How do you find all the zeros of $f(x)=x^4+5x^3+5x^2-5x-6$ ?

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Answer 1 · 2016-03-15T22:38:21

Look at coefficient sums and divide by the factors found to simplify the problem and find zeros:

$x=1$ , $x=-1$ , $x=-2$ and $x=-3$

First note that the sum of the coefficients is zero.

That is: $1+5+5-5-6 = 0$

So $f(1) = 0$ and $(x-1)$ is a factor:

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

Next note the if you reverse the signs of the terms of the remaining cubic factor with odd degree then the sum of the coefficients is zero.

That is $-1+6-11+6 = 0$

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

$x^3+6x^2+11x+6 = (x+1)(x^2+5x+6)$

Then note that $2+3 = 5$ and $2 xx 3 = 6$ , so the remaining quadratic factor factorises as follows:

$x^2+5x+6 = (x+2)(x+3)$

Putting this all together, we find:

$f(x) = (x-1)(x+1)(x+2)(x+3)$

with zeros $x=1$ , $x=-1$ , $x=-2$ and $x=-3$

How do you find all the zeros of f(x)=x^4+5x^3+5x^2-5x-6f(x)=x^4+5x^3+5x^2-5x-6?