Find the real zeros of P(x)=x^4-7x^3+14x^2-3x-9?

Question

The answers are 3, 1 plus or minus the square root of 5 over 2. I factored it into (x-3)(X^3-4x^2+2x+3) but having trouble factoriing the 2nd factor. Please explain your answer because I will die if you don't help me!

Answer 1

P(x)=(x^2-x-1)*(x-3)^2
so there are 3 distinct real roots:
3 (twice)
(1+sqrt(5))/2
(1-sqrt(5))/2

Answer 2

Try looking at some of the many answers people have already given you instead of reposting the same question four times.