Find a degree 3 polynomial with real coefficients having zeros 1 and 1 − 2 i and a lead coefficient of 1. Write P in expanded form. P ( x ) = .

Answer 1

since P(x) = (x-x1)(x-x2)(x-x3),

x1=1, x2=1-2i, since the complex zeros always appear in pares, and the pare is symmetric about x-axis, so the other complex zero is x3=1+2i, therefore we have:

P(x) = (x-1)(x-1+2i)(x-1-2i) = (x-1)(x²-2x+5) = x³ - 3x² + 7x - 5

Answer 2

The critical piece of knowledge that enables one to do this problem is that complex roots ALWAYS appear as conjugate pairs. So if 1 - 2i is a zero, then 1 + 2i is also a zero. We then have all three of the zeros for the cubic function (by the Fundamental Theorem of Algebra, each 3rd degree polynomial has in total 3 zeros).

To find the polynomial with zeros b, c, and d we just multiply together the associated binomials so:

P(x) = a(x - b)(x - c)(x - d). Note that the simplest version of this polynomial (in my opinion) is when a = 1. That makes the polynomial for this case:

P(x) = (x - 1)[x - (1 - 2i)][x - (1 + 2i)]

P(x) = (x - 1)(x - 1 + 2i)(x - 1 - 2i)

To multiply these, a generic rectangle is useful. I'll first multiply the two complex parts:

Notice that we get terms that zero out. 2ix cancels with -2ix and 2i cancels with -2i. We can also simplify -4i² since i² = -1, then -4i² = -4(-1) = 4 and we combine like terms to get x² - 2x + 5.

Then multiply (x² - 2x + 5) by (x - 1) in the same way to get: P(x) = x³ - 3x² + 7x - 5