How do you find all the zeros of f(x)= x^4 + 3x^3 - 4x^2 - 12x with its multiplicities?

Jul 16, 2016

Zeros: $0 , 2 , - 2 , - 3$

Explanation:

$f \left(x\right) = {x}^{4} + 3 {x}^{3} - 4 {x}^{2} - 12 x$

Note that all of the terms are divisible by $x$ and the ratio of the first and second terms is the same as that between the third and fourth terms. So this will factor by grouping...

${x}^{4} + 3 {x}^{3} - 4 {x}^{2} - 12 x$

$= x \left(\left({x}^{3} + 3 {x}^{2}\right) - \left(4 x + 12\right)\right)$

$= x \left({x}^{2} \left(x + 3\right) - 4 \left(x + 3\right)\right)$

$= x \left({x}^{2} - 4\right) \left(x + 3\right)$

$= x \left({x}^{2} - {2}^{2}\right) \left(x + 3\right)$

$= x \left(x - 2\right) \left(x + 2\right) \left(x + 3\right)$

Hence zeros:

$0 , 2 , - 2 , - 3$