find the real and complex zeros of the following function f(x)=x^3-5x^2+8x-6

Question

i need the work on how each 7 steps are done

Answer 1

First we list the rational zeroes: 1, -1, 2, -2, 3, -3, 6, -6.

If any of these are actual zeroes we only need one to find all the zeroes. 

x=1: 1-5+8-6=-2

x=-1: -1-5-8-6=-20

x=2: 8-20+16-6=-2

x=-2: -8-20-16-6=-50

x=3: 27-45+24-6=0, so x=3 is a zero.

Now we use synthetic division to reduce the cubic to a quadratic:

3 | 1 -5  8 -6

     1  3 -6

     1 -2  2  0 = x²-2x+2.

To solve x²-2x+2=0 we can write this as x²-2x+1+1=0, (x+1)²=-1, x+1=±i.

Therefore x=-1+i and x=-1-i. The zeroes are 3, -1+i, -1-i.