Question

factor: (greatest coefficient first. if coefficients are the same, least valued constant first.)
$5x^2 + 19x + 12$
$= a (bx + c) (dx + e)$
$= \boxed{\quad} (\boxed{\quad} x + \boxed{\quad})(\boxed{\quad} x + \boxed{\quad})$

Explanation:

Step1: Find factors of \(5\times12\)

We need two numbers that multiply to \(5\times12 = 60\) and add up to \(19\). The numbers are \(15\) and \(4\) since \(15\times4 = 60\) and \(15 + 4=19\).

Step2: Rewrite the middle term

Rewrite \(19x\) as \(15x + 4x\) in the quadratic: \(5x^{2}+15x + 4x+12\).

Step3: Group and factor

Group the first two and last two terms: \((5x^{2}+15x)+(4x + 12)\). Factor out common factors from each group: \(5x(x + 3)+4(x + 3)\). Now factor out \((x + 3)\): \((5x + 4)(x + 3)\). Since the leading coefficient is \(5\) (and we can consider \(a = 1\) here as the factoring gives \((5x + 4)(x + 3)\) which matches \(a(bx + c)(dx + e)\) with \(a = 1\), \(b = 5\), \(c = 4\), \(d = 1\), \(e = 3\)).

Answer:

\(1\) \((5x + 4)\) \((x + 3)\) (filling the boxes as \(1\), \(5\), \(4\), \(1\), \(3\) respectively in the order of the boxes: first box \(1\), second \(5\), third \(4\), fourth \(1\), fifth \(3\))