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

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Answer 1 · 2016-03-22T02:53:45

$x = - 8, - 1, 1$ $x=-8,-1,1$

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

$1 + 8 - 1 - 8 = 0$ $1+8-1-8=0$

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

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

$\frac{x^{3} + 8 x^{2} - x - 8}{x - 1} = x^{2} + 9 x + 8$ $(x^3+8x^2-x-8)/(x-1)=x^2+9x+8$

To find the remaining zeros of

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

Factor the quadratic.

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

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

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

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

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

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)=x3+8x2−x−8f(x) = x^3+ 8x^2- x - 8?