Sure! Let's factor the polynomial [tex]\(x^2 + 3x - 18\)[/tex] completely.
To factor [tex]\(x^2 + 3x - 18\)[/tex], we first need to find two numbers that multiply to the constant term, [tex]\(-18\)[/tex], and add up to the coefficient of the linear term, [tex]\(3\)[/tex].
1. The polynomial is in the form [tex]\(ax^2 + bx + c\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = -18\)[/tex].
2. We need two numbers that multiply to [tex]\(ac = 1 \times -18 = -18\)[/tex] and add up to [tex]\(b = 3\)[/tex].
Let's list the pairs of factors of [tex]\(-18\)[/tex]:
- [tex]\((1, -18)\)[/tex] adds to [tex]\(-17\)[/tex]
- [tex]\((-1, 18)\)[/tex] adds to [tex]\(17\)[/tex]
- [tex]\((2, -9)\)[/tex] adds to [tex]\(-7\)[/tex]
- [tex]\((-2, 9)\)[/tex] adds to [tex]\(7\)[/tex]
- [tex]\((3, -6)\)[/tex] adds to [tex]\(-3\)[/tex]
- [tex]\((-3, 6)\)[/tex] adds to [tex]\(3\)[/tex]
We can see that [tex]\((-3, 6)\)[/tex] are the numbers we are looking for because they multiply to [tex]\(-18\)[/tex] and add up to [tex]\(3\)[/tex].
3. Now, we write the middle term [tex]\(3x\)[/tex] as [tex]\(-3x + 6x\)[/tex]:
[tex]\[ x^2 + 3x - 18 = x^2 - 3x + 6x - 18 \][/tex]
4. Group the terms:
[tex]\[ (x^2 - 3x) + (6x - 18) \][/tex]
5. Factor each group:
[tex]\[ x(x - 3) + 6(x - 3) \][/tex]
6. Notice that [tex]\((x - 3)\)[/tex] is a common factor:
[tex]\[ (x - 3)(x + 6) \][/tex]
So, the polynomial [tex]\(x^2 + 3x - 18\)[/tex] factors to:
[tex]\[ (x - 3)(x + 6) \][/tex]
Therefore, the correct answer is:
A. [tex]\((x-3)(x+6)\)[/tex]
