How do you find all the zeros of $f(x)=2x^3+5x^2+4x+1$ ?

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Answer 1 · 2016-08-03T22:06:39

$f(x)$ has zeros: $-1, -1, -1/2$

$f(x) = 2x^3+5x^2+4x+1$

Note that:

$f(-1) = -2+5-4+1 = 0$

So $x=-1$ is a zero and $(x+1)$ a factor:

$2x^3+5x^2+4x+1$

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

Substituting $x=-1$ in the remaining quadratic, we find:

$2x^2+3x+1 = 2-3+1 = 0$

So $x=-1$ is a zero again and $(x+1)$ a factor:

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

The final zero is $x = -1/2$

How do you find all the zeros of f(x)=2x^3+5x^2+4x+1f(x)=2x^3+5x^2+4x+1?