1. Obviously x = 0 is one of the zeros of this polynomial. So work with
12x^3 + x^2 - 4x - 2. Now the factors must begin with 1,2,3,4,6, or 12; and must end with 1 or 2. So the possible zeros are +/-2, +/-1, +/-1/2, +/-1/3, +/-1/4, +/-1/6, +/-1/12, +/-2/3.
2. Because the coefficient of the highest power is the largest coefficient, the whole...
show more
1. Obviously x = 0 is one of the zeros of this polynomial. So work with
12x^3 + x^2 - 4x - 2. Now the factors must begin with 1,2,3,4,6, or 12; and must end with 1 or 2. So the possible zeros are +/-2, +/-1, +/-1/2, +/-1/3, +/-1/4, +/-1/6, +/-1/12, +/-2/3.
2. Because the coefficient of the highest power is the largest coefficient, the whole numbers are unlikely solutions, so I'll start with x = +/-1/2. In this case, the absolute values of the terms would be
3/2, 1/4, 2, 2, and no combination of signs will make them balance.
Next I'll try x = +/- 1/4. The absolute values of the terms would be
3/8, 1/16, 1, 2, and no combination of signs will make them balance.
Next I'll try x = +/- 1/3. The absolute values of the terms would be
4/9, 1/9, 4/3, 2, and these MIGHT balance...if x = -1/3 you'd have
-4/9 + 1/9 + 4/3 - 2...nope, doesn't work. If x = +1/3 you'd have
4/9 + 1/9 - 4/3 - 2, also doesn't work.
Now I'll try x = +/- 2/3. The absolute values of the terms would be
32/9, 1/9, 8/9, 2 Still won't balance with any sign combination.
At this point I got lazy and went to a graphing calculator and found that the only real zero is around 0.71262, and cannot match any of the possible rational zeros.
So there are no rational zeros (except for x=0).
3. If you divide 12x^3 + x^2 - 4x - 2 by the quantity
(x - 0.71262), you'll get a quadratic whose coefficients are not exactly right, but that quadratic can then be solved to find approximate values for the imaginary zeros.