Respuesta :

dalendrk
[tex]2a^2+3a+1\\\\x=2;\ y=3;\ z=1\\\\\Delta=y^2-4xz\\if\ \Delta \ \textgreater \ 0\ then\ a_1=\dfrac{-y-\sqrt\Delta}{2x}\ and\ a_2=\dfrac{-y+\sqrt\Delta}{2x}\\\\\Delta=3^2-4\cdot2\cdot1=9-8=1 \ \textgreater \ 0\\\sqrt\Delta=\sqrt1=1\\\\a_1=\dfrac{-3-1}{2\cdot2}=\dfrac{-4}{4}=-1;\ a_2=\dfrac{-3+1}{2\cdot2}=\dfrac{-2}{4}=-\dfrac{1}{2}\\\\2a^2+3a+1=2(a+1)\left(a+\dfrac{1}{2}\right)=(a+1)(2a+1)\\\\other\ method\\\\2a^2+3a+1=2a^2+2a+a+1=2a(a+1)+1(a+1)\\=(a+1)(2a+1)[/tex]

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