Question

question 9 of 25
what is the factorization of the trinomial below?
$x^2 - 3x - 18$

a. $(x + 2)(x - 9)$
b. $(x + 3)(x - 6)$
c. $(x + 9)(x - 2)$
d. $(x + 6)(x - 3)$

Explanation:

Step1: Recall factoring trinomial \(x^2 + bx + c\)

We need two numbers that multiply to \(c = - 18\) and add up to \(b=-3\).

Step2: Find the two numbers

Let the numbers be \(m\) and \(n\) such that \(m\times n=-18\) and \(m + n=-3\).
We check pairs of factors of \(- 18\):

• For \(m = 3\) and \(n=-6\): \(3\times(-6)=-18\) and \(3+( - 6)=-3\)

Step3: Factor the trinomial

Using the formula \(x^{2}+bx + c=(x + m)(x + n)\) where \(m = 3\) and \(n=-6\), we get \(x^{2}-3x - 18=(x + 3)(x-6)\)

Answer:

B. \((x + 3)(x - 6)\)