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The zeros of f(x) algebraically

The zeros of f(x) algebraically-example-1

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Answer:

The zeros are 4, -6, and 1.

Explanation:

Given f(x) = x³ + x² - 26x + 24

(x - 4) is a factor of f(x). That means it is a zero of f(x).

To find the remaining factors algebraically, we take out the factor (x - 4) from f(x).

That is,
$ f(x) = x^3 + x^2 - 26x + 24 $


$ \implies x^3 - 4x^2 + 5x^2 - 20x - 6x + 24 $

Taking
$ x^2 $ out, we have:


$ = x^2(x^2 - 4) + 5x(x - 4) - 6(x - 4) $

Taking (x - 4) common out, we have:


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


$ = (x - 4)(x^2 + 6x - x - 6) $


$ = (x - 4)\{x(x + 6) -1(x + 6)\} $


$ = (x - 4)(x + 6)(x - 1) $

This means the zeros are 4, -6, & 1.

User Jaimie Sirovich
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