How do you find the zeros of $f(x) = x^3 − 5x^2 − 3x + 15$ ?

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Answer 1 · 2016-08-12T16:18:57

$f(x)$ has zeros $+-sqrt(3)$ and $5$

$f(x) = x^3-5x^2-3x+15$

Notice that the ratio of the first and second terms is the same as the ratio between the third and fourth terms.

So this cubic will factor by grouping:

$x^3-5x^2-3x+15$

$=(x^3-5x^2)-(3x-15)$

$=x^2(x-5)-3(x-5)$

$=(x^2-3)(x-5)$

$=(x-sqrt(3))(x+sqrt(3))(x-5)$

Hence zeros:

$+-sqrt(3)$ and $5$

How do you find the zeros of f(x) = x^3 − 5x^2 − 3x + 15f(x) = x^3 − 5x^2 − 3x + 15?