Problem 64 - A Graphical Approach to College Algebra

Question

For each polynomial, at least one zero is given. Find all others analytically. $$P(x)=2 x^{3}+8 x^{2}-11 x-5 ; \quad-5$$

Step-by-Step Solution

Answer

The zeros are \(-5\), \(\frac{1 + \sqrt{3}}{2}\), and \(\frac{1 - \sqrt{3}}{2}\).

Step 1: Understand the Given Zero

The problem states that one of the zeros of the polynomial is \(-5\). This means that \(P(-5) = 0\). We will use this fact to simplify the polynomial and find other zeros.

Step 2: Perform Polynomial Division

To factor the polynomial, we divide \(P(x)\) by \(x + 5\), because \(-5\) is a zero. Perform synthetic or long division to divide \(2x^3 + 8x^2 - 11x - 5\) by \(x + 5\) and find the quotient.

Step 3: Complete Polynomial Division

By performing synthetic division, we obtain the quotient \(2x^2 - 2x - 1\). This tells us that:\[ P(x) = (x + 5)(2x^2 - 2x - 1) \]

Step 4: Solve the Quadratic Equation

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

Step 5: Apply the Quadratic Formula

Substitute the values into the quadratic formula:\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \times 2 \times (-1)}}{2 \times 2} \]\[ x = \frac{2 \pm \sqrt{4 + 8}}{4} \]\[ x = \frac{2 \pm \sqrt{12}}{4} \]\[ x = \frac{2 \pm 2\sqrt{3}}{4} \]\[ x = \frac{1 \pm \sqrt{3}}{2} \]

Step 6: Conclusion

The quadratic \(2x^2 - 2x - 1\) has zeros at \(x = \frac{1 + \sqrt{3}}{2}\) and \(x = \frac{1 - \sqrt{3}}{2}\). Thus, the complete set of zeros for the polynomial \(P(x)\) is \(-5\), \(\frac{1 + \sqrt{3}}{2}\), and \(\frac{1 - \sqrt{3}}{2}\).

Key Concepts

Polynomial Division

Polynomial division is a method used to divide polynomials, similar to how long division is applied in arithmetic. It helps simplify complex polynomial expressions by breaking them down into more manageable parts. In the context of finding polynomial zeros, if you know one zero, you can use it to divide the polynomial and reduce its degree. This makes finding other zeros easier.

In the problem provided, we know that \(-5\) is a zero of the polynomial. This means \(x + 5\) is a factor of the polynomial. To find other zeros, we perform polynomial division on \(P(x) = 2x^3 + 8x^2 - 11x - 5\) by \(x + 5\).

There are two main methods for polynomial division:

• **Long Division**: This method involves dividing the polynomials similar to long division of numbers, by repeatedly dividing, multiplying, and subtracting.
• **Synthetic Division**: A shortcut applicable when dividing by linear factors like \(x + a\). It's efficient and reduces the complexity, providing the quotient quickly.

Once the division completes, you get a quotient and sometimes a remainder. In our problem, the quotient is \(2x^2 - 2x - 1\), and the remainder is zero, confirming \(x + 5\) as a factor.

Quadratic Equation

A quadratic equation is a second-degree polynomial of the form \(ax^2 + bx + c = 0\). Solving a quadratic equation is a critical step in finding the zeros of polynomials, especially when you have reduced a higher-degree polynomial using division.

After performing polynomial division in the problem, the polynomial is factored into \(P(x) = (x + 5)(2x^2 - 2x - 1)\). Here, \(2x^2 - 2x - 1\) is the quadratic equation that needs solving to find other zeros.

Quadratic equations can be solved using several methods:

• **Factoring**: If the equation can be easily broken into simpler binomials, this is often the quickest method.
• **Completing the Square**: A method where you make the quadratic a perfect square trinomial.
• **Quadratic Formula**: A universal method that works for any quadratic equation.

In our example, the quadratic equation \(2x^2 - 2x - 1 = 0\) isn’t easily factorable by inspection, so we use the quadratic formula.

Quadratic Formula

The quadratic formula is a powerful tool in algebra for solving any quadratic equation of the form \(ax^2 + bx + c = 0\). It provides the solution directly using the coefficients of the equation. The formula is:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

This formula can solve any quadratic equation and is especially useful when the equation does not easily factor.

For the equation \(2x^2 - 2x - 1 = 0\) from our problem:

• **Identify coefficients**: Here, \(a = 2\), \(b = -2\), and \(c = -1\).
• **Substitute into the formula**: Plug these values into the quadratic formula to find the solutions.
• **Calculate**: You determine the discriminant \(b^2 - 4ac\) to decide on the nature of the roots. Here, the discriminant is positive, suggesting two distinct real roots.

The solutions to the quadratic, and thus the remaining zeros of the original polynomial, are \(x = \frac{1 + \sqrt{3}}{2}\) and \(x = \frac{1 - \sqrt{3}}{2}\). These results, together with the given zero \(-5\), provide the complete set of zeros.