Problem 41 - College Algebra with Corequisite Support

Question

For the following exercises, find all complex solutions (real and non-real). $$ x^{3}-8 x^{2}+25 x-26=0 $$

Step-by-Step Solution

Answer

The complex solutions are x = 2, x = 3 + 2i, and x = 3 - 2i.

Step 1: Identify Possible Rational Roots

Use the Rational Root Theorem to list possible rational roots. The theorem suggests that any rational root, when expressed in its simplest form, should be a factor of the constant term divided by a factor of the leading coefficient. Here, the constant term is -26 and the leading coefficient is 1, so the possible rational roots are ±1, ±2, ±13, ±26.

Step 2: Test Possible Roots Using Synthetic Division

Apply synthetic division to test the potential roots listed in Step 1. Start with x = 1 and check if it results in a remainder of 0: For x = 1: - 1 | 1 -8 25 -26 - | 1 -7 18 ------------------ 1 -7 18 -8 The remainder is -8, not 0, so x = 1 is not a root. Continue testing with x = 2. For x = 2: - 2 | 1 -8 25 -26 - | 2 -12 26 ------------------ 1 -6 13 0 Remainder is 0, thus x = 2 is a root of the polynomial.

Step 3: Factor the Polynomial

We have determined x = 2 is a root. Thus, (x - 2) is a factor. The quotient obtained from synthetic division in Step 2 is x^2 - 6x + 13. Thus, we can express the polynomial as (x - 2)(x^2 - 6x + 13).

Step 4: Find Complex Roots of the Quadratic

Now solve the quadratic equation x^2 - 6x + 13 = 0 using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, a = 1, b = -6, and c = 13. Substitute these values into the formula:\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \times 1 \times 13}}{2 \times 1} \]\[ x = \frac{6 \pm \sqrt{36 - 52}}{2} \]\[ x = \frac{6 \pm \sqrt{-16}}{2} \]\[ x = \frac{6 \pm 4i}{2} \]\[ x = 3 \pm 2i \]Thus, the solutions to the equation are x = 3 + 2i and x = 3 - 2i.

Step 5: State All Solutions

The complete solution set of the equation consists of the real root found in Step 2 and the complex roots found in Step 4. Therefore, the solutions are x = 2, x = 3 + 2i, and x = 3 - 2i.

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a handy tool used in algebra to find potential rational solutions for polynomial equations. It states that any potential rational root, in its simplest form, is a factor of the constant term divided by a factor of the leading coefficient. In our example, the polynomial is \(x^3 - 8x^2 + 25x - 26 = 0\).
To apply the theorem here, we first identify the constant term, which is -26, and the leading coefficient, which is 1.

• Factors of -26 are ±1, ±2, ±13, and ±26.
• Factors of 1 are simply ±1.

To find potential rational roots, we divide the factors of the constant term by the factors of the leading coefficient, yielding the possible roots: ±1, ±2, ±13, and ±26. These values are where we start testing to find actual rational roots.

synthetic division

Synthetic division is a simplified method of dividing polynomials, especially when dealing with roots of a polynomial equation. It is particularly useful for testing potential roots identified by the Rational Root Theorem.
In synthetic division, we first arrange the polynomial coefficients. For this polynomial, \(x^3 - 8x^2 + 25x - 26\), the coefficients are 1, -8, 25, and -26.

• We test each candidate from the Rational Root Theorem, starting with x = 1.
• The process involves multiplying and adding coefficients, checking if the remainder is zero.
• If the remainder is zero, that x-value is a root.

After testing with synthetic division, x = 1 is not a root (remainder is -8). When we test x = 2, the remainder is 0, indicating 2 is indeed a root. Therefore, \(x - 2\) is a factor of the polynomial. The quotient from synthetic division gives us a simpler polynomial to further factor or solve.

quadratic formula

Once we have confirmed a rational root with synthetic division, we can reduce the original polynomial and solve the resulting quadratic equation using the quadratic formula. In our case, after finding x = 2 as a root, the quotient left is \(x^2 - 6x + 13\).
The quadratic formula is applicable to equations of the form \(ax^2 + bx + c = 0\), and it allows us to find the roots as follows:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For \(x^2 - 6x + 13 = 0\), the values are: a = 1, b = -6, and c = 13. Substituting these into the formula results in:\[ x = \frac{6 \pm \sqrt{36 - 52}}{2} \]\[ x = \frac{6 \pm \sqrt{-16}}{2} \]\[ x = \frac{6 \pm 4i}{2} \]\[ x = 3 \pm 2i \]Thus, the equation has two complex roots, x = 3 + 2i and x = 3 - 2i.

complex roots

Complex roots arise when the discriminant \((b^2 - 4ac)\) in the quadratic formula is negative. In simpler terms, it's when there’s no real number that can square to produce a negative number. This leads us to imaginary numbers.
In the example of \(x^2 - 6x + 13 = 0\), the discriminant turns out to be -16, a negative number. Consequently, the square root of -16 is not a real number, leading to a pair of complex solutions involving the imaginary unit \(i\), where \(i^2 = -1\).
From our findings using the quadratic formula, the complex roots are:

• x = 3 + 2i
• x = 3 - 2i

Complex roots in quadratic equations typically occur in conjugate pairs. The presence of these pairs indicates that the graph of the quadratic equation wouldn’t intersect with the x-axis. For the full solution, these complex roots complement the real root obtained earlier, giving the complete set of solutions to the polynomial equation.