Problem 18 - College Algebra

Question

a. List all possible rational roots. b. Use synthetic division to test the possible rational roots and find an actual root. c. Use the root from part (b) and solve the equation. $$ x^{3}-5 x^{2}+17 x-13=0 $$

Step-by-Step Solution

Answer

The roots of the cubic equation \(x^{3}-5x^{2}+17x-13\)=0 are \(x = 1\), \(x= 2 + 3i\) and \(x= 2 - 3i\).

Step 1: List All Possible Rational Roots

Use the Rational Root Theorem which states that the possible rational roots are given by the factors of the constant term divided by the factors of the leading coefficient. Here, the constant term is -13 and the leading coefficient is 1. The factors of -13 are \(\pm1, \pm13\) and the factors of 1 are \(\pm1\). So, the possible rational roots are \(\pm1, \pm13\).

Step 2: Use Synthetic Division to Find an Actual Root

Set up a synthetic division for each possible root and perform the synthetic division. If the remainder is zero, then the tested root is a root of the polynomial. Start by testing \(\pm1\) and \(\pm13\). Try 1 first, you find that the remainder is 0, which means x=1 is a root of the equation.

Step 3: Solve the Equation

Now that we have one root \(x=1\), we need to find the remaining roots. So, by using synthetic division, we can rewrite the given equation as \((x-1)(x^2-4x+13)=0\). Now, solve \(x^2 - 4x + 13 = 0\). To solve the quadratic equation, use the quadratic formula \(x = [-b \pm sqrt(b^2 - 4ac)] / (2a)\). Substituting the constants a=1, b=-4 and c=13, you get \(x = 2 \pm sqrt(4 -52)\). Since the discriminant is negative and square root of a negative number is an imaginary number. Therefore, the roots are \(x = 2 \pm 3i\).

Step 4: Summary of the Roots

So, the roots of the equation \(x^{3}-5x^{2}+17x-13\)=0 are real root \(x = 1\) and complex roots \(x= 2 \pm 3i\).

Key Concepts

Synthetic Division

Synthetic division is a handy method for dividing polynomials, especially when you want to quickly determine whether a given value is a root of a polynomial equation. By simplifying the division process to focus only on coefficients, synthetic division saves time and avoids messy algebra.

To use synthetic division, you start by writing down the coefficients of the polynomial you want to divide. For example, in the polynomial equation \(x^{3} - 5x^{2} + 17x - 13 = 0\), the coefficients are \([1, -5, 17, -13]\). Next, you select a potential root to test. In our exercise, we test \(x = 1\). Place 1 to the left of the coefficients.

• Bring down the first coefficient (1) as it is.
• Multiply the root you're testing (1) by the number just brought down and write the result under the next coefficient. Then, add this result to the next coefficient (-5), giving a new total (-4).
• Repeat this process for each coefficient.
• Continue until you reach the end, and find the remainder.

If the remainder is 0, you've found a root. For \(x = 1\), the remainder is 0, confirming that 1 is a real root of the polynomial. This successful division means the polynomial can be rewritten as \((x - 1)(x^2 - 4x + 13)\). This new form is easier to handle when searching for additional roots.

Polynomial Equations

Polynomial equations like \(x^{3} - 5x^{2} + 17x - 13 = 0\) involve terms with varying powers of the variable \(x\). The highest power indicates the degree of the polynomial. In our case, it is a cubic equation since the highest power is 3.

Solving these equations means finding the values of \(x\) that make the equation true. Depending on the degree, polynomial equations can have multiple solutions, which include both real and complex numbers.

• A cubic polynomial equation, like ours, can have up to three roots.
• The roots can be real or complex, and some could be repeated or unique.

Using the Rational Root Theorem, you can initially find potential rational roots by dividing factors of the constant term by factors of the leading coefficient. Each discovered root reduces the degree of the polynomial, making subsequent solutions simpler. As we found in our calculation, after identifying \(x = 1\) as a real root, we reduced our polynomial to \((x - 1)(x^2 - 4x + 13)\), which led us to solve for the remaining roots.

Complex Roots

Complex roots arise in polynomial equations when the equation cannot be solved entirely using real numbers. These roots appear in conjugate pairs for polynomials with real coefficients, meaning if \(a + bi\) is a root, then \(a - bi\) must also be a root. This is due to the requirement that the coefficients remain real.

In our exercise, after using synthetic division to eliminate the real root, we were left with a quadratic equation: \(x^2 - 4x + 13 = 0\). Solving this with the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) revealed that the discriminant \(b^2 - 4ac\) was negative, specifically \(16 - 52 = -36\).

• A negative discriminant indicates the presence of complex roots.

When calculating the square root of a negative number, you incorporate the imaginary unit \(i\), where \(i = \sqrt{-1}\). Therefore, our solutions for the quadratic become \(x = 2 \pm 3i\), making \(x = 2 + 3i\) and \(x = 2 - 3i\) the complex roots of the polynomial. These complex roots complete the set of solutions for this polynomial equation together with the earlier found real root, \(x = 1\).