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

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Answer 1 · 2016-09-14T15:01:12

$x=1$

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

$x=1$ is zero. Divide $2x^3-3x^2+1$ at x-1. It will turn: $2x^2+x+1$

$2x^2+x+1=0$ have not rational zeros (discriminant<0)

Answer 2 · 2016-09-14T15:09:31

The Zeroes of $f(x)$ are $1,1,-1/2$ .

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

Here, the sum of the co-effs. $=2-3+1=0.$

Hence, $(x-1)$ is a factor of $f(x)$ . Rewriting $f(x)$ as,

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

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

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

In the quadr., we have the same story $[2-1-1=0]$ !

$:. f(x)=(x-1){ul(2x^2-2x)+ul(x-1)}$

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

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

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

Clearly, the Zeroes of $f(x)$ are $1,1,-1/2$ .

Enjoy Maths.!

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