Question

factor.
$7y^2 + 3y - 4$

Explanation:

Step1: Find two numbers

We need two numbers that multiply to \(7\times(-4)= -28\) and add up to \(3\). The numbers are \(7\) and \(-4\) since \(7\times(-4)= -28\) and \(7 + (-4)=3\).

Step2: Rewrite the middle term

Rewrite \(3y\) as \(7y - 4y\) in the expression \(7y^{2}+3y - 4\), so we get \(7y^{2}+7y - 4y - 4\).

Step3: Group and factor

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

Answer:

\((7y - 4)(y + 1)\)