How do you find all the zeros of $f(x)=x^5+4x^4+5x^3$ ?

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Answer 1 · 2016-02-26T07:13:45

$f(x) = x^5+4x^4+5x^3 =x^3(x+2-i)(x+2+i)$

Hence the zeros of $f(x)$ are $x=0$ and $x=-2+-i$ .

First separate out the common factor $x^3$ .

Then complete the square and use the difference of squares identity:

$a^2-b^2 = (a-b)(a+b)$

with $a=x+2$ and $b=i$ , as follows:

$f(x) = x^5+4x^4+5x^3$

$=x^3(x^2+4x+5)$

$=x^3(x^2+4x+4+1)$

$=x^3((x+2)^2-i^2)$

$=x^3((x+2)-i)((x+2)+i)$

$=x^3(x+2-i)(x+2+i)$

Hence the zeros of $f(x)$ are $x=0$ and $x=-2+-i$ .

How do you find all the zeros of f(x)=x^5+4x^4+5x^3f(x)=x^5+4x^4+5x^3?