P(x)=x^4-4x^3-2x^2+12x+9=0 P(3)=3^4-4*3^3-2*3²+12*3+9=81-108-18+36+9=0 P(-1)=(-1)^4 - 4*(-1)^3 -2*(-1)²+12*(-1)+9=1+4-2-12+9=0 P'(x)=4x^3-12x²-4x+12 P'(-1)=4*(-1)^3-12*(-1)²-4*(-1)+12=-4-12+4+12=0 (-1) is a double root Ok 3,-1,-1 are roots.
If the 4th root is not a real but a complex (a+ib), its conjugate will be also a root , there would be 5 roots and not 4
So, the 4th root is real and equal to 3 ( a double root)