Question

factor.

$4y^2 + 12y - 7$

Explanation:

Step1: Find two numbers for ac method

For quadratic \(ax^2 + bx + c\) (here \(a = 4\), \(b = 12\), \(c=-7\)), calculate \(ac=4\times(-7)= - 28\). We need two numbers that multiply to \(-28\) and add to \(12\). The numbers are \(14\) and \(-2\) (since \(14\times(-2)=-28\) and \(14 + (-2)=12\)).

Step2: Rewrite the middle term

Rewrite \(12y\) as \(14y-2y\): \(4y^{2}+14y - 2y-7\).

Step3: Group and factor

Group the first two and last two terms: \((4y^{2}+14y)+(-2y - 7)\). Factor out common factors from each group: \(2y(2y + 7)-1(2y + 7)\).

Step4: Factor out the common binomial

Factor out \((2y + 7)\): \((2y + 7)(2y - 1)\).

Answer:

\((2y + 7)(2y - 1)\)