Answer :
[tex]ax^2+bx+c\\\\\Delta=b^2-4ac\\\\if\ \Delta > 0\ then\ ax^2+bx+c=a(x-x_1)(x-x_2)\ where\ x_{1;2}=\frac{-b\pm\sqrt\Delta}{2a}\\\========================================[/tex]
[tex]-35x^2-41x-12\\\\a=-35;\ b=-41;\ c=-12\\\\\Delta=(-41)^2-4\cdot(-35)\cdot(-12)=1681-1680=1 > 0\\\\\sqrt\Delta=\sqrt1=1\\\\x_1=\frac{-(-41)-1}{2\cdot(-35)}=\frac{41-1}{-70}=\frac{40}{-70}=-\frac{4}{7}\\\\x_2=\frac{-(-41)+1}{2\cdot(-35)}=\frac{41+1}{-70}=\frac{42}{-70}=-\frac{21}{35}=-\frac{3}{5}\\\\therefore:\\\\\boxed{-35x^2-41-12=-35\left(x+\frac{4}{7}\right)\left(x+\frac{3}{5}\right)}[/tex]
[tex]-35x^2-41x-12\\\\a=-35;\ b=-41;\ c=-12\\\\\Delta=(-41)^2-4\cdot(-35)\cdot(-12)=1681-1680=1 > 0\\\\\sqrt\Delta=\sqrt1=1\\\\x_1=\frac{-(-41)-1}{2\cdot(-35)}=\frac{41-1}{-70}=\frac{40}{-70}=-\frac{4}{7}\\\\x_2=\frac{-(-41)+1}{2\cdot(-35)}=\frac{41+1}{-70}=\frac{42}{-70}=-\frac{21}{35}=-\frac{3}{5}\\\\therefore:\\\\\boxed{-35x^2-41-12=-35\left(x+\frac{4}{7}\right)\left(x+\frac{3}{5}\right)}[/tex]
