[tex]4x^2+25x+6=0\\\\Method\ 1:\\\\4x^2+25x+6=0\\4x^2+x+24x+6=0\\x(4x+1)+6(4x+1)=0\\(4x+1)(x+6)=0\iff4x+1=0\ or\ x+6=0\\\\4x+1=0\ \ \ |subtract\ 1\ from\ both\ sides\\4x=-1\ \ \ \ |divide\ both\ sides\ by\ 4\\\boxed{x=-\frac{1}{4}}\\\\x+6=0\ \ \ \ |subtract\ 6\ from\ both\ sides\\\boxed{x=-6}\\\\therefore\\\\4x^2+25x+6=4\left(x+\dfrac{1}{4}\right)(x+6)[/tex]
[tex]Method\ 2:\\\\4x^2+25x+6=0\\a=4;\ b=25;\ c=6\\\\\Delta=b^2-4ac\\\\\Delta=25^2-4\cdot4\cdot6=625-96=529 \ \textgreater \ 0\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{529}=23\\\\x_1=\dfrac{-25-23}{2\cdot4}=\dfrac{-48}{8}=-6\\\\x_2=\dfrac{-25+23}{2\cdot8}=\dfrac{-2}{8}=-\dfrac{1}{4}\\\\therefore\\\\4x^2+25x+6=4(x+6)\left(x+\dfrac{1}{4}\right)[/tex]
