SOLUTION: Find a polynomial function of least degree having only one real coefficients, a leading coefficient of 1, and roots 1-square root 3 , 1+square root 3, and 6-i .

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Find a polynomial function of least degree having only one real coefficients, a leading coefficient of 1, and roots 1-square root 3 , 1+square root 3, and 6-i .       Log On


   



Question 1023290: Find a polynomial function of least degree having only one real coefficients, a leading coefficient of 1, and roots 1-square root 3 , 1+square root 3, and 6-i .
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The roots we know of are the real roots
x%5B1%5D=1-sqrt%283%29 , and x%5B2%5D=1%2Bsqrt%283%29 ,
and the complex root x%5B3%5D=6-i .
For the polynomial to have only real coefficients,
it must also have the conjugate complex root x%5B4%5D=6%2Bi .
So, the polynomial of least degree, with a leading coefficient of red%281%29 is
.

P%28x%29=%28x-1%2Bsqrt%283%29%29%2A%28x-1-sqrt%283%29%29%2A%28x-6%2Bi%29%2A%28x-6-i%29

P%28x%29=%28%28x-1%29%5E2-%28sqrt%283%29%29%5E2%29%2A%28%28x-6%29%5E2-i%5E2%29
P%28x%29=%28%28x%5E2-2x%2B1%29-3%29%2A%28%28x%5E2-12x%2B36%29-%28-1%29%29
P%28x%29=%28x%5E2-2x-2%29%2A%28%28x%5E2-12x%2B36%29%2B1%29
P%28x%29=%28x%5E2-2x-2%29%2A%28x%5E2-12x%2B37%29
From here on, it is just busywork.
That is the part where I get bored and distracted, and make mistakes, so check my math from this point on.
P%28x%29=x%5E2%2A%28x%5E2-12x%2B37%29-2x%2A%28x%5E2-12x%2B37%29-2%2A%28x%5E2-12x%2B37%29
P%28x%29=%28x%5E4-12x%5E3%2B37x%5E2%29-%282x%5E3-24x%5E2%2B74x%29-%282x%5E2-24x%2B74%29
P%28x%29=x%5E4-12x%5E3%2B37x%5E2-2x%5E3%2B24x%5E2-74x-2x%5E2%2B24x-74
P%28x%29=x%5E4-14x%5E3%2B59x%5E2-50x-74