Three roots of the polynomial equation x4 - 4x3 - 2x2 + 12x + 9 = 0 are 3, - 1 and - 1. Explain why the fourth root must be a real number. Find the fourth root.

By the complex conjugate root theorem, complex roots "come in pairs", so there can't be an odd number of complex solutions. So the fourth root must be the real one.

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