Alg3.6.2

From Exampleproblems

Solve for $y: y^2+7y+12=0\,$

Multiplying the coefficient of $y^2\,$ and the constant gives $12\,$ The different factors for $12\,$ are $(4,3)(-4,-3)(12,1)(-12,-1)(6,2)(-6,-2)\,$ The combination which suits the coefficient of $+7\,$ is $(4,3)\,$ .

Now splitting the $y\,$ coefficient,the equation looks like $y^2+4y+3y+12=0\,$

Taking out the common in the first two and the last two terms $y(y+4)+3(y+4)=0\,$

Taking out the common terms in the above $(y+3)(y+4)=0\,$

Final solution is $(y+3)=0\,$ or $(y+4)=0\,$

$y=-3\,$ or $y=-4\,$