Question 1121240: How to find a polynomial function with rational coefficients so that p(x)=0 has the given roots 2+i and 1-sqrt5
I tried (x-(2+i))(x-(2-i))(x-(1-sqrt5))(x-(1+sqrt5)) and got x^4-6x^3+9x^2+6x-20, which does not seem to include 2+i as a root. Am I doing it right?
Found 2 solutions by ikleyn, Boreal:
Answer by ikleyn(52803)   (Show Source):
You can put this solution on YOUR website! .
This step
(x-(2+i))(x-(2-i))(x-(1-sqrt5))(x-(1+sqrt5))
was absolutely correct.
I did not check yours further expression, since it is just a technique you must do on your own.
Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! At the end, I got what you got.
If one raises 2+i to the 4th power, the answer is -7+24 i
to the third power is 2+ 11i
squared is 3+4i
The constants, after using the exponents, are -7-12+27+12-20
The i s will be 24-66+36+6. You did it right
 Your answer
another way, longer, uses the quadratic formula to find where 2+i and 2-i work
a=1
b=-4, because will divide by 2
sqrt(b^2-4ac)=2i, and when that is divided by 2, will give 1i as a result. sqrt(16-20) does that, and since 4ac=20, c must equal 5
One polynomial is x^2-4x+5
The other polynomial has
a=1
b=-2, so when divided by 2 will get -1
sqrt(b^2-4ac) needs to be 2sqrt(5), because we are dividing the numerator by 2
b^2-4ac=20
4-4c=20
-4c=16
c=-4
x^2-2x-4 is the other polynomial
product is x^4-2x^3-4x^2
==========-4x^3+8x^2+16x
===============5x^2-10x-20
or x^4-6x^3+9x^2+6x-20
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