SOLUTION: How do I find the third degree polynomial equation with rational coefficients that has the given numbers as roots 1) 5, 2i 2)-7, i 3)6, 3-2i

Question 699685: How do I find the third degree polynomial equation with rational coefficients that has the given numbers as roots
1) 5, 2i

2)-7, i
3)6, 3-2i
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
How do I find the third degree polynomial equation with rational coefficients that has the given numbers as roots
1) 5, 2i
-2i must also be a root.
f(x) = (x-5)(x-2i)(x+2i)
= (x-5)(x^2+4)
---------
2)-7, i
-i must also be a root.
f(x) = (x+7)(x-i)(x+i)
= (x+7)(x^2+1)
--------------------
3)6, 3-2i
3+2i mult also be a root.
f(x) (x-6)[(x-3)+2i][(x-3)-2i]
= (x-6)[(x-3)^2+4]
= (x-6)(x^2-6x+13)
=======================
Cheers,
Stan H.
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Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

If a polynomial equation with rational coefficients has a complex or irrational zero, then the conjugate of that zero is also a zero of the equation. The conjugate of a complex number is

If is a zero of a polynomial equation, then is a factor of the polynomial.

For your first problem, , , and are the three linear factors of the third degree polynomial. Multiply them together. Hint: Multiply the two complex factors first, recalling that the product of two complex conjugates is the sum of two squares.

John

Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it

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