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.

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

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.