Identify all of the root(s) of g(x) = (x² - 3x - 4)(x² - 4x + 29).

Solution:

Given: Function g(x) = (x² - 3x - 4)(x² - 4x + 29)

Let us first consider x² - 3x - 4

By splitting the middle term

= x² + 1x - 4x - 4

= x(x + 1) - 4(x + 1)

= (x + 1)(x - 4)

Now let us consider x² - 4x + 29

First group the terms with variable on LHS and move the constant on the other side

x² - 4x = -29

Add 4 on both sides

x² - 4x + 4 = -29 + 4

x² - 4x + 4 = -25

It can be written as perfect squares

(x - 2)² = -25

We know that

i =√1

Take square root on both sides

x - 2 = ± 5i

x = 2 ± 5i

So we get,

x = 2 + 5i and x = 2 - 5i

g(x) = (x + 1)(x - 4)(2 + 5i)(2 - 5i)

Therefore, the roots are (x + 1), (x - 4), (2 + 5i), and (2 - 5i).

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Identify all of the root(s) of g(x) = (x² - 3x - 4)(x² - 4x + 29).

Summary:

All the root(s) of g(x) = (x² - 3x - 4)(x² - 4x + 29) are (x + 1), (x - 4), (2 + 5i), and (2 - 5i).