Question

How do you use synthetic division to find factors for `x^3 + 2x^2 - 5x - 6`?

Answer 1

Noting that `(x-2)` is a factor [see explanation], we can use synthetic division to reduce `x^3+2x-5x-6` to a more easily factored expression.
Complete factors: `(x-2)(x+3)(x+1)`

Explanation:

Part 1: The initial Factor
If `x^3+2x-5x-6 = 0`
then setting the negative terms on one side and the positive terms on the other:
`color(white)("XX")x^3+2x^2=5x+6`
then testing a few values:

#{: (x,color(white)("XX"),x^3+2x^2,color(white)("XX"),5x+6), (0,color(white)("XX"),0,color(white)("XX"),6), (1,color(white)("XX"),3,color(white)("XX"),11), (2,color(white)("XX"),16,color(white)("XX"),16) :}#
So if `x=2` then `x^3+2x^2-5x-6 =0`
`rarr (x-2)` is a factor of `x^3+2x^2-5x-6`

Part 2: Use of synthetic division

So
`color(white)("XX")x^2+2x^2-5x-6 = (x-2)(x^2+4x+3)`

Part 3: Factoring the remaining quadratic
By observation or using the quadratic formula we can factor:
`color(white)("XX")x^2+4x+3=(x+3)(x+1)`

Part 4: Summarize results
`x^3+2x^2-5x-6 = (x-2)(x+3)(x+1)`