SOLUTION: Find all the real values of x for which the function: f(x) = 18x^3 - 33x^2 + 20x – 4 becomes zero.

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Question 41623: Find all the real values of x for which the function: f(x) = 18x^3 - 33x^2 + 20x – 4 becomes zero.
Answer by psbhowmick(489) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29+=+18x%5E3+-+33x%5E2+%2B+20x - 4

By trial we put x = 1%2F2.
We find that f%281%2F2%29+=+0.
So, by the Remainder theorem, (2x - 1) must be a factor of f(x).

Now, let us arrrange f(x) in such way that (2x - 1) can be taken as a factor.
f%28x%29+=+18x%5E3+-+33x%5E2+%2B+20x+%96+4
= +9x%5E2%282x-1%29+-+12x%282x-1%29+%2B+4%282x-1%29
= %282x-1%29%289x%5E2-12x%2B4%29
= %282x-1%29%28%283x%29%5E2+-+2%2A%283x%29%2A2+%2B+2%5E2%29
= %282x-1%29%283x-2%29%5E2

Hence the function becomes zero when 2x - 1 = 0 i.e. x = 1%2F2 or when 3x - 2 = 0 i.e. x = 2%2F3.