Identify a cubic polynomial with integer coefficients that has 2^(1/3) + 4^(1/3) as a root.
I tried solving it but my polynomial ended up having non-integer coefficients. I just can't think of a way to find it so that my polynomial has integer coefficients. Help, please!
recall that 2+4 = (∛2^3 + ∛4^3) = (∛2 + ∛4)(∛4 - ∛8 + ∛16)
See what you can do with that. You might try using perfect cubes to see what's going on.
Thanks for the reply.
I hadn't considered the idea of using perfect cubes. How would I use perfect cubes, though? I'm still confused.
Never mind, I found an answer. Thanks for the help though!
To find a cubic polynomial with integer coefficients that has 2^(1/3) + 4^(1/3) as a root, we can use a technique called polynomial factorization.
Step 1: Consider the equation x = 2^(1/3) + 4^(1/3).
Step 2: Raise both sides of the equation to the power of 3 to eliminate the cube roots. (x^3 = (2^(1/3) + 4^(1/3))^3)
Step 3: Simplify the right side of the equation using the binomial theorem. (x^3 = 2 + 3 * 2^(1/3) * 4^(1/3) + 3 * (2^(1/3))^2 * (4^(1/3))^2 + (4^(1/3))^3)
Step 4: Simplify further using exponent rules for powers of 2 and 4. (x^3 = 2 + 3 * 2^(1/3) * 4^(1/3) + 3 * 2^(2/3) * 4^(2/3) + 4)
Step 5: Rewrite the equation as a polynomial equation by subtracting x^3 from both sides. (x^3 - x^3 = 2 + 3 * 2^(1/3) * 4^(1/3) + 3 * 2^(2/3) * 4^(2/3) + 4 - x^3)
Simplifying the equation further, we get:
0 = 2 + 3 * 2^(1/3) * 4^(1/3) + 3 * 2^(2/3) * 4^(2/3) + 4 - x^3
0 = 6 * 2^(1/3) * 4^(1/3) + 3 * 2^(2/3) * 4^(2/3) + 6
0 = 6 * (2 * 4)^(1/3) + 3 * (2 * 4)^(2/3) + 6
0 = 6 * (8)^(1/3) + 3 * (8)^(2/3) + 6
0 = 6 * 2 + 3 * 4 + 6
0 = 12 + 12 + 6
0 = 30
Therefore, we have obtained an equation with integer coefficients: x^3 - 30 = 0.
Thus, the cubic polynomial with integer coefficients that has 2^(1/3) + 4^(1/3) as a root is x^3 - 30 = 0.