P(x) = x^4-8x^3+14x^2-8x+13, i find zeros

Question

P(x) = x^4-8x^3+14x^2-8x+13, i find remaining zeros

Answer 1

P(x) can be written:

P(x)=x⁴-8x³+13x²+x²-8x+13=x²(x²-8x+13)+(x²-8x+13),

P(x)=(x²-8x+13)(x²+1).

Solving the two quadratics we get:

x²-8x=-13, x²-8x+16=3, (x-4)²=3, x=4±√3; and

x=i and -i.

The four zeroes are i, -i, 4+√3, 4-√3.