SOLUTION: Please write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and 2,2i, 4-[6] as zeros. The brackets [] are for square roots. Th

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Please write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and 2,2i, 4-[6] as zeros. The brackets [] are for square roots. Th      Log On


   



Question 1085271: Please write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and 2,2i, 4-[6] as zeros. The brackets [] are for square roots.
Thank you!

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
TIP: The conventional way to write sqrt%286%29 is sqrt(6).

1) For each real or complex zero p of a polynomial f%28x%29 ,
there is a factor %28x-p%29 in the factorization of the polynomial.
So, if f%28x%29 has 2 , 2i, and 4-sqrt%286%29 as zeros,
the factors %28x-2%29 , %28x-2i%29 , and %28x-%284-sqrt%286%29%29%29=%28x-4%2Bsqrt%286%29%29 are in the factorization of f%28x%29 .

2) If a polynomial f%28x%29 has real coefficients and has a non-real complex number b%2Bi%2Ac for example) as a zero,
the conjugate complex number, b-i%2Ac is also a zero of the polynomial,
and
if f%28x%29 has rational coefficients and has an irrational number such as d-e%2Asqrt%28f%29 as a zero,
the conjugate irrational number, d%2Be%2Asqrt%28f%29 , is also a zero of the polynomial.
In other words, irrational zeros and non-real complex zeros come as conjugate pairs.
So, along with 2i , f%28x%29 also has -2i as a zero,
and along with 4-sqrt%286%29 , f%28x%29 also has 4%2Bsqrt%286%29 as a zero.

Then, the factors %28x-%28-2i%29%29=%28x%2B2i%29 and
%28x-%284%2Bsqrt%286%29%29%29=%28x-4-sqrt%286%29%29 are also in the factorization of f%28x%29 .

3) Besides at least one factor of the form %28x-p%29 for every zero,
the factorization of a polynomial with rational coefficients,
also has a constant factor a , which will be the leading coefficient.

So,
f%28x%29=1%28x-2%29%28x-2i%29%28x%2B2i%29%28x-4%2Bsqrt%286%29%29%28x-4-sqrt%286%29%29
Simplifying,
f%28x%29=%28x-2%29%28x%5E2-%282i%29%5E2%29%28%28x-4%29%5E2-%28sqrt%286%29%29%5E2%29
f%28x%29=%28x-2%29%28x%5E2%2B4%29%28%28x-4%29%5E2-6%29
f%28x%29=%28x-2%29%28x%5E2%2B4%29%28x%5E2-8x%2B16-6%29
f%28x%29=%28x-2%29%28x%5E2%2B4%29%28x%5E2-8x%2B10%29
If we really must, we multiply to get
f%28x%29=%28x-2%29%28x%5E4-32x%5E3%2B14x%5E2-32x%2B40%29 , and
f%28x%29=x%5E5-10x%5E4%2B30x%5E3-60x%5E2%2B104x-80%29 .