Question 350342: Find all zeros of the polynomial:
P(x) = x^3 − 7x^2 + 17x − 15
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! 
To find the zeros we need to factor. GCF factoring, factoring by patterns, trinomial factoring and factoring by grouping all fail on P(x). All that is left is factoring bytrial and error of the possible rational roots.
The possible rational roots are all fractions, positive and negative, of factors of the constant term (in this case the -15) over the factors of the leading coefficient (in this case the 1 in front of the  ). Therefore the possible rational roots of P(x) are 1, -1, 3, -3, 5, -5, 15 and -15.
If we are clever we can save some time. Looking at P(x) and noticing the alternating signs of the coefficients, we can rule out all the negative roots. (Think about it. If x is negative, then won't all 4 terms turn out negative? And if they're all negative they cannot add up to zero.)
We can also rule out 1 as a root. Since 1 to any power is 1 then each term of P(x) will work out to be the coefficient. And we can quickly see that 1 - 7 + 17 - 15 will not be zero. So all we have left to try are 3, 5 and 15. I find that the quickest way to test a root is with synthetic division. I hope you f\have learned this. Trying 3 as a root:
3 | 1 -7 17 -15
---- 3 -12 15
---------------
1 -4 5 0
Because of the zero remainder, this shows that (x-3) is a factor abd that 3 is a root. It also show that the other factor is  . So now
One root is 3. The other two roots will come from . Since this quadratic will not factor easily, we will use the Quadratic Formula to find them:
Simplifying:
This makes the 3 roots: 3, 2+i and 2-i.
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