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

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Answer 1 · 2017-02-01T23:51:09

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

Given:

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

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

$2 - 3 + 1 = 0$ $2-3+1 = 0$

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

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

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

$2 - 1 - 1 = 0$ $2-1-1 = 0$

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

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

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

How do you find all zeroes for f(x)=2x3−3x2+1f(x)=2x^3-3x^2+1?