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

Sagot :

[tex]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=\frac{-b-\sqrt\Delta}{2a};\ y_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{1296}=36\\\\y_1=\frac{4-36}{2\cdot80}=\frac{-32}{160}=-\frac{1}{5};\ y_2=\frac{4+36}{2\cdot80}=\frac{40}{160}=\frac{1}{4}[/tex]


[tex]80y^2-4y-4=80(y+\frac{1}{5})(y-\frac{1}{4})\\\\\\80(y+\frac{1}{5})(y-\frac{1}{4})=4\cdot5\cdot4\cdot(y+\frac{1}{5})(y-\frac{1}{4})=4(5y+1)(4y-1)[/tex]




[tex]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)[/tex]
kate200468
[tex] 80y^2-4y-4=4(20y^2-5y+4y-1)=4[5y(4y-1)-(4y-1)]=\\ \\=4(4y-1)(5y-1)[/tex]
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