given `x^3-kx^2-6x+8` find `k` such that x+2 will be a factor:

We use synthetic division; if x+2 is a factor , x=-2 is a zero:

-2 | 1 -k -6 8

------------------------

1 -k-2 6k-2 -4k+12

So if -2 is a zero ( (x+2) a factor) then -4k+12=0 or k=3.

**Thus the value we need is k=3**

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If k=3 we have:

`x^3-3x^2-6x+8=(x+2)(x-4)(x-1)` , verifying that x+2 is a factor.

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