How do you find all zeros with multiplicities of $f (x) = x^{3} - 2 x^{2} - 5 x + 6$ $f(x)=x^3-2x^2-5x+6$ ?

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Answer 1 · 2017-08-22T19:21:27

The zeros of $f (x)$ $f(x)$ are $1, 3, - 2$ $1, 3, -2$ , each with multiplicity $1$ $1$ .

Given:

$f (x) = x^{3} - 2 x^{2} - 5 x + 6$ $f(x) = x^3-2x^2-5x+6$

Note that the sum of the coefficients is $0$ $0$ . That is:

$1 - 2 - 5 + 6 = 0$ $1-2-5+6 = 0$

Hence we find $f (1) = 0$ $f(1) = 0$ and $(x - 1)$ $(x-1)$ is a factor:

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

To factor the remaining quadratic find a pair of factors of $6$ $6$ which differ by $1$ $1$ . The pair $3, 2$ $3, 2$ works, so we find:

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

So the zeros of $f (x)$ $f(x)$ are $1, 3, - 2$ $1, 3, -2$ , each with multiplicity $1$ $1$ .

How do you find all zeros with multiplicities of f(x)=x3−2x2−5x+6f(x)=x^3-2x^2-5x+6?