Find the zeros, HELP!!

x^3 + 3x^2  + 7x  - 11   =  0

By Vieta

Sum of the roots  = -3 / 1  =   -3

Product of the roots = -(-11) /  1 =  11

$x^3 + 3x^2 + 7x − 11 = 0$

simplifies to:

$(x-1)(x^2+4x+11)=0$

and 

$x^2+4x+11=0$

which gives an imaginary number

but for the complex part, 

you get the roots of x as:

$x=\frac{-4±\sqrt{4^2-4*1*11}}{2}$

whic becomes

$x=2±\sqrt{-7}$

so the roots of 

$x^3 + 3x^2 + 7x − 11 = 0$

are

$x = 1$

and

$x=2±\sqrt{-7}$

In response to your question:

A good way to do this is by the rational root theorem.

For any polynomial, $a_nx^n+a_{n-1}x^{n-1}\dots a_1x^1+a_0$, the rational roots can all be expressed as:$\pm\frac{\text{factor of }a_0}{\text{factor of }a_n}$.

Therefore in this polynomial, the only possible rational roots are $\pm 1, \pm11$. (factors of a₀ are 1, 11 and the only factor of a_n is 1)

Plug these values into the function, to see which ones produce zero. $\pm1$ is a really, really common ones, so always try those first.

Remainder theorem is always helpful. Remember if for a function $f(x)$, $f(a)=0$, then $x-a$ is a factor of $f(x)$.

Here is the workthrough of the first step:

Use rational root theorem, plug in 1 to $f(x)={x}^{3}+3x^{2}+7x-11$.

${1}^{3}+3*{1}^{2}+7*1-11=0$.

Therefore we know because $f(1) = 0$, then x-1 is a factor of $f(x)$, so we can factor it out.