solution(s) to each equation. $$ 2 x^{3}-3 x^{2}+4 x+3=0 $$

Step-by-Step Solution

Answer

The real solution is \( x = \frac{1}{2} \). The quadratic has no real solutions.

Step 1: Identify Possible Rational Zeros

According to the Rational Zero Theorem, any rational solution of a polynomial equation \( a_nx^n + a_{n-1}x^{n-1} + ... + a_0 = 0 \) must be of 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 \( a_n \). For the equation \( 2x^3 - 3x^2 + 4x + 3 = 0 \), the constant term is 3, and the leading coefficient is 2. The factors of 3 are \( \pm 1, \pm 3 \), and the factors of 2 are \( \pm 1, \pm 2 \). Possible rational zeros are \( \pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2} \).

Step 2: Test Possible Zeros

Substitute each possible rational zero into the polynomial equation to determine if it is indeed a zero. \( f(x) = 2x^3 - 3x^2 + 4x + 3 \). Start by testing \( x = 1 \):### Calculation\( f(1) = 2(1)^3 - 3(1)^2 + 4(1) + 3 = 2 - 3 + 4 + 3 = 6 \)### ConclusionSince \( f(1) eq 0 \), \( x = 1 \) is not a zero. Continue this process with each possible rational zero.

Step 3: Finding the Correct Zero

Continue testing each potential zero:1. Test \( x = -1 \): \[ f(-1) = 2(-1)^3 - 3(-1)^2 + 4(-1) + 3 = -2 - 3 - 4 + 3 = -6 \] \( f(-1) eq 0 \). Thus, \( x = -1 \) is not a zero.2. Test \( x = \frac{1}{2} \): \[ f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right)^2 + 4\left(\frac{1}{2}\right) + 3 = \frac{2}{8} - \frac{3}{4} + 2 + 3 = 0 \] \( f\left(\frac{1}{2}\right) = 0 \). Thus, \( x = \frac{1}{2} \) is a solution.

Step 4: Polynomial Division

Since \( x = \frac{1}{2} \) is a zero, divide the polynomial by \( 2x - 1 \) using synthetic division to find the remaining quadratic. Performing the division:### Calculation- Coefficients: 2, -3, 4, 3- Use \( \frac{1}{2} \) for synthetic division: - Drop 2 down - Multiply \( 2 \times \frac{1}{2} = 1 \), add to -3 gives -2 - Multiply \( -2 \times \frac{1}{2} = -1 \), add to 4 gives 3 - Multiply \( 3 \times \frac{1}{2} = \frac{3}{2} \), add to 3 gives 0 confirming division success since remainder is 0.- The quotient polynomial is \( 2x^2 - 2x + 3 \).

Step 5: Solving the Quadratic

Find the roots of the quadratic equation \( 2x^2 - 2x + 3 = 0 \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -2 \), and \( c = 3 \).### Calculation\[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \times 2 \times 3}}{2 \times 2}\]\[x = \frac{2 \pm \sqrt{4 - 24}}{4} = \frac{2 \pm \sqrt{-20}}{4} \]Because the discriminant \(-20\) is negative, the quadratic has no real solutions.

Key Concepts

Polynomial Equations

Polynomial equations are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. A general polynomial equation in one variable looks like this:

• The highest power, known as the degree of the polynomial, determines the behavior and complexity of the equation.
• Coefficients and constant terms are real numbers that influence the solution set and nature of the polynomial.
• In our example, the polynomial equation is given as: \[ 2x^3 - 3x^2 + 4x + 3 = 0 \]Having a degree of 3, which means it's a cubic equation.

Understanding polynomial equations is fundamental as they appear in various mathematical problems and applications. Solving them can involve techniques like factoring, graphing, or applying the Rational Zero Theorem to find possible rational solutions.

Synthetic Division

Synthetic division is a simplified form of polynomial division, particularly useful when dividing a polynomial by a linear factor of the form \( x - c \). It's an efficient way to determine if a given number is a zero of the polynomial, which is to say it divides the polynomial without a remainder. The steps involved are:

• Write down the coefficients of the polynomial.
• Use the potential zero as the divisor.
• Perform operations to determine if there's a remainder.

In our case, once \( x = \frac{1}{2} \) was found as a zero of \[ 2x^3 - 3x^2 + 4x + 3 \], synthetic division confirmed it, resulting in a simpler quadratic polynomial: \[ 2x^2 - 2x + 3 \].This method is not only faster but also helps simplify polynomials for easier solving.

Quadratic Formula

The quadratic formula is a reliable tool for solving quadratic equations, which are second-degree polynomials of the form \( ax^2 + bx + c = 0 \). The formula given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Allows you to find solutions by computing specific values:

• \( a \), \( b \), and \( c \) are coefficients from the polynomial.
• The "plus-minus" symbol \( \pm \) indicates there can be two solutions based on the sign, which are called roots of the equation.

For the quadratic \[ 2x^2 - 2x + 3 \], this formula is applied to determine if real-number solutions exist and unveil the nature of these solutions.

Discriminant

The discriminant is a component of the quadratic formula, given by \( b^2 - 4ac \), that provides insights into the nature of the roots of a quadratic equation without actually solving it. Here's why it matters:

• If the discriminant is positive, there are two distinct real roots.
• If it's zero, there is exactly one real root, meaning the quadratic "touches" the x-axis.
• A negative discriminant, as seen in \( 2x^2 - 2x + 3 \) that had a discriminant of \(-20\), means no real roots exist, only complex ones are possible.

Understanding the discriminant helps anticipate the kinds of solutions a quadratic might have and shapes how we approach solving the equation.