Question

factor: (greatest coefficient first. if coefficients are the same, least valued constant first.)
$x^2 + 3x - 18$
$= a (bx + c) (dx + e)$
$= \boxed{\quad} (\boxed{\quad} x + \boxed{\quad})(\boxed{\quad} x + \boxed{\quad})$

Explanation:

Step1: Identify the form

The quadratic is \(x^2 + 3x - 18\), which is in the form \(ax^2+bx+c\) with \(a = 1\), \(b = 3\), \(c=-18\).

Step2: Find two numbers

Find two numbers that multiply to \(a\times c=1\times(-18)= - 18\) and add up to \(b = 3\). The numbers are \(6\) and \(- 3\) since \(6\times(-3)=-18\) and \(6+( - 3)=3\).

Step3: Rewrite the middle term

Rewrite the quadratic as \(x^2+6x - 3x-18\).

Step4: Factor by grouping

Group the terms: \((x^2 + 6x)+(-3x - 18)=x(x + 6)-3(x + 6)\).

Step5: Factor out the common binomial

Factor out \((x + 6)\) to get \((x + 6)(x-3)\). Since \(a = 1\) (the coefficient of \(x^2\) is \(1\)), the factored form is \(1(x + 6)(x-3)\).

Answer:

\(1\) \((x + 6)\) \((x-3)\) (filling the boxes as: first box \(1\), second box \(1\), third box \(6\), fourth box \(1\), fifth box \(- 3\))