Problem 6 - A Graphical Approach to College Algebra

Question

Find a cubic polynomial in standard form with real coefficients, having the given zeros. Let the leading coefficient be \(1 .\) Do not use a calculator. 0 and \(4-3 i\)

Step-by-Step Solution

Answer

The polynomial is \(f(x) = x^3 - 8x^2 + 25x\).

Step 1: Understand the nature of zeros

If one zero of a polynomial is a complex number (like \(4 - 3i\)), then its complex conjugate (\(4 + 3i\)) is also a zero since coefficients are real numbers.

Step 2: List all zeros

The given zeros are \(0\), \(4 - 3i\), and \(4 + 3i\).

Step 3: Form the factors for each zero

For zero \(0\), it is \(x\). For zero \(4 - 3i\), it is \((x-(4-3i))\). For zero \(4 + 3i\), it is \((x-(4+3i))\).

Step 4: Write the polynomial as a product of factors

The polynomial can be written as \(f(x) = x(x-(4-3i))(x-(4+3i))\).

Step 5: Expand the complex conjugates

First, expand \((x-(4-3i))(x-(4+3i))\) using the formula \((a-b)(a+b) = a^2 - b^2\). Here, \(a = x-4\) and \(b = 3i\).Thus, \((x-4)^2 - (3i)^2 = x^2 - 8x + 16 + 9 = x^2 - 8x + 25\).

Step 6: Multiply by the remaining factor

Now, multiply \(x\) by the result from Step 5: \(f(x) = x(x^2 - 8x + 25)\).

Step 7: Distribute to fully expand

Expand by distribution: \(f(x) = x(x^2) - x(8x) + x(25) = x^3 - 8x^2 + 25x\).

Step 8: Write in standard form

The polynomial is already in standard form: \(f(x) = x^3 - 8x^2 + 25x\).

Key Concepts

Complex Conjugates

When working with cubic polynomials with real coefficients, the presence of complex numbers among the zeros is quite exciting! Here's the deal: if a polynomial has real coefficients and one of its zeros is a complex number, say \(4 - 3i\), then its complex conjugate must also be a zero. So, in our case, \(4 + 3i\) is automatically a zero too. This happens because complex roots always come in pairs when you're dealing with polynomials with real coefficients. This rule keeps everything in the polynomial real and balanced.

• For each complex zero \(a + bi\), there's a conjugate zero \(a - bi\).
• Conjugates ensure polynomial coefficients stay real.
• This is vital for forming real-valued polynomials from complex roots.

Polynomial Zeros

Let's talk about zeros. In any polynomial, zeros, or roots, are the values of \(x\) for which the polynomial equals zero. They're like the heartbeat of polynomials as they help in forming the polynomial itself through their associated factors. For example, if we know that \(0\), \(4 - 3i\), and \(4 + 3i\) are zeros of our cubic polynomial, then we can easily express the polynomial as a product of its factors. Each zero corresponds to a factor of the polynomial:

• The zero \(0\) corresponds to the factor \(x\).
• The zero \(4 - 3i\) corresponds to \((x - (4 - 3i))\).
• The zero \(4 + 3i\) corresponds to \((x - (4 + 3i))\).

Understanding zeros allows you to reconstruct the polynomial through these essential building blocks.

Standard Form Expansion

Once you have the factors resulting from the zeros of a polynomial, it's time to expand them to get the polynomial in standard form. The standard form is the most familiar one: it displays the polynomial as a sum of terms arranged in descending order of their degree. Our journey with the polynomial starts with multiplying the factors associated with complex conjugates. Use the identity \((a-b)(a+b) = a^2 - b^2\) to simplify their product. In our polynomial, \((x - (4-3i))(x - (4+3i))\) expands to \(x^2 - 8x + 25\).

• Always apply the identity \((a-b)(a+b) = a^2 - b^2\) for complex factors.
• After simplifying, multiply the result by the remaining factor, \(x\), to expand further.

Next, distribute \(x\) across the terms of the result to get a polynomial \(x^3 - 8x^2 + 25x\). This expanded polynomial is already in standard form, showcasing coefficients of each term in an orderly fashion.