Polynomial of lowest degree with lowest possible integer coefficients, and with zeros 7-5i and 0 (multiplicity 2)

Answer 1

If we have a polynomial function with zeros 7 - 5i and 0 with a multiplicity of 2, we must also have a zero at 7 + 5i because of the rule for conjugates. The lowest degree polynomial we can have is 4. Therfore, we know the following:

x = 7 - 5i

x = 7 + 5i

x = 0 (multiplicity of 2)

Let's solve for the equation now:

f(x) = x²(x - 7 + 5i)(x - 7 - 5i)

f(x) = x²(x² - 7x - 5ix - 7x + 49 + 35i + 5ix - 35i - 25i²)

f(x) = x²(x² - 14x + 49 + 25) We use the rule that i² = -1 to get this as well

f(x) = x²(x² - 14x + 74)

f(x) = x⁴ - 14x³ + 74x²

Answer 2

A key to this problem is the realization that complex number zeros come in conjugate pairs. So if "7 - 5i" is a zero, then "7 + 5i" must also be a zero.

Forming binomial factors merely requires us to add "x -" in front of each zero so the polynomial would be:

P(x) = a(x - 0)(x - 0)[x - (7 - 5i)][x - (7 + 5i)] where "a" is any multiplier but to keep it simple, let's let a = 1.

Multiplying the complex binomials together we get:

[x - (7 - 5i)][x - (7 + 5i)]

(x - 7 + 5i)(x - 7 - 5i)

I like to use an "area model" type rectangle to organize the multiplication of binomials with more than 2 elements like this:

Combining and simplifying we get:

x² - 14x + 49 - 25i² but since i² = -1, we get x² - 14x + 49 + 25 = x² - 14x + 74

Then multiplying by x² [simplified version of (x - 0)(x - 0)] we get:

P(x) = x⁴ - 14x³ + 74x²