How do you find all the zeros of $f (x) = x^{3} - 8 x^{2} + 19 x - 12$ $f(x)= x^3 - 8x^2 +19x - 12$ with its multiplicities?

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Answer 1 · 2016-08-11T23:03:16

$f (x)$ $f(x)$ has zeros $1$ $1$ , $3$ $3$ and $4$ $4$ all with multiplicity $1$ $1$ .

$f (x) = x^{3} - 8 x^{2} + 19 x - 12$ $f(x) = x^3-8x^2+19x-12$

First note that the sum of the coefficients is zero. That is:

$1 - 8 + 19 - 12 = 0$ $1-8+19-12 = 0$

Hence $f (1) = 0$ $f(1) = 0$ , $x = 1$ $x=1$ is a zero and $(x - 1)$ $(x-1)$ a factor:

$x^{3} - 8 x^{2} + 19 x - 12$ $x^3-8x^2+19x-12$

$= (x - 1) (x^{2} - 7 x + 12)$ $=(x-1)(x^2-7x+12)$

To factor the remaining quadratic, note that $3 + 4 = 7$ $3+4=7$ and $3 \times 4 = 12$ $3xx4=12$ .

So we find:

$x^{2} - 7 x + 12 = (x - 3) (x - 4)$ $x^2-7x+12 = (x-3)(x-4)$

with associated zeros $x = 3$ $x=3$ and $x = 4$ $x=4$ .

How do you find all the zeros of f(x)= x^3 - 8x^2 +19x - 12f(x)=x3−8x2+19x−12f(x)= x^3 - 8x^2 +19x - 12 with its multiplicities?