Math......?

Question

Can [x-8] MULTIPLIED BY [2-X] ever be equal to x to the power of 3-10x to the power of 2+16x+12=0.

Answer 1

I am assuming that the =0 at the end is a mistake. If not (if the =0 really is there) then hustolemyname is correct.
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(x-8)(2-x) = 2x -16 -x^2 +8x = -x^2 + 10x -16

x^3 -10x^2+16x+12 ?=? -x^2 + 10x -16

Add x^2 -10x +16 to both sides:

x^3 - 9x^2 +6x +28 = 0

Let call this function f(x)

Because of the x^3 (x cubed), when x is very large and negative, the absolute value of x^3 will be much bigger than the absolute value of the rest: therefore f(x) will be less than x.

when x is positive and large, there will be a point when the value of x^3 will be much bigger than the value of 9x^2: therefore f(x) will be greater than 0.

By continuity, there has to be at least one point where f(x) = 0.

At that value of x (where f(x) = 0), you must have

(x-8)(2-x) = x^3 -10x^2+16x+12

The answer is 'yes'

Answer 2

no .. becuase (x-8)(2-x) is a numeric expression and your rhs is a boolean statement (which has the solution x=0 by the way)