SOLUTION: Suppose that a polynomial function of degree 5 with rational coefficients has the numbers {{{ 1/2 }}} , {{{ 2+ sqrt ( 7 ) }}} , {{{ 1-3i }}} as zeros. Find other zeros. I am not

Algebra ->  Rational-functions -> SOLUTION: Suppose that a polynomial function of degree 5 with rational coefficients has the numbers {{{ 1/2 }}} , {{{ 2+ sqrt ( 7 ) }}} , {{{ 1-3i }}} as zeros. Find other zeros. I am not      Log On


   



Question 288826: Suppose that a polynomial function of degree 5 with rational coefficients has the numbers +1%2F2+ , +2%2B+sqrt+%28+7+%29+ , +1-3i+ as zeros. Find other zeros.
I am not positive what Degree 5 means, and although I could find the zeros through trial and error, I know there is some process a bit easier, could you show me?

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
A degree of 5 means the highest exponent in the polynomial is 5. The degree also tells you how many roots/zeros it has.

With rational coefficients, the zeros with square roots will come in conjugate pairs: (p+q) and (p-q). This is true because when you multiply (p+q)(p-q) you get p%5E2-q%5E2 which is an expression of perfect squares. This means the square roots will disappear (which needs to happen if there are rational coefficients).

You are given one rational zero, 1/2, and two zeros with square roots:
2%2Bsqrt%287%29
and
1-3i (Remember that "i" is a square root! It is sqrt%28-1%29.)
There are 5 zeros in a polynomial of degree 5 so there two missing zeros. They will be the other "half" of the respective conjugate pairs:
2-sqrt%287%29
and
1+3i