find polynomial f(x) with real coefficients having the given degree and zeros.

Question

degree 4, zero:1, multiplicity: 2,2i

Answer 1

zeros=roots = 1, 1, 2i, -2i

polynomial = (x-1)^2(x^2 +4)

= (x^2 -2x +1)(x^2 +4)

= x^4 -2x^3 +5x^2 -8x +4

or any multiple such as double

= 2x^4 -4x^3 +10x^2 -16x +8

Answer 2

It's difficult to decipher, but I'm assuming you're conveying that f(x) has a root of 1, with that root of multiplicity 2. In addition, it seems you're saying 2i is also a root.

If 2i is a root (and f(x) has real coefficients), then the Conjugate Pairs Theorem states that -2i is also a root. Then, since x = -1 is of multiplicity 2:

f(x) = a(x-1)²(x-2i)(x+2i), for some real leading coefficient, a. If there is a constraint (e.g. f(x) passes through a point (2,3)), then we might solve for a. Otherwise, if the question allows ANY function that meets the previous criteria, we can assume a = 1 for simplicity.

Also, your questions may require the polynomial to be in standard (rather than factored) form. In this case, we would expand the 4th-degree polynomial completely.

f(x) = (x-1)²(x-2i)(x+2i) [if a=1]

f(x) = (x²-2x+1)(x²+2ix-2ix-4i²)

f(x) = (x²-2x+1)(x²+4)

f(x) = x⁴+4x²-2x³-8x+x²+4

f(x) = x⁴-2x³+5x²-8x+4

Answer 3

It has one more zero -2i;

f(x) = a(x - 1)²(x - 2i)(x + 2i) = a(x - 2)²(x² + 4), where a ≠ 0 real number