How do you find the rest of the zeros given one of the zero $c = \frac{2}{3}$ $c=2/3$ and the function $f (x) = 3 x^{3} + 4 x^{2} - x - 2$ $f(x)=3x^3+4x^2-x-2$ ?

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Answer 1 · 2016-12-03T19:03:57

The zeros of $f (x)$ $f(x)$ are $\frac{2}{3}$ $2/3$ (with multiplicity $1$ $1$ ) and $- 1$ $-1$ with multiplicity $2$ $2$

The given zero corresponds to a linear factor $(3 x - 2)$ $(3x-2)$ , so $f (x)$ $f(x)$ will divide evenly as:

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

We can recognise the remaining quadratic factor as the perfect square trinomial:

$x^{2} + 2 x + 1 = {(x + 1)}^{2}$ $x^2+2x+1 = (x+1)^2$

So the other zeros are both $- 1$ $-1$

How do you find the rest of the zeros given one of the zero c=23c=2/3 and the function f(x)=3x3+4x2−x−2f(x)=3x^3+4x^2-x-2?