Tom K. answered • 05/01/20
Knowledgeable and Friendly Math and Statistics Tutor
Because signs alternate, we can only have positive roots. Since the leading power is odd, we have at least one real root. Thus, we have either 1 or 3 positive roots (as all coefficients are real; imaginary coefficients come in pairs). Since the leading coefficient is 1 and the constant is 32, all rational roots must be factors of 32: 1, 2, 4, 8, 16, or 32.
We can now just try these different factors. We could just substitute each of these numbers into the equation and see which one works. Note that once we find one, we can divide/use synthetic division, then either use the quadratic formula, factor, or use some other method to find the other two.
It is pretty easy to see that 1 does not work (3 even and one odd coefficient means an odd sum, or just add).
2^3 - 10*2^2 + 32 * 2 - 32 = 8 - 40 + 64 - 32 = 72 - 72 = 0
2 is a root.
Use synthetic division
2 1 -10 32 -32
2 -16
1 -8 16
We now must factor x^2 - 8x + 16
Clearly, this is (x - 4)^2
Thus, our solution is (x-2)(x-4)(x-4) or (x-2)(x-4)^2
An alternative solution would be to graph the polynomial. We would see that 2 and 4 are zeroes, and the polynomial rises on both sides of 4, telling us that this is a double root.
