Hello,
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)
x^4-4x^3-2x²+12x+9=0
==> x^4+x^3-5x^3-5x²+3x²+3x+9x+9=0
==>x^3(x+1)-5x²(x+1)+3x(x+1)+9(x+1)=0
==>(x+1)(x^3-5x²+3x+9)=0
==>(x+1)(x^3+x²-6x²-6x+9x+9)=0
==>(x+1)[x²(x+1)-6x(x+1)+9(x+1)]=0
==>(x+1)²(x²-6x+9)=0
==>(x+1)²(x-3)²=0
P(x)=(x+1)²*(x-3)²
