# How do you find all the zeros of f(x) = x^4 + 2x^3 + 22x^2 + 50x - 75?

Feb 26, 2016

#### Answer:

Zeros of ${x}^{4} + 2 {x}^{3} + 22 {x}^{2} + 50 x - 75$ are $- 3$ and $1$

#### Explanation:

To find all the zeros of $P \left(X\right) = {x}^{4} + 2 {x}^{3} + 22 {x}^{2} + 50 x - 75$, first find a factor of the independent term $75$ such as

$\left\{1 , - 1 , 3 , - 3 , 5 , - 5 , 15 , - 15 , \ldots .\right\}$ for which $P \left(x\right) = 0$.

It is apparent that for $x = 1$ as well as for $x = - 3$, $P \left(x\right) = 0$. Hence $1$ and $- 3$ are two zeros of x^4−4x^3−35x^2+6x+144

Hence, $\left(x - 1\right) \left(x + 3\right)$ or ${x}^{2} + 2 x - 3$ divides $P \left(x\right)$. Dividing latter by former, we get

$\frac{{x}^{4} + 2 {x}^{3} + 22 {x}^{2} + 50 x - 75}{{x}^{2} + 2 x - 3}$ = ${x}^{2} + 25$

As the RHS cannot be factorized in real factors there are no further zeros.

Hence zeros of ${x}^{4} + 2 {x}^{3} + 22 {x}^{2} + 50 x - 75$ are $- 3$ and $1$