Form a polynomial f(x) with real coefficients having the given degree and zeros. Degree​ 4; ​ zeros: 3−5i; −1 multiplicity 2

Answer 1

Since you are given that f(x) has a complex root,3+5i, you can conclude that the conjugate,3-5i, is also a root.( Complex Roots always come in pairs). You are also given that -1 is a double root, and that f(x) has degree four. You now have all your roots. You can factor

f(x) = (x+1)²_. (x - (3+5i)).(x-(3-5i)) which expands to

= ( x + 1 ) ². ( x² -6x- 36)

Answer 2

Recall that a polynomial with real coefficients with a complex root will also have the complex conjugate root.

So, given that 3 - 5i is a root, its conjugate, 3 + 5i, must also be a root. The polynomial also has a double root (multiplicity = 2) at x = -1.

So factored form will look like this: f(x) = (x + 1)²(x - (3 - 5i))(x - (3 + 5i))

That should be an acceptable form, but we can make it look a little tidier by multiplying the complex factors:

f(x) = (x + 1)²(x² - 6x + 34)

Multiplying those factors out to leave f(x) in standard form is also possible, and left as an exercise for the asker.