Question

How do you factor 80y^2-4y-4? or is it prime?

Answer 1

`80y^2-4y-4=4(20y^2-5y+4y-1)=4[5y(4y-1)-(4y-1)]=\\ \\=4(4y-1)(5y-1)`

Answer 2

`80y^2-4y-4\\\\a=80;\ b=-4;\ c=-4\\\\\Delta=b^2-4ac\\\\\Delta=(-4)^2-4\cdot80\cdot(-4)=16+1280=1296 > 0\\\\y_1=(-b-\sqrt\Delta)/(2a);\ y_2=(-b+\sqrt\Delta)/(2a)\\\\\sqrt\Delta=√(1296)=36\\\\y_1=(4-36)/(2\cdot80)=(-32)/(160)=-(1)/(5);\ y_2=(4+36)/(2\cdot80)=(40)/(160)=(1)/(4)`



`80y^2-4y-4=80(y+(1)/(5))(y-(1)/(4))\\\\\\80(y+(1)/(5))(y-(1)/(4))=4\cdot5\cdot4\cdot(y+(1)/(5))(y-(1)/(4))=4(5y+1)(4y-1)`





`80y^2-4y-4=80y^2-4y-16y+16y-4=80y^2-20y+16y-4\\\\=20y(4y-1)+4(4y-1)=(4y-1)(20y+4)`