Answer:
[tex]x_1=-2,\ x_2=x_3=3\\ \\(x+2)(x-3)^2=0[/tex]
Step-by-step explanation:
Consider the equation
[tex]x^3-4x^2-3x+18=0[/tex]
The zeros of the equation can be the divisors of 18:
[tex]\pm 1,\ \pm2,\ \pm3,\ \pm 6,\ \pm 9,\ \pm 18[/tex]
Check them:
[tex]1^3-4\cdot 1^2-3\cdot 1+18=1-4-3+18=10\neq 0\\ \\(-1)^3-4\cdot (-1)^2-3\cdot (-1)+18=-1-4+3+18=16\neq 0\\ \\2^3-4\cdot 2^2-3\cdot 2+18=8-16-6+18=4\neq 0\\ \\(-2)^3-4\cdot (-2)^2-3\cdot (-2)+18=-8-16+6+18=0[/tex]
So, x=-2 is zero. Now
[tex]x^3-4x^2-3x+18\\ \\=x^3+2x^2-6x^2-12x+9x+18\\ \\=x^2(x+2)-6x(x+2)+9(x+2)\\ \\=(x+2)(x^2-6x+9)\\ \\=(x+2)(x-3)^2[/tex]
Thus, x=3 is the second root (of multiplicity 2)
