How do you find the zeros of f(x) = x^3 + 4x^2 - 25x - 100?

Mar 23, 2016

There are three zeros:
$x = - 5 \text{ or " x = 5 " or } x = - 4$.

Explanation:

For example, here, you can notice that $25$ and $100$ are both dividable by $25$, so you could try to to factor $25$ and find the following factorization:

${x}^{3} + 4 {x}^{2} - 25 x - 100 = {x}^{2} \left(x + 4\right) - 25 \left(x + 4\right) = \left({x}^{2} - 25\right) \left(x + 4\right)$

Now you can use the identity ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$ to factorize further:

$\ldots = \left(x + 5\right) \left(x - 5\right) \left(x + 4\right)$

Now,

$f \left(x\right) = 0$

$\iff {x}^{3} + 4 {x}^{2} - 25 x - 100 = 0$

$\iff \left(x + 5\right) \left(x - 5\right) \left(x + 4\right) = 0$

A product is equal to zero if one or more factors is/are equal to zero:

$\iff x + 5 = 0 \text{ or " x-5 = 0 " or } x + 4 = 0$

$\iff x = - 5 \text{ or " x = 5 " or } x = - 4$

Mar 23, 2016

$\left\{\begin{matrix}x = - 5 \\ x = 5 \\ x = - 4\end{matrix}\right.$

Explanation:

Factor by grouping:

${x}^{3} + 4 {x}^{2} - 25 x - 100$

$= \left({x}^{3} + 4 {x}^{2}\right) - \left(25 x + 100\right)$

$= {x}^{2} \left(x + 4\right) - 25 \left(x + 4\right)$

$= \left({x}^{2} - 25\right) \left(x + 4\right)$

$= \left(x + 5\right) \left(x - 5\right) \left(x + 4\right)$

This gives the solutions

$\left\{\begin{matrix}x = - 5 \\ x = 5 \\ x = - 4\end{matrix}\right.$