Question

1. ( x^{3}+2x^{2}+9x + 18 = 0 )

Explanation:

Step1: Group the terms

We group the first two terms and the last two terms:
\( (x^3 + 2x^2) + (9x + 18) = 0 \)

Step2: Factor out the GCF from each group

Factor out \( x^2 \) from the first group and \( 9 \) from the second group:
\( x^2(x + 2) + 9(x + 2) = 0 \)

Step3: Factor out the common binomial factor

Now we can factor out \( (x + 2) \) from both terms:
\( (x + 2)(x^2 + 9) = 0 \)

Step4: Set each factor equal to zero and solve

For \( x + 2 = 0 \), we get \( x = -2 \).
For \( x^2 + 9 = 0 \), we have \( x^2 = -9 \), so \( x = \pm 3i \) (where \( i \) is the imaginary unit, \( i^2 = -1 \)).

Answer:

The solutions are \( x = -2 \), \( x = 3i \), and \( x = -3i \).