A polynomial of minimum degree has rational coefficients and has the roots: x1=−1−√5;x2=1+2i so there are x3=−1+√5 and x4=1−2i. I need to find the polynomial equation.
I tried to use (x−x1)(x−x2)(x−x3)(x−x4) but the calculations are too "heavy" and too long.There is an easier method to solve this?
Right answer: x4−3x2+18x−20
Answer
Just to give a different approach, from x1+x2+x3+x4=0 and x1x2x3x4=−20, we know the answer has the form
P(x)=x4+ax2+bx−20
Now
P(1)=(2−√5)(2+√5)(−2i)(2i)=(4−5)(4)=−4
and
P(−1)=(−√5)(√5)(−2−2i)(−2+2i)=(−5)(4+4)=−40
so 1+a+b−20=−4 and 1+a−b−20=−40, or
a+b=15a−b=−21
from which we find 2a=−6 and 2b=36, so a=−3 and b=18.
Whether this is "easier" than computing the two quadratic factors of P(x) and simply multiplying them together is unclear.
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