find all zeros of f(x)=x^4-3x^3+4x^2-6x+4

Question

please show how its done. i have math finals coming up and have NO idea what i am doing. THANK YOU!!

Answer 1

f(1)=1-3+4-6+4=0 so x=1 is a zero. Also f(2)=16-24+16-12+4=0 so x=2 is another zero.

Using synthetic division let's divide by the zero x=1:

1 | 1 -3  4 -6    4

     1  1 -2  2 | -4

     1 -2  2 -4 | 0 = x³-2x²+2x-4. Since 1-2+2-4 is not zero, x=1 is not another zero.

Nor is x=-1, since -1-2-2-4 is not zero. Let's try x=2: 8-8+4-4=0 so x=2 is a zero, and we can divide by this:

2 | 1 -2 2  -4

     1  2 0 | 4

     1  0 2 | 0 = x²+2, which has complex zeroes: -i√2 and i√2.

The zeroes are 1, 2, -i√2 and i√2.