Certainly! Let's factor the polynomial [tex]\(x^2 + 3x - 18\)[/tex] completely.
To factor a quadratic polynomial of the form [tex]\(ax^2 + bx + c\)[/tex], we look for two numbers that multiply to [tex]\(c\)[/tex] (the constant term, which is [tex]\(-18\)[/tex] in this case) and add up to [tex]\(b\)[/tex] (the coefficient of [tex]\(x\)[/tex], which is [tex]\(3\)[/tex] in this case).
Here are the steps:
1. Identify the coefficients:
- [tex]\(a = 1\)[/tex] (coefficient of [tex]\(x^2\)[/tex])
- [tex]\(b = 3\)[/tex] (coefficient of [tex]\(x\)[/tex])
- [tex]\(c = -18\)[/tex] (constant term)
2. Find two numbers that multiply to [tex]\(-18\)[/tex] and add up to [tex]\(3\)[/tex]:
- The pairs of factors of [tex]\(-18\)[/tex] are:
- [tex]\( (1, -18) \)[/tex]
- [tex]\((-1, 18)\)[/tex]
- [tex]\( (2, -9) \)[/tex]
- [tex]\((-2, 9)\)[/tex]
- [tex]\( (3, -6) \)[/tex]
- [tex]\((-3, 6)\)[/tex]
- Out of these pairs, [tex]\((6, -3)\)[/tex] gives us:
- [tex]\(6 \times (-3) = -18\)[/tex] (product is [tex]\(-18\)[/tex])
- [tex]\(6 + (-3) = 3\)[/tex] (sum is [tex]\(3\)[/tex])
3. Write the polynomial in factored form using these numbers:
- So, [tex]\(x^2 + 3x - 18\)[/tex] can be factored as [tex]\((x - 3)(x + 6)\)[/tex].
Therefore, the completely factored form of the polynomial [tex]\(x^2 + 3x - 18\)[/tex] is:
[tex]\[
(x - 3)(x + 6)
\][/tex]
So the correct answer is:
C. [tex]\((x - 3)(x + 6)\)[/tex]
