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

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Answer 1 · 2017-02-01T23:51:09

The zeros of $f(x)$ are: $1$ , $1$ , $-1/2$

Given:

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

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

$2-3+1 = 0$

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

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

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

$2-1-1 = 0$

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

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

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

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