g(x)=x^4-5x^3+7x^2+3x-10

Question

Help find the zeros to the equation pretty please

Answer 1

Look for rational zeroes first. The magnitude of these will be factors of 10: 1, 2, 5, 10.

We can have positive or negative values of these so let's start with 1 and -1. Even powers of x will give us positive coefficient values: 1-5+7+3-10=-10 so 1 is not a zero (root); 1+5+7-3-10=0 so -1 is a zero.

Using synthetic division we can divide by the zero:

-1|1 -5   7    3  -10

    1 -1   6 -13 | 10

    1 -6 13 -10 |  0 = x³-6x²+13x-10.

Now we look for zeroes of the cubic. Neither 1 nor -1 are zeroes, so try 2:

8-24+26-10=0, so 2 is another zero. Divide by it:

2|1 -6 13  -10

   1  2  -8 | 10

   1 -4   5 |  0 = x²-4x+5 has no real zeroes (it has complex zeroes).

We can find the complex zeroes:

x²-4x=-5,

x²-4x+4=-5+4=-1,

(x-2)²=-1,

x-2=±i,

x=2±i.

We now have all 4 zeroes: -1, 2, 2+i, 2-i.