How do you find all the zeros of $f(x) = x^3 - 8x^2 - x + 8$ ?

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Answer 1 · 2016-07-02T11:42:56

Factor by grouping to find zeros:

$x=1$ , $x=-1$ , $x=8$

$f(x) = x^3-8x^2-x+8$

Notice that the ratio of the first and second terms is the same as that between the third and fourth. So this cubic factors by grouping:

$x^3-8x^2-x+8$

$=(x^3-8x^2)-(x-8)$

$=x^2(x-8)-1(x-8)$

$=(x^2-1)(x-8)$

$=(x-1)(x+1)(x-8)$

Hence zeros: $x=1$ , $x=-1$ , $x=8$

How do you find all the zeros of f(x) = x^3 - 8x^2 - x + 8f(x) = x^3 - 8x^2 - x + 8?