How do you find all zeros with multiplicities of $f (x) = x^{3} + 4 x^{2} - 11 x + 6$ $f(x)=x^3+4x^2-11x+6$ ?

Question

2522 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-04T01:14:44

The zeros of $f (x)$ $f(x)$ are $1$ $1$ with multiplicity $2$ $2$ and $- 6$ $-6$ with multiplicity $1$ $1$ .

Given:

$f (x) = x^{3} + 4 x^{2} - 11 x + 6$ $f(x) = x^3+4x^2-11x+6$

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

$1 + 4 - 11 + 6 = 0$ $1+4-11+6 = 0$

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

$x^{3} + 4 x^{2} - 11 x + 6 = (x - 1) (x^{2} + 5 x - 6)$ $x^3+4x^2-11x+6 = (x-1)(x^2+5x-6)$

The sum of the coefficients of the remaining quadratic is also $0$ $0$ , so $x = 1$ $x=1$ is a zero again and $(x - 1)$ $(x-1)$ a factor:

$x^{2} + 5 x - 6 = (x - 1) (x + 6)$ $x^2+5x-6 = (x-1)(x+6)$

So the remaining zero is $x = - 6$ $x=-6$ .

So the zeros of $f (x)$ $f(x)$ are $1$ $1$ with multiplicity $2$ $2$ and $- 6$ $-6$ with multiplicity $1$ $1$ .

How do you find all zeros with multiplicities of f(x)=x3+4x2−11x+6f(x)=x^3+4x^2-11x+6?