Problem 27 - Precalculus Mathematics for Calculus

Question

Factor the polynomial completely, and find all its zeros. State the multiplicity of each zero. $$P(x)=x^{3}+x^{2}+9 x+9$$

Step-by-Step Solution

Answer

Zeros: \( x = -1, 3i, -3i \) with multiplicity 1 for each.

Step 1: Identify potential rational roots using the Rational Root Theorem

The Rational Root Theorem suggests that potential rational roots are factors of the constant term 9 divided by factors of the leading coefficient, which is 1. Therefore, potential roots could be \( \pm 1, \pm 3, \pm 9 \).

Step 2: Use synthetic division to find a root

Perform synthetic division with potential rational roots to find a root that results in a remainder of zero. Attempt division starting with \( x = -1 \), which successfully divides the polynomial \( P(x) \) and leaves a remainder of 0, confirming \( x = -1 \) is a root.

Step 3: Write down the reduced polynomial

After performing synthetic division using \( x = -1 \), the polynomial reduces to \( x^2 + 9 \). So, \( P(x) = (x + 1)(x^2 + 9) \).

Step 4: Factor the quadratic polynomial

Identify that \( x^2 + 9 \) is a sum of squares and factor it using complex numbers: \( x^2 + 9 = (x + 3i)(x - 3i) \).

Step 5: Write the complete factored form of the polynomial

Combine all factors to get \( P(x) = (x + 1)(x + 3i)(x - 3i) \). This is the fully factored form of the polynomial.

Step 6: Identify and state all zeros and their multiplicities

The zeros of the polynomial are \( x = -1, x = 3i, x = -3i \) with multiplicity 1 for each root as they appear only once in the factorization.

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a powerful tool for finding simple roots of a polynomial without resorting to guessing and checking. This theorem states that any potential *rational* root of a polynomial is a fraction \( \frac{p}{q} \), where \( p \) is a factor of the constant term, and \( q \) is a factor of the leading coefficient. In the case of the polynomial \( P(x) = x^3 + x^2 + 9x + 9 \), our constant term is 9 and the leading coefficient is 1.
Hence, the set of potential rational roots can be determined by:

• Factors of the constant term: \( \pm 1, \pm 3, \pm 9 \)
• Factors of the leading coefficient: \( \pm 1 \)

Since the leading coefficient is 1, our potential rational roots are simply the factors of 9, which are \( \pm 1, \pm 3, \pm 9 \). To determine actual roots, we can substitute these values into the polynomial or use synthetic division to check if they yield a remainder of zero, signifying a successful root.

Complex Roots

Complex roots occur in pairs and are found when a polynomial cannot be factored completely using real numbers. This happens often when the polynomial includes a sum of squares, like \(x^2 + 9\). Real numbers can't express the square roots of negative numbers, so we introduce the imaginary unit \(i\), where \(i^2 = -1\).
Through this concept, we can express the root of \(x^2 + 9 = 0\) as:

• Set \(x^2 = -9\)
• Take the square root of both sides: \(x = \pm 3i \)

In the problem above, the quadratic term \(x^2 + 9\) produces the complex roots \(x = 3i\) and \(x = -3i\). It's important to always use complex conjugates like these, ensuring that even though they aren’t visible on the real number line, they exist to maintain the algebraic integrity of the factorization.

Synthetic Division

Synthetic division is a simplified method for dividing a polynomial by a linear factor of the form \(x - a\). This process is much quicker than long division and is particularly useful in determining if a candidate from the Rational Root Theorem is indeed a root. Let's break down this process with an example using the polynomial \(P(x) = x^3 + x^2 + 9x + 9\) and the candidate root \(x = -1\):

• Write down the coefficients of the polynomial: 1, 1, 9, 9
• Place the candidate \(-1\) to the left
• Bring the lead coefficient down unchanged
• Multiply this lead coefficient by the candidate and add the result to the next coefficient
• Repeat the multiply-add process across all coefficients

The remainder being zero confirms \(x = -1\) is a valid root. Hence, we're left with a reduced polynomial \(x^2 + 9\) to further factor. Synthetic division thus helps in streamlining the process of polynomial division and discovering roots efficiently.