Question

Find all integer roots of x³+x²-3x+9=0

Answer 1

Answer:

     -3 is the only integer root

    It occurs with multiplicity 1.

Step-by-step explanation:

If x³+x²-3x+9=0 and x is an integer, then x divides 9.  So the only possibilities are ±1, ±3, ±9.

Putting x=1 gives x³+x²-3x+9 = 1+1-3+9 ≠ 0, so 1 is not a root.

Putting x=-1 gives x³+x²-3x+9 = -1+1+3+9 ≠ 0, so -1 is not a root.

Putting x=3 gives x³+x²-3x+9 = 27+9-9+9 ≠ 0, so 3 is not a root.

Putting x=-3 gives x³+x²-3x+9 = -27+9+9+9 = 0, so -3 is a root.

Factoring out (x+3) gives

  x³+x²-3x+9 = (x+3)(x²-2x+3)

so any other roots would have to be roots of x²-2x+3.  Roots of this quadratic must divide 3, so must be ±1 or ±3.  We have already ruled out 1, -1, and 3.

Putting x=-3 gives x²-2x+3 = 9+6+3 ≠ 0, so -3 is not a root.

So x²-2x+3 has no integer roots.

It follows that -3 is the only integer root of x³+x²-3x+9, and it occurs with multiplicity one.