Alg3.6.1

From Exampleproblems

Solve for $x: 2x^2-5x+2=0\,$

Multiplying the coefficient of $x^2\,$ and constant gives you 4, Different combinations of factors for 4 are $(4,1),(2,2),(-4,-1),(-2,-2)\,$

Take the third combination $(-4,-1)\,$ , which gives on addition -5. Now splitting the $x\,$ term into two terms. $2x^2-4x-1x+2=0\,$

Now taking out the common terms in the first two and the next two gives $2x(x-2)-1(x-2)=0\,$

$(2x-1)(x-2)=0\,$

$(2x-1)=0\,$ or $(x-2)=0\,$

The solution is $2x=1=0\,$ or $x=2\,$

The final solution is $x=1/2\,$ or $x=2\,$