How do you find the rest of the zeros given one of the zero $c=-1$ and the function $f(x)=x^5+2x^4-12x^3-38x^2-37x-12$ ?

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Answer 1 · 2016-08-01T06:15:04

The zeros of $f(x)$ are $x=-1$ with multiplicity $3$ , $x=4$ and $x=-3$ .

$f(x) = x^5+2x^4-12x^3-38x^2-37x-12$

$f(-1) = -1+2+12-38+37-12 = 0$

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

$x^5+2x^4-12x^3-38x^2-37x-12=(x+1)(x^4+x^3-13x^2-25x-12)$

Substituting $x=-1$ in the remaining quartic factor, we get:

$1-1-13+25-12 = 0$

So the quartic also has $x = -1$ as a zero and $(x+1)$ as a factor:

$x^4+x^3-13x^2-25x-12 = (x+1)(x^3-13x-12)$

Substituting $x = -1$ in the remaining cubic factor, we get:

$-1+13-12 = 0$

So the cubic also has $x = -1$ as a zero and $(x+1)$ as a factor:

$x^3-13x-12 = (x+1)(x^2-x-12)$

To factor the remaining quadratic, note that $4*3 = 12$ and $4-3=1$ , so:

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

hence zeros $x=4$ and $x=-3$

So the zeros of $f(x)$ are $x=-1$ with multiplicity $3$ , $x=4$ and $x=-3$ .

How do you find the rest of the zeros given one of the zero c=-1c=-1 and the function f(x)=x^5+2x^4-12x^3-38x^2-37x-12f(x)=x^5+2x^4-12x^3-38x^2-37x-12?