# How do you find all the zeros of P(X) = x^4 - 4x^3 - 35x^2 + 6x + 144?

##### 1 Answer
Feb 26, 2016

Zeros of x^4−4x^3−35x^2+6x+144 are $- 3$, $2$ and $8$

#### Explanation:

To find all the zeros of P(X)=x^4−4x^3−35x^2+6x+144, first find a factor of the independent term $144$ such as

$\left\{1 , - 1 , 2 , - 2 , 3 , - 3 , 4 , - 4 , 6 , - 6 , 8 , - 8 , \ldots .\right\}$ for which $P \left(x\right) = 0$.

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

Hence, $\left(x - 2\right) \left(x + 3\right)$ or ${x}^{2} + x - 6$ divides $P \left(x\right)$. Dividing latter by former, we get

(x^4−4x^3−35x^2+6x+144)/(x^2+x-6) = ${x}^{2} - 5 x - 24$

As the RHS can be factorized to $\left(x - 8\right) \left(x + 3\right)$, $8$ and $- 3$ are other two zeros of x^4−4x^3−35x^2+6x+144. But $- 3$ has come earlier.

Hence zeros of x^4−4x^3−35x^2+6x+144 are $- 3$, $2$ and $8$.