SOLUTION: a.list all Zeroes b.use synthetic division f(x)=x^3-4x^2+8x-5

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Question 716674: a.list all Zeroes
b.use synthetic division
f(x)=x^3-4x^2+8x-5
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=x%5E3-4x%5E2%2B8x-5
a. To find the zeros of a polynomial of degree 2 or more we factor the polynomial. (Although in the special case of Quadratics we can use the Quadratic Formula instead of factoring.) The greatest common factor is 1 (which we rarely factor out). There are too many terms for any of the factoring patterns or for trinomial factoring. I don't see how to factor the polynomial by grouping so we are left with factoring by trial and error of the possible rational roots.

The possible rational roots of a polynomial are all the ratios, positive and negative, that can be formed using a factor of the constant term (at the end) in the numerator over a factor of the leading coefficient (at the front). The contant term in our polynomial is 5. (Actually it is -5 but since we are going to use all positive and negative ratios it doesn't matter if we use the "-".) The factors of 5 are just 1 and 5. The leading coefficient (in front of x%5E3) is 1 which just has 1's for factors. This makes the possible rational roots of our polynomial:
+1/1 or +5/1
which reduce to:
+1 or +5

Now we see if any of these actually are roots/zeros of the polynomial. Since powers of 1 are very simple we can easily check that mentally. But since the problem tells you to use synthetic division, we will do so:
1  |   1   -4   8   -5
----        1  -3    5
      -----------------
       1   -3   5    0
The remainder, in the lower right corner is zero. This means that (x-1) is a factor and that 1 is a root/zero of the polynomial. Not only that but the rest of the bottom row tells us what the other factor will be. "1 -3 5" translates into x%5E2-3x%2B5. So f(x) in (partially) factored form is:
f%28x%29+=+%28x-1%29%28x%5E2-3x%2B5%29

The remaining zeros of the polynomial will be zeros of the second factor. That factor will not factor further, no matter which factoring techniques we use! But it is quadratic so we can use the formula:
x+=+%28-%28-3%29+%2B-+sqrt%28%28-3%29%5E2-4%281%29%285%29%29%29%2F2%281%29
Simplifying...
x+=+%28-%28-3%29+%2B-+sqrt%289-4%281%29%285%29%29%29%2F2%281%29
x+=+%28-%28-3%29+%2B-+sqrt%289-20%29%29%2F2%281%29
x+=+%28-%28-3%29+%2B-+sqrt%28-11%29%29%2F2%281%29
x+=+%283+%2B-+sqrt%28-11%29%29%2F2
which is short for:
x+=+%283+%2B+sqrt%28-11%29%29%2F2 or x+=+%283+-+sqrt%28-11%29%29%2F2
With the negative inside the square root, these roots/zeros will be complex numbers. So we should express them in standard a + bi form:
x+=+%283+%2B+sqrt%28-1%2A11%29%29%2F2 or x+=+%283+-+sqrt%28-1%2A11%29%29%2F2
x+=+%283+%2B+sqrt%28-1%29%2Asqrt%2811%29%29%2F2 or x+=+%283+-+sqrt%28-1%29%2Asqrt%2811%29%29%2F2
x+=+%283+%2B+i%2Asqrt%2811%29%29%2F2 or x+=+%283+-+i%2Asqrt%2811%29%29%2F2
x+=+3%2F2+%2B+i%2Asqrt%2811%29%2F2 or x+=+3%2F2+-+i%2Asqrt%2811%29%2F2
x+=+3%2F2+%2B+%28sqrt%2811%29%2F2%29%2Ai or x+=+3%2F2+%2B+%28-sqrt%2811%29%2F2%29%2Ai

These two, plus the zero of 1 we found earlier, are the three zeros of your polynomial function.