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

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Answer 1 · 2016-08-14T13:25:20

$f(x)$ has zeros $-5$ and $+-i$

Since the ratio of the first and second terms is the same as that between the third and fourth terms, this cubic will factor by grouping.

So we find:

$x^3+5x^2+x+5$

$=x^2(x+5)+1(x+5)$

$=(x^2+1)(x+5)$

$=(x^2-i^2)(x+5)$

$=(x-i)(x+i)(x+5)$

Hence zeros: $+-i$ and $-5$

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