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

Answer 1

The Fundamental Theorem of Algebra says that a degree 4 polynomial will have a total number of zeros (real and complex) of 4. There are only 3 listed: -3 + 2i, 5, and 5 (5 with a multiplicity of 2 means it appears twice in the list of zeros).

The key to this problem is to realize that if "-3 + 2i" is a zero, then the conjugate "-3 - 2i" is also going to be a zero.

So we can write the polynomial in factored form as f(x) = a(x - (-3 + 2i))(x - (-3 - 2i))(x - 5)(x - 5) where "a" is any constant. In this case, because they require any polynomial, let's just let a = 1

f(x) = (x - (-3 + 2i))(x - (-3 - 2i))(x - 5)(x - 5)

f(x) = (x + 3 - 2i)(x + 3 + 2i)(x - 5)(x - 5) ←you can use any method to multiply this out if it is required (they don't really specify the form of the answer)

f(x) = (x² + 6x + 13)(x² - 10x + 25)

f(x) = x⁴ - 4x³ - 22x² + 20x + 325

Answer 2

By conjugate zeros theorem (- 3 - 2i) is also zero. So, f(x) = a(x - 5)²(x + 3 - 2i)(x + 3 + 2i) =

a((x - 5)²(x² + 6x + 13), where a ≠ 0 is leading coefficient no data uniquily identify it