Can you help me with 2 tricky algebra problems

1. Let there be a cubic polynomial x^3+bx+c=0, where b and c are integers. If 5+sqrt3 is one root, find the other roots of the equation, and the values of b and c. I forgot, is the conjugate of a term like 5+sqrt3 also a root or is that only for complex numbers?
2. Give an example of a polynomial with integer coefficients and has the root r=3-(i)(2)^{1/4}
I'm kind of stumped on this one.

Can someone please help?

1. Because the coefficients are INTEGERS yes, the conjugate 5 - sqrt(3) must also be a root. So we have two roots:

5+sqrt(3), 5-sqrt(3), and we know that r1 + r2 + r3 = 0 (because a=0). Therefore the third root must be -10.

The three roots are 5+sqrt(3), 5-sqrt(3), and -10. You can multiply out to get the values of b and c.

2. Okay - if the polynomial has integer coefficients then first take the complex conjugate:

r1 = 3 - i (2^(1/4))
r2 = 3 + i(2^(1/4))

r1*r2 = 9 - i sqrt(2)

So we also need 9+i sqrt(2) as a root. So I would multiply out the following:

(x^2 - (9 - i sqrt(2))) (x^2 - (9 + i sqrt(2)))
= x^4 - 18 x^2 + (81 - 2i^2)
= x^4 - 18 x^2 + 83

Q.1:

1) Here one root being 5+√3, the other root is 5-√3 [Since irrational and complex roots occur only in pairs]. As it is 3rd degree polynomial, it has to have 3 roots, and let the third root be α.

2) Taking the polynomial as (x^3) + {(0)(x^2)} + bx + c,
Sum of roots = α + (5+√3) + (5-√3) = -{coefficent of x^3/coefficent of x^2} = 0/1 = 0

==> α = -10

Thus the polynomial is p(x) = (x + 10)(x - 5 - √3)(x - 5 + √3)

Expanding and simplifying, p(x) = (x^3) - 78x + 220 = (x^3) + bx + c

==> b = -78 and c = 220

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