Problem 19 - College Algebra

Question

For the following exercises, use the Factor Theorem to find all real zeros for the given polynomial function and one factor. $$ x^{3}+3 x^{2}+4 x+12 ; x+3 $$

Step-by-Step Solution

Answer

The real zero is \(-3\).

Step 1: Apply the Factor Theorem

To find the zeros of the polynomial, we apply the Factor Theorem. It states that if \(x-c\) is a factor of a polynomial \(f(x)\), then \(f(c) = 0\). We are given \(x+3\) as a factor, which implies \(x = -3\) is a zero. We will verify that \(f(-3) = 0\).

Step 2: Verify the given zero

Evaluate the polynomial \(f(x) = x^3 + 3x^2 + 4x + 12\) at \(x = -3\): \[ f(-3) = (-3)^3 + 3(-3)^2 + 4(-3) + 12 = -27 + 27 - 12 + 12 = 0. \] Since \(f(-3) = 0\), \(x = -3\) is indeed a zero of the polynomial.

Step 3: Perform Synthetic Division

Use synthetic division to divide the polynomial by \(x + 3\). Set up \(-3\) outside the division box, and place the coefficients \(1, 3, 4, 12\) inside.Perform the division:1. Bring down the leading coefficient \(1\).2. Multiply by \(-3\) and add to the next coefficient: \(3 \rightarrow (3 + (-3 \times 1) = 0)\).3. Repeat the process for the next coefficients: \(4 \rightarrow (4 + (-3 \times 0) = 4)\) and \(12 \rightarrow (12 + (-3 \times 4) = 0)\).The resulting quotient is \(x^2 + 0x + 4\).

Step 4: Solve the Quotient Polynomial

The quotient polynomial from synthetic division is \(x^2 + 4\). To find its zeros, set it equal to zero: \(x^2 + 4 = 0\). Solve for \(x\):\[ x^2 = -4 \]\[ x = \pm \sqrt{-4} = \pm 2i \]The zeros of \(x^2 + 4\) are \(2i\) and \(-2i\), which are complex, not real.

Key Concepts

Polynomial Zeros

Polynomial zeros are values of the variable that make the polynomial equal to zero. For instance, in a polynomial function \(f(x)\), the values \(x = c\) that satisfy \(f(c) = 0\) are called zeros. Zeros can be real or complex numbers.

Finding zeros is crucial because they represent the points where the graph of the polynomial crosses or touches the x-axis. To determine the zeros:

• Express the polynomial in a factorable format.
• Use methods such as factoring, synthetic division, or the quadratic formula to solve.
• Check if the solutions are real or complex.

In the given exercise, applying the Factor Theorem helped identify \(x = -3\) as a zero because \(x + 3\) is a factor. The polynomial \(x^3 + 3x^2 + 4x + 12\) evaluated at \(x = -3\) equals zero, confirming \(-3\) as a zero.

Synthetic Division

Synthetic division is a simpler alternative to the traditional long division method for polynomials, especially useful when dividing by a linear factor. It's an effective process for both finding zeros and simplifying polynomials. Here's how it works:

• The divisor must be in the form \(x - c\).
• Write down only the coefficients of the polynomial.
• Bring down the leading coefficient as is.
• Multiply by \(c\) (the zero obtained from the factor) and add to the next coefficient, repeating this for all terms.

Using synthetic division with the original polynomial \(x^3 + 3x^2 + 4x + 12\) and the factor \(x + 3\), we determined the quotient polynomial \(x^2 + 4\). This quotient is vital for further analysis of the original polynomial's behavior and additional zeros.

Complex Numbers

Complex numbers are numbers that include a real and an imaginary component. The imaginary unit \(i\), defined as \(i = \sqrt{-1}\), allows for the square root of negative values, which are not possible within the real numbers. A complex number is typically written in the form \(a + bi\), where \(a\) and \(b\) are real numbers.

Complex numbers often appear as zeros of polynomial equations that cannot be factored over the real numbers. For example, the polynomial \(x^2 + 4\) does not intersect the x-axis in the real coordinate plane, but its zeros are \(2i\) and \(-2i\), purely imaginary numbers. This occurs because setting \(x^2 + 4 = 0\) leads to \(x^2 = -4\), whose solutions are imaginary zeros \(x = \pm \sqrt{-4} = \pm 2i\). Using these concepts allows students to broaden their understanding and solve more complex polynomials efficiently.