How do you find all the zeros of $f(x) = x^3+ 8x^2- x - 8$ ?

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Answer 1 · 2016-03-22T02:53:45

$x=-8,-1,1$

Notice that the coefficients of each term, when added, equal $0$ :

$1+8-1-8=0$

This means that $f(1)=0$ , so $1$ is a zero of the function and $x-1$ is a factor.

Use polynomial long division or synthetic division to determine that the other two zeros can be found from:

$(x^3+8x^2-x-8)/(x-1)=x^2+9x+8$

To find the remaining zeros of

$x^2+9x+8=0$

Factor the quadratic.

$(x+1)(x+8)=0$

Thus, if either of these terms equals $0$ ,

$x+1=0" "=>" "x=-1$

$x+8=0" "=>" "x=-8$

So, the function's three zeros are located at $x=-8,-1,1$ .

We can check a graph of the function:

graph{x^3+8x^2-x-8 [-12, 4, -33.4, 85.3]}

How do you find all the zeros of f(x) = x^3+ 8x^2- x - 8f(x) = x^3+ 8x^2- x - 8?