Question

The zeros of f(x) algebraically

Answer 1

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.