Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form.

7, -11, and 2 + 6i

f(x) = x4 - 53x2 + 468x - 3080

f(x) = x4 - 9x3 - 42x2 + 234x - 3080

f(x) = x4 - 117x2 + 468x - 3080

f(x) = x4 - 9x3 + 42x2 - 234x + 3080

Question

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form.

7, -11, and 2 + 6i

 f(x) = x4 - 53x2 + 468x - 3080

 f(x) = x4 - 9x3 - 42x2 + 234x - 3080

 f(x) = x4 - 117x2 + 468x - 3080

 f(x) = x4 - 9x3 + 42x2 - 234x + 3080

anonymous · Answer

zeros are $7, -11, 2 + 6i$ and therefore also $2-6i$ the conjugate

anonymous · Answer

the quadratic with zeros at $7$ and $-11$ is $(x-7)(x+11)$

anonymous · Answer

the quadratic with zeros at $2+6i$ and $2-6i$ is easiest found by working backwards 
$$x=2+6i\
x-2+6i\
(x-2)^2=(6i)^2=-36\
x^2-4x+4=-36\
x^2-4x+40=0$$

anonymous · Answer

okay do i multiply them with the two conjugates
?

anonymous · Answer

now your job is to multiply out 
$$(x-7)(x+11)(x^2-4x+40)$$

anonymous · Answer

you do not multiply by the two conjugates, you multiply by the quadratic that has those two zeros

anonymous · Answer

okayy so its the first answer

anonymous · Answer

f(x) = x4 - 53x2 + 468x - 3080

anonymous · Answer

yes, it is