Real zeros?

Question

I am having a hard time finding the real zeros for

15x^3-102x^2+207x-108

Can someone please help?

Answer 1

 Take out the common factor 3: 5x³-34x²+69x-36.

Start by finding rational zeroes. To do this 36 would be the product of the three zeroes, that is, the cubic can be expressed in the form (5x-a)(x-b)(x-c) producing -abc as the constant term. A cubic has at least one real zero.

36=2²3², so it has only two prime factors: 2 and 3. It's possible that one or both of these are factors in either positive or negative form.

Try x=2 first: cubic evaluates to 6, reject this. Try x=3 next: cubic evaluates to 0, so x=3 is a zero. Divide by this zero using synthetic division:

3 | 5 -34  69   -36

      5   15 -47 | 36

      5 -19  12 |  0 = 5x²-19x+12=(5x-4)(x-3),

which gives us two more real zeroes: x=⅘ and 3.

So we have a duplicate zero (x=3) and x=⅘.