# What are all the possible rational zeros for f(x)=2x^3+5x^2+4x+1 and how do you find all zeros?

Sep 11, 2016

the zeroes are. $- 1 , - 1 , - \frac{1}{2}$.

#### Explanation:

$f \left(x\right) = 2 {x}^{3} + 5 {x}^{2} + 4 x + 1$

Note that, the sum of the co-effs. of odd-powered terms of

$x$=2+4=6, and, that of even-powered$= 5 + 1 = 6.$.

$\therefore \left(x + 1\right) \text{ is a factor of } f \left(x\right)$.

Now, $f \left(x\right) = 2 {x}^{3} + 5 {x}^{2} + 4 x + 1$

$= \underline{2 {x}^{2} + 2 {x}^{2}} + \underline{3 {x}^{2} + 3 x} + \underline{x + 1}$

$= 2 {x}^{2} \left(x + 1\right) + 3 x \left(x + 1\right) + 1 \left(x + 1\right)$

$= \left(x + 1\right) \left(2 {x}^{2} + 3 x + 1\right)$

$\left(x + 1\right) \left\{\underline{2 {x}^{2} + 2 x} + \underline{x + 1}\right\}$

$= \left(x + 1\right) \left\{2 x \left(x + 1\right) = 1 \left(x + 1\right)\right\}$

=(x+1){(x+1))(2x+1)}

$= {\left(x + 1\right)}^{2} \left(2 x + 1\right)$

Hence, the zeroes are. $- 1 , - 1 , - \frac{1}{2}$.