In this case we need to test the remainder `r = -1`.

`R= f(r)`

`= f(-1) ` `= (-1)^3+ 2(-1)^2- 5(-1) - 6` `= -1 + 2 + 5 - 6` `= 0`

`= f(-1) `

`= (-1)^3+ 2(-1)^2- 5(-1) - 6`

`= -1 + 2 + 5 - 6`

`= 0`

Therefore, since `f(-1) = 0`, we conclude that `(x + 1)` is a factor of `f(x)`.

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