Problem 13 - A Graphical Approach to College Algebra

Question

For each polynomial, one or more zeros are given. Find all remaining zeros. \(P(x)=x^{3}-x^{2}-4 x-6 ; \quad 3\) is a zero.

Step-by-Step Solution

Answer

The zeros are 3, \(-1 + i\), and \(-1 - i\).

Step 1: Use the Given Zero

It is given that 3 is a zero of the polynomial \(P(x) = x^3 - x^2 - 4x - 6\). This means that \(P(3) = 0\). We will use synthetic division with the zero 3 to help factor the polynomial.

Step 2: Perform Synthetic Division

Set up the synthetic division using 3 as the zero on the left and the coefficients of the polynomial \([1, -1, -4, -6]\) on the top row. 1. Bring down the first coefficient 1. 2. Multiply 3 by 1 (the number just brought down), giving 3, and write it under the next coefficient. 3. Add -1 and 3, resulting in 2. 4. Multiply 3 by 2, giving 6, and write it under the next coefficient. 5. Add -4 and 6, resulting in 2. 6. Multiply 3 by 2, giving 6, and write it under -6. 7. Add -6 and 6 to get 0. The remainder is 0, confirming that 3 is indeed a zero, and the reduced polynomial (quotient) is \(x^2 + 2x + 2\).

Step 3: Find Remaining Zeros from Quotient

Now solve the quadratic equation \(x^2 + 2x + 2 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 2\), and \(c = 2\).

Step 4: Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac\):\[2^2 - 4(1)(2) = 4 - 8 = -4\]The discriminant is -4, which indicates two complex zeros.

Step 5: Solve for Complex Zeros

Use the quadratic formula to find the complex zeros:\[x = \frac{-2 \pm \sqrt{-4}}{2(1)} = \frac{-2 \pm 2i}{2} = -1 \pm i\]Thus, the other two zeros are \(-1 + i\) and \(-1 - i\).

Key Concepts

Synthetic Division

Did you know that synthetic division is a quick way to divide a polynomial by a linear factor of the form \(x - c\)? It’s super efficient because you only need the coefficients of the polynomial and the zero of the linear factor. Let’s see how this works with our example polynomial \(P(x) = x^3 - x^2 - 4x - 6\).

We already have a known zero, \(3\), meaning \(P(3) = 0\). To divide the polynomial by \(x - 3\), we follow these steps:

• Start with the coefficients: \([1, -1, -4, -6]\).
• Place \(3\) to the left (this is our divisor).
• Bring down the first coefficient, \(1\).
• Multiply \(3\) by the number just brought down, \(1\), and write the product item below the next coefficient.
• Add -1 and 3, then repeat this process for all coefficients.

After this neat side-by-side arithmetic, you end up with \(0\) as a remainder, confirming \(3\) is indeed a zero, leaving us with a simpler quadratic: \(x^2 + 2x + 2\). This new polynomial we get is called the quotient.

Quadratic Formula

Faced with a quadratic equation like \(x^2 + 2x + 2 = 0\), you can use the quadratic formula to quickly find the zeros. This formula is super handy, especially when the quadratic equation does not easily factor. The quadratic formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In our problem, the coefficients are \(a = 1\), \(b = 2\), and \(c = 2\).

Before plugging into the formula, calculate the discriminant \(b^2 - 4ac\):

• \(2^2 = 4\)
• \(4(1)(2) = 8\)
• So, \(4 - 8 = -4\)

The discriminant tells us if the roots will be real or complex. A negative value indicates that the zeros are complex. Learn to embrace the quadratic formula; it reveals so much about the nature of solutions!

Complex Zeros

What happens when the discriminant is negative, like our \(-4\)? This means we deal with complex numbers. Complex zeros arise when dealing with numbers that include the imaginary unit \(i\), defined as \(\sqrt{-1}\). So, when resolving \[x = \frac{-2 \pm \sqrt{-4}}{2}\],it begins by evaluating \(\sqrt{-4}\) as \(2i\).

Let's continue:

• \(x = \frac{-2 \pm 2i}{2}\)
• Breaking it into parts gives \(-1 \pm i\)

These results, \(-1 + i\) and \(-1 - i\), are our complex zeros. They often appear in conjugate pairs when working with polynomials with real coefficients. Mastering complex zeros unlocks understanding of more sophisticated polynomial solutions!