
Andy C. answered 07/07/18
Tutor
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Math/Physics Tutor
zero is a solution since -3x can be factored out
the remaining polynomial is
x^3 -7x^2 + 16x -12
By rational root theorem the possible rational zeros are {+-1 ,+-2, +-3, +-4, +-6, +-12}
2 is a zero since 2^3 - 7(2)^2 + 16(2) - 12 = 8 - 28 + 32 - 12 = -20 +32-12 = 12 - 12 = 0
Synthetic division says:
2 | 1 -7 16 -12
2 -10 12
_______________________
1 -5 6 0
the quadratic that remains when (x-2) is factored is:
x^2 - 5x + 6 = ( x - 3 )(x - 2 )
So the four zeros are 0,2,2,and 3
Since x=2 has multiplicity 2, it will "Bounce" at x=2
All that remains is to set up a sign table for the following intervals:
x<0
0 < x < 2
2 < x < 3
x>3
where this function is negative, positive, positive, and negative, respectively.
the max occurs around (0.557,8.5) and another local max at (2.693,1.191)