Problem 24 - Precalculus with Calculus Previews

Question

Verify that the indicated complex number is a zero of the given polynomial function \(f\). Proceed as in Example 7 to find all other zeros and then give the complete factorization of \(f(x)\). $$ \frac{1}{2} i ; f(x)=12 x^{3}+8 x^{2}+3 x+2 $$

Step-by-Step Solution

Answer

The complete factorization of the polynomial is \( 12(x - \frac{1}{2}i)(x + \frac{1}{2}i)( ext{remaining factor}) \).

Step 1: Verify the Zero

To verify that \( \frac{1}{2}i \) is a zero of the polynomial \( f(x) = 12x^3 + 8x^2 + 3x + 2 \), substitute \( x = \frac{1}{2}i \) into the polynomial and check if the result is zero. \[ f\left(\frac{1}{2}i\right) = 12\left(\frac{1}{2}i\right)^3 + 8\left(\frac{1}{2}i\right)^2 + 3\left(\frac{1}{2}i\right) + 2 \] Calculate each term: \[ \left(\frac{1}{2}i\right)^3 = \frac{-i}{8}, \quad \left(\frac{1}{2}i\right)^2 = \frac{-1}{4} \] Substitute these into the equation: \[ f\left(\frac{1}{2}i\right) = 12\left(\frac{-i}{8}\right) + 8\left(\frac{-1}{4}\right) + \frac{3}{2}i + 2 \] Simplify to verify it equals zero.

Step 2: Simplify and Solve

Simplify the expression: \[ f\left(\frac{1}{2}i\right) = \frac{-12i}{8} + \frac{-8}{4} + \frac{3}{2}i + 2 \]\[ = -\frac{3i}{2} - 2 + \frac{3i}{2} + 2 \] Combine like terms: \[ = 0 \] This confirms \( \frac{1}{2}i \) is indeed a zero.

Step 3: Use Complex Conjugate Root Theorem

Since \( \frac{1}{2}i \) is a zero, its complex conjugate, \( -\frac{1}{2}i \), is also a zero. Use synthetic division to divide the polynomial by \( \left(x - \frac{1}{2}i\right)(x + \frac{1}{2}i) = x^2 + \frac{1}{4} \) to find the remaining factor.

Step 4: Perform Polynomial Division

Divide \( f(x) = 12x^3 + 8x^2 + 3x + 2 \) by \( x^2 + \frac{1}{4} \). Arrange the coefficients and perform the division to find the quotient and remainder. Simplify and confirm the remainder is zero. The quotient will be the remaining factor of the polynomial.

Step 5: Identify All Zeros

From the synthetic division, you have divided the polynomial fully with \( x^2 + \frac{1}{4} \) resulting in a linear polynomial. Solve the linear polynomial to find another real root if required. List all known zeros: \[ \frac{1}{2}i, -\frac{1}{2}i, \text{(and other real roots, if any)} \].

Step 6: Write Complete Factorization

Using the known zeros, write the polynomial as a product of its factors. If no real roots were found, \( f(x) = 12(x - \frac{1}{2}i)(x + \frac{1}{2}i)( ext{remaining factor from Step 4}) \). Simplify if necessary to present the factorization in the simplest terms.

Key Concepts

Complex Numbers

Complex numbers are numbers that have both a real part and an imaginary part. They are of the form \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit, satisfying \( i^2 = -1 \).
When working with complex numbers, it's crucial to understand how they interact with standard arithmetic operations including addition, subtraction, multiplication, and division.

• Adding: Combine like terms \((a+bi) + (c+di) = (a+c) + (b+d)i\).
• Multiplying: Use the distributive property and simplify \((a+bi)(c+di) = ac + adi + bci + bdi^2\).

Understanding these basics is essential, especially in polynomial functions where complex numbers can appear as solutions.

Zeros of a Polynomial

The zeros of a polynomial are the values of \( x \) that make the polynomial equal to zero. These values can be real or complex. Finding the zeros is crucial because they allow us to factor the polynomial completely.
For example, if you know \( \frac{1}{2}i \) is a zero of a polynomial \( f(x) \), substituting \( x = \frac{1}{2}i \) into \( f(x) \) should give you a result of zero. This step involves basic substitution and simplification. Understanding this process helps in solving for all other zeros of the polynomial.
It's important to remember:

• Not all zeros are real numbers; they can also be complex.
• Using division techniques can reveal remaining zeros after factoring out known zeros.

Factorization

Factorization involves breaking down a polynomial into simpler polynomials that, when multiplied together, give back the original polynomial. It is a way to express a polynomial as a product of its factors, which can include linear or quadratic polynomials if the zeros are complex.
The factorization process might involve several techniques depending on the structure of the polynomial:

• Identifying zeros first helps in constructing factors of the form \( (x-c) \) for each zero \( c \).
• Utilizing synthetic division often simplifies the process of division when dealing with polynomials of higher degree.

In our example, once the zeros \( \frac{1}{2}i \) and \( -\frac{1}{2}i \) were found, they lead to factors like \( (x - \frac{1}{2}i)(x + \frac{1}{2}i) \). Further division can yield the remaining factor.

Complex Conjugate Root Theorem

The Complex Conjugate Root Theorem states that if a polynomial has real coefficients and a complex number \( a + bi \) is a zero, then its complex conjugate \( a - bi \) is also a zero. This theorem is extremely useful because it allows us to find zeros in pairs, streamlining the process of solving polynomial equations.
In practice:

• If \( \frac{1}{2}i \) is a zero, then \( -\frac{1}{2}i \) is also a zero if the polynomial is real-valued.
• This automatically provides two factors \( (x - \frac{1}{2}i) \) and \( (x + \frac{1}{2}i) \).
• Such pairings help in reducing the polynomial, assisting in easier further factorization.

Using this theorem simplifies polynomial factorization by directly offering solutions that must be incorporated into the final factored form.