Question 975180: Find a polynomial function of lowest degree with rational coefficients and the following as some of its zeros: 0,-√3,1-i.
(A) f(x)=x(x+√3)(x-(1-i))
(B) f(x)=x^4-3x^2+x-1
(C) f(x)=x(x+√3)(x-(1-i))(x-(1+i))
(D) f(x)= x^5-2x^4-x^3+6x^2-6x
(E) none of the above
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Both irrational and complex zeros are always present in conjugate pairs. Which is to say that if  is rational and  is irrational and  is a zero of a polynomial, then  must also be a zero of the same polynomial. Likewise, if  is a complex zero, then  is also a complex zero for the given polynomial.
Since you have five zeros, one rational, a pair of irrationals, and a pair of complex, the lowest degree polynomial that you can have is a degree 5 polynomial -- so says the Fundamental Theorem of Algebra.
You only have one 5th degree polynomial in your list. However, in order to be certain that the answer is indeed D rather than E, you are going to have to multiply
\left(x\ +\ \sqrt{3}\right)\left(x\ -\ \left(1\ -\ i\right)\right)\left(x\ -\ \left(1\ +\ i\right)\right))
and collect like terms to determine if answer D is indeed correct. I'll get you started.
Note that the product of a pair of conjugates is always the difference of two squares, and recall that  , so multiplying the two irrational factors together and the two complex factors we get:
(x^2\ -\ 2x\ +\ 2))
I'll leave it to you to finish.
John
My calculator said it, I believe it, that settles it
|
|
|
