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

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Answer 1 · 2016-08-03T22:06:39

$f (x)$ $f(x)$ has zeros: $- 1, - 1, - \frac{1}{2}$ $-1, -1, -1/2$

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

Note that:

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

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

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

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

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

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

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

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

The final zero is $x = - \frac{1}{2}$ $x = -1/2$

How do you find all the zeros of f(x)=2x^3+5x^2+4x+1f(x)=2x3+5x2+4x+1f(x)=2x^3+5x^2+4x+1?