Question

Question #9dcdb

Answer 1

Let us find all zeros of `f(x)=x^3-7x+6`.

By inspection, we see that

`f(1)=(1)^3-7(1)+6=0`

So, `x=1` is one of the zeros, which means that `f(x)` has the factor `(x-1)`.

By factoring out `(x-1)`,

`f(x)=(x-1)(x^2+x-6)`

by factoring out a bit further,

`f(x)=(x-1)(x-2)(x+3)=0`

Hence, the zeros of `f` are `x=-3, 1, 2`.

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I hope that this was helpful.