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College algebra asked before not answered?
3.What is the lowest-degree polynomial with integer coefficients and the roots 6, i, and –i?
I have the answer as x^3-6x^2+x-6

1.
If 1 and –6 are two of the roots of 6x4 + 31 x3 − 33x2 − 16x + 12, then what are the other two?
I have -1 and 6
2.
Which of these rational numbers does the Rational Roots Theorem say can't be a solution to 14x7 + 13x5 − 19x4 + 7x2 − 6x − 21 = 0?
3 days ago - 1 day left to answer.

a(x - 6)(x - i)(x + i)
= a(x - 6)(x^2 + 1)
= a(x^3 - 6x^2 + x - 6), where a is any non-zero integer.

(6x^4 + 31x^3 - 33x^2 - 16x + 12) / ((x - 1)(x + 6))
= 6x^2 + x - 2
= 6x^2 + 4x - 3x - 2
= 2x(3x + 2) - (3x + 2)
= (3x + 2)(2x - 1)
(3x + 2)(2x - 1) = 0
=> 3x + 2 = 0, 3x = -2, x = -2/3
=> 2x - 1 = 0, 2x = 1, x = 1/2
So the remaining roots are -2/3 and 1/2.

Divisors of 14 are ±1, ±2, ±7, and ±14.
Divisors of -21 are ±1, ±3, ±7, and ±21.
So possible rational roots are ±1, ±1/2, ±1/7, ±1/14, ±3, ±3/2, ±3/7, ±3/14, ±7, ±7/2, ±21, and ±21/2.

1) (x - 6)(x - i)(x + i) =
(x - 6)(x^2 + 1) =
x^3 - 6x^2 + x - 6

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