# What are all the possible rational zeros for f(x)=x^3+x^2-5x+3 and how do you find all zeros?

Nov 23, 2016

1,1, -3

#### Explanation:

Substituting x=1 makes f(x)= 1+1-5+3 =0 hence x=1 is one of the rational zeros.

Since x=1 is a zero, x-1 would be a factor of f(x). therefore divide f(x) by x-1 by using long or synthetic division.

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

Now for finding other zeros of f(x), put ${x}^{2} + 2 x - 3 = 0$ and solve for x.

This quadratic equation can be solved by factorisation or using quadratic formula.

Factorisation is quite easy. Writing 2x as 3x-x, the quadratic expression becomes ${x}^{2} + 3 x - x - 3 = 0$

$x \left(x + 3\right) - 1 \left(x + 3\right) = 0$

(x-1)(x+3)=0 giving x= 1, -3

Thus all the zeros of f(x) are now known. These are, 1( which is a repeated zero) and -3