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

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Answer 1 · 2016-07-09T21:46:15

Find zeros: $x = 1$ $x=1$ , $x = - 1$ $x=-1$ , $x = - 2$ $x=-2$ and $x = - 3$ $x=-3$

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

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

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

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

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

Next note that if you reverse the signs of the coefficients of odd degree in the remaining cubic, then the sum is $0$ $0$ . That is:

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

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

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

To factor the remaining quadratic, note that $2 + 3 = 5$ $2+3 = 5$ and $2 \times 3 = 6$ $2xx3=6$ .

Hence:

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

giving zeros $x = - 2$ $x=-2$ and $x = - 3$ $x=-3$

How do you find all the zeros of f(x)=x^4+5x^3+5x^2-5x-6f(x)=x4+5x3+5x2−5x−6f(x)=x^4+5x^3+5x^2-5x-6?