# What are all the possible rational zeros for f(x)=x^3+13x^2+38x+24 and how do you find all zeros?

##### 1 Answer
Sep 16, 2017

Zeros are $- 3 , - 9.123 , - 0.877$

#### Explanation:

Possible rational zeros are found as under ,

Factors of constant term $24$ are +-( 1,2,3,4,6,8,12,24)/

factor of leading coefficient $1$ is $\pm 1$ , so possible

rational zeros are $\pm \left(1 , 2 , 3 , 4 , 6 , 8 , 12 , 24\right)$. By trying long divisions

$- 3$ suits only . $f \left(x\right) = {x}^{3} + 13 {x}^{2} + 38 x + 24$ or

$f \left(x\right) = {x}^{3} + 3 {x}^{2} + 10 {x}^{2} + 30 x + 8 x + 24$ or

$f \left(x\right) = {x}^{2} \left(x + 3\right) + 10 x \left(x + 3\right) + 8 \left(x + 3\right)$ or

$f \left(x\right) = \left(x + 3\right) \left({x}^{2} + 10 x + 8\right)$, for $\left({x}^{2} + 10 x + 8\right)$

$x = \frac{- 10 \pm \sqrt{100 - 32}}{2}$ or

$x \approx - 9.123 , x \approx - 0.877 \left(3 \mathrm{dp}\right) \therefore$

$f \left(x\right) = {x}^{3} + 13 {x}^{2} + 38 x + 24 = \left(x + 3\right) \left(x + 9.123\right) \left(x + 0.877\right)$

Zeros are $- 3 , - 9.123 , - 0.877$ [Ans]