f(x)=6x^5-41x^?4+93x^3-35x^2-45x+22

Question

find the zeros of this function

Answer 1

Try rational zeroes first. Factors of 6 are 1,2,3,6 and factors of 22 are 1,11,2,22. Rational zeroes to try are ±1, ±½, ±⅓, ±⅙, ±11, ±11/2, ±11/3, ±11/6, ±2, ±2/11, etc.

f(1)=6-41+93-35-45+22=0 so x-1 is a factor and 1 is a zero.

Synthetic division:

1 | 6 -41  93 -35 -45 |  22

     6    6 -35  58  23 | -22

     6 -35  58  23 -22 |   0

g(x)=6x⁴-35x³+58x²+23x-22.

When g(½)=3/8-35/8+29/2+23/2-22=-4+26-22=0, so 2x-1 is another factor, x=½ is a zero.

Synthetic division:

½ | 6 -35  58 23 | -22

      6    3 -16 21 |  22

      6 -32  42 44 |    0

h(x)=6x³-32x²+42x+44

h(-⅔)=-16/9-128/9-28+44=-16-28+44=0, so 3x+2 is a factor and x=-⅔.

Synthetic division:

-⅔ | 6 -32  42 |  44

       6   -4  24 | -44

       6 -36  66 |    0

j(x)=6x²-36x+66=6(x²-6x+11). The quadratic has only complex solutions so we have already found the real zeroes: -⅔, ½, 1. The remaining two zeroes are complex: 

x²-6x=-11, x²-6x+9=-2, (x-3)²=-2, x=3±i√2.