# How do you find all zeroes for f(x)=2x^3-3x^2+1?

Feb 1, 2017

The zeros of $f \left(x\right)$ are: $1$, $1$, $- \frac{1}{2}$

#### Explanation:

Given:

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

First note that the sum of the coefficients is $0$. That is:

$2 - 3 + 1 = 0$

Hence $f \left(1\right) = 0$ and $\left(x - 1\right)$ is a factor:

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

Note that the sum of the coefficients of the remaining quadratic is also zero:

$2 - 1 - 1 = 0$

So $x = 1$ is a zero again and $\left(x - 1\right)$ a factor again:

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

From the last linear factor we can see that the remaining zero is $x = - \frac{1}{2}$