Question

Factor 7(x − 3)2 − 4(x − 3) − 3 completely.

Answer 1

`7(x-3)^2-4(x-3)-3\\\\substitute\ t=x-3,\ then\ we\ have\\\\7t^2-4t-3\\\\a=7;\ b=-4;\ c=-3\\\\\Delta=b^2-4ac\\\\\Delta=(-4)^2-4\cdot7\cdot(-3)=16+84=100\\\\t_1=(-b-\sqrt\Delta)/(2a);\ t_2=(-b+\sqrt\Delta)/(2a)\\\\\sqrt\Delta=√(100)=10\\\\t_1=(-(-4)-10)/(2\cdot7)=(4-10)/(14)=(-6)/(14)=-(3)/(7)\\\\t_2=(-(-4)+10)/(2\cdot7)=(4+10)/(14)=(14)/(14)=1\\\\7t^2-4t-3=7\left(t+(3)/(7)\right)(t-1)=(7t+3)(t-1)\\\\=(7(x-3)+3)(x-3-1)=(7x-21+3)(x-4)\\\\=\boxed{(7x-18)(x-4)}`