Question

Find a polynomial f(x) with integer coefficients such that one root f(x) = 0 is x = √2 + ³√2

Answer 1

I don't know how to go about solving this problem, can anyone show me the steps?

Answer 2

Okay, uhmmm x1 = (-b - sqrt(D))/(2a), x2 = (-b + sqrt(D))/(2a)

Answer 3

Where D = b^2 - 4ac
so you have:
x2 = (-b + sqrt(b^2-4ac))/(2a)

Answer 4

Or,
x2 = (-b/(2a)) + sqrt(b^2 - 4ac)/(2a)

Answer 5

Which means that
(-b/(2a)) = sqrt(2)
and
sqrt(b^2-4ac)/(2a) = cuberoot(2)

Answer 6

-b = 2a*sqrt(2)

sqrt(8a^2 - 4ac)/(2a) = cuberoot(2)
(8a^2-4ac)/(4a^2) = 2^(2/3)
(2a - c) = 2^(2/3) a

Answer 7

We need to like plug in something for c or a, i think. i gotta go now, @satellite73 @amistre64 @experimentX @estudier can one of you finish this? Thanks :)

Answer 8

Based on your work, you're assuming its a quadratic. Why are you allowed to make that assumption? Since there is a cube root, wouldn't it be a third degree polynomial?