How do you find all the zeros of $f(x)=x^3+x^2-5x+3$ ?

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Answer 1 · 2016-05-19T22:57:05

$x^3+x^2-5x+3 = (x-1)(x-1)(x+3)$

$f(x) = x^3+x^2-5x+3$

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

$1+1-5+3 = 0$

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

$x^3+x^2-5x+3 = (x-1)(x^2+2x-3)$

Again note that the sum of the coefficients of $x^2+2x-3$ is $0$ , so there is another factor $(x-1)$

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

Putting it all together:

$x^3+x^2-5x+3 = (x-1)(x-1)(x+3)$

How do you find all the zeros of f(x)=x^3+x^2-5x+3 f(x)=x^3+x^2-5x+3?