How do you find all the zeros of $f (x) = x^{3} - 6 x^{2} - x + 6$ $f(x) = x^3 - 6x^2 - x + 6$ ?

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Answer 1 · 2016-03-23T00:24:39

Factor by grouping to find zeros: ${ (x=1), (x=-1), (x=6) :}$

Factor by grouping:

$x^3-6x^2-x+6$

$=(x^3-6x^2)-(x-6)$

$=x^2(x-6)-1(x-6)$

$=(x^2-1)(x-6)$

$=(x-1)(x+1)(x-6)$

So the zeros are:

${ (x=1), (x=-1), (x=6) :}$

How do you find all the zeros of f(x) = x^3 - 6x^2 - x + 6 f(x)=x3−6x2−x+6f(x) = x^3 - 6x^2 - x + 6 ?