Find a polynomial function f(x) of degree 2 with real coefficients that has 5-i as one of its zeros.

Answer 1

If one of the zeroes is 5-i, this means the factor is (x-5-i). In order for all the coefficients to be real, the "i" has to be removed from the equation. This means 5-i has to be multiplied by its counter, 5+i, since (a+b)(a-b) = a²-b², so (5-i)(5+i) is equal to 25 - i², or 25 +1, which gets rid of the i. Now that we know the two factors are (x-5-i) and (x-5+i), we can multiple the two factors to come to the final equation:

x²-5x+26

Answer 2

Hi Jillian,

This can be a confusing problem, but it is made far simpler if you understand some basics about quadratic functions.

First, you can solve any quadratic equation by using the Quadratic Formula.

You know, the whole x = (-b ± √(b²-4ac))/(2a)

The quadratic formula isn't nearly as easy as factoring, but not every quadratic factors, so we have this formula to fall back on. Only solutions that are not rational (fractions) do we need to use the quadratic formula on. Realize though, that they would always come in pairs. Due to the ± in the fomula.

Therefore, if 5-i is a zero, then automatically 5 + i is a zero as well. Therefore, you could write the solutions to the quadratic equation that has these zeros as x = 5±i

We can use this solution to go backwards to find what the quadratic equation looks like:

x = 5±i

x-5 = ±i (Subtract 5 from each side)

(x-5)²=(±i)² (Square both sides)

x² - 10x + 25 = i² (Expand each side)

x² - 10x + 25 = -1 (Replace i² with -1)

x²- 10x + 26 = 0 (Add 1 to each side)

This then is your quadratic that has 5 ± i as zeros. I hope this helps!