In Exercises 47–48, find an nth-degree polynomial function with real coefficients satisfying the given conditions. Verify the real zeros and the given function value. n = 3; 2 and 2 - 3i are zeros; f(1) = -10

Verified step by step guidance

1. Identify the given zeros of the polynomial. Since the polynomial has real coefficients and one zero is complex (2 - 3i), its complex conjugate (2 + 3i) must also be a zero. So, the zeros are 2, 2 - 3i, and 2 + 3i.
2. Write the polynomial in factored form using the zeros: \(f(x) = a(x - 2)(x - (2 - 3i))(x - (2 + 3i))\), where \(a\) is a real number coefficient to be determined.
3. Simplify the product of the complex conjugate factors: \((x - (2 - 3i))(x - (2 + 3i))\) can be expanded using the difference of squares formula for complex conjugates: \((x - 2)^2 - (3i)^2\).
4. Calculate the simplified quadratic factor: \((x - 2)^2 - (3i)^2 = (x - 2)^2 - (-9) = (x - 2)^2 + 9\). So the polynomial becomes \(f(x) = a(x - 2)((x - 2)^2 + 9)\).
5. Use the given function value \(f(1) = -10\) to find \(a\). Substitute \(x = 1\) into the polynomial and solve for \(a\): \(f(1) = a(1 - 2)((1 - 2)^2 + 9) = -10\).

Key Concepts

Polynomial Zeros and Their Multiplicity

Zeros of a polynomial are the values of x that make the polynomial equal to zero. For an nth-degree polynomial, there are n zeros (counting multiplicities). Understanding zeros helps in constructing the polynomial by forming factors like (x - zero).

Complex Conjugate Root Theorem

If a polynomial has real coefficients and a complex zero a + bi, then its conjugate a - bi is also a zero. This ensures the polynomial remains with real coefficients. For example, if 2 - 3i is a zero, then 2 + 3i must also be a zero.

Using Given Function Values to Find Leading Coefficient

After forming the polynomial from its zeros, the leading coefficient can be found by substituting a given x-value and its corresponding function value into the polynomial. This step ensures the polynomial satisfies all given conditions.