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How do you find the zeros of the polynomial function with equation $f (x) = x^{3} + 7 x^{2} - 4 x - 28$ $f(x) = x^3 +7x^2 - 4x - 28$ ?

1 Answer

May 4, 2016

Using the Factor Theorem (hoping or integer factors) we find zeros at $x \in {- 2, + 2, - 7}$ $x in {-2,+2,-7}$

Explanation:

According to the Factor Theorem if $1 x^{3} + 7 x^{2} - 4 x - 28$ $color(blue)(1)x^3+7x^2-4x-color(red)(28)$ as rational zeros
they must be factors of $\frac{28}{1}$ $color(red)(28)/color(blue)(1)$

The possible factors are therefore
$XXX {\pm 1, \pm 2, \pm 4, \pm 7, \pm 14 \pm 28}$ $color(white)("XXX"){+-1,+-2,+-4,+-7,+-14+-28}$

Testing these factors:

we find three zeros (as given in the "Answer" above.

Since the given expression is of degree 3 and we have 3 zeros, these 3 zeros represent all possible solutions.

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