Bob T. answered • 11/02/14

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Hello, Michelle,

My approach to this problem is through use of the factor theorem.

p(x) = x^3 – 3x^2 + 4x – 12

Finding the zeroes of the function p = 0 means that p(x) = 0, and therefore, x^3 – 3x^2 + 4x – 12 = 0

We would have a line of the coefficients 1, -3, 4, and -12;

Next, place the 3, representing the factor (x-3):

3 1 -3 4 -12 for x^3 – 3x^2 + 4x – 12

3 0 12

1 0 4 0 for x^2 + 0 x^1 + 4, or x^2 + 4

3 0 12

1 0 4 0 for x^2 + 0 x^1 + 4, or x^2 + 4

From here, we established that the factor (x - 3) is a factor and that (x^2 + 4) is a factor.

x - 3 = 0 x^2 + 4 = 0

x = 3

From this latter part, we have something

*similar*to the special polynomial or quadratic product x^2 - 4. However, this product is

*not*one of the special products we know, nor do we have this factor as a part of our solution.

From this, we subtract 4 on both sides, which implies that

x^2 = -4; x = √(-4); x = ±2i

([-2]i)^2 = ([-2]^2)(i^2) = (4)(-1) = -4

(2i)^2 = (2^2)(i^2) = (4)(-1) = -4

Therefore, we have two imaginary solutions ±2i

Remember: Every imaginary or complex solutions come in pairs. The imaginary part is in the form x±c, where c , in our case, is ±2i, where the real part is 0: 0±2i

⇒ (implies [that]) x = {0-2i, 0+2i, 3}, or x = {-2i, 2i, 3}.

Remember also that IF x is only in the REAL numbers, then ±2i would not be a part of the solution. That is, if x is in only the set of real numbers, and x = {3}.