Problem 44 - Precalculus with Calculus Previews

Question

Solve the given equation if the indicated number \(c\) is a zero of the function \(f\). $$ \begin{array}{l} x^{4}-3 x^{3}-8 x^{2}-10 x=0 ; f(x)=x^{4}-3 x^{3}-8 x^{2}- 10 x, c=5 \end{array} $$

Step-by-Step Solution

Answer

The solutions are \(x = 0, 5, -1 + i, -1 - i\).

Step 1: Verify that c is a Zero

First, confirm that the given value \(c = 5\) is indeed a zero of the function \(f(x) = x^4 - 3x^3 - 8x^2 - 10x\). Substitute \(x = 5\) into the function and ensure the result is zero: \(f(5) = 5^4 - 3(5)^3 - 8(5)^2 - 10(5) = 625 - 375 - 200 - 50 = 0\). Thus, 5 is a zero.

Step 2: Perform Synthetic Division

Now, use synthetic division to divide the polynomial \(f(x) = x^4 - 3x^3 - 8x^2 - 10x\) by \(x - 5\). Set up the division with 5: \(1, -3, -8, -10, 0\). The remainder should be zero as 5 is a root.

Step 3: Calculate Using Synthetic Division

Write the coefficients \(1, -3, -8, -10, 0\) down. Bring down the 1, multiply it by 5, add to the next coefficient: repeat this process for all coefficients: \[\begin{array}{c|cccc}5 & 1 & -3 & -8 & -10 & 0 \ & & 5 & 10 & 10 & 0 \\hline & 1 & 2 & 2 & 0 & 0 \\end{array}\] The result is \(x^3 + 2x^2 + 2x\) with a remainder of 0.

Step 4: Solve the Reduced Polynomial

The resulting polynomial is \(x^3 + 2x^2 + 2x = 0\). Factor out the common term \(x\): \(x(x^2 + 2x + 2) = 0\). Solve the factor \(x = 0\), and then solve the quadratic \(x^2 + 2x + 2 = 0\).

Step 5: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = 1\), \(b = 2\), \(c = 2\). Calculate the discriminant: \(b^2 - 4ac = 4 - 8 = -4\). Thus, the solutions are complex: \(x = \frac{-2 \pm 2i}{2}\), simplifying to \(x = -1 \pm i\).

Step 6: Write All Solutions

Thus, the solutions of the equation are \(x = 0\), \(x = 5\), \(x = -1 + i\), and \(x = -1 - i\).

Key Concepts

Synthetic Division

Synthetic division is a simplified form of polynomial division, similar to long division but with a streamlined process. It's especially useful when you want to divide a polynomial by a linear factor of the form \(x - c\). In this case, we confirm that \(c = 5\) is a zero of the polynomial \(f(x)\). This means when divided by \(x - 5\), the remainder should be zero.

• Firstly, list the coefficients of the polynomial \(f(x) = x^4 - 3x^3 - 8x^2 - 10x\) as \(1, -3, -8, -10, 0\).
• Place \(5\) on the left side of the synthetic division setup. The first coefficient, which is \(1\), is brought down unchanged.
• Next, multiply this coefficient by \(5\) and add it to the next coefficient.
• Continue this process of multiplying and adding across all coefficients.
• The row below the division line gives you the reduced polynomial.

In the example, synthetic division yielded \(x^3 + 2x^2 + 2x = 0\), confirming the zero and allowing us to solve for remaining roots.

Quadratic Formula

The quadratic formula is a powerful tool for solving any quadratic equation of the form \(ax^2 + bx + c = 0\). It's useful when factoring isn't straightforward. The formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our exercise, the quadratic that needs to be solved is \(x^2 + 2x + 2 = 0\). Here, \(a = 1\), \(b = 2\), and \(c = 2\).

• Calculate the discriminant: \(b^2 - 4ac\). For this equation, that's \(4 - 8 = -4\).
• The negative discriminant indicates complex roots, so expect solutions involving imaginary numbers.
• Plug the values into the quadratic formula to find the roots.

Ultimately, the solutions are \(x = -1 \pm i\), showing how the quadratic formula helps solve equations even when roots aren't real.

Complex Roots

Complex roots arise when solving equations where the discriminant is negative, such as in the quadratic formula. This gives us solutions that aren't on the real number line, but rather in the complex plane.

• Complex numbers are of the form \(a + bi\), where \(i\) is the imaginary unit, satisfying \(i^2 = -1\).
• In our example, the solutions \(x = -1 \pm i\) arise due to a negative discriminant of \(-4\).

These solutions are not just academic but have real-world applications in fields like engineering and physics, where systems sometimes need to consider dimensions beyond just the real numbers. Complex roots provide a way to handle these complexities.