what is 2t^3+8t^2-24t in factor form
\(2t^3+8t^2-24t=0\\2t(t^2+4t-12)=0\\2t=0\ \vee\ t^2+4t-12=0\)\(t^2+4t-12=0\\a=1\ ;\ b=4\ ;\ c=-12\\\Delta=b^2-4ac\\\Delta=4^2-4\cdot1\cdot(-12)=16+48=64\\\sqrt\Delta=\sqrt{64}\)\(t_1=\frac{-b-\sqrt\Delta}{2a}\ \vee\ t_2=\frac{-b+\sqrt\Delta}{2a}\\t_1=\frac{-4-8}{2\cdot1}\ \vee\ t_2=\frac{-4+8}{2\cdot1}\\t_1=\frac{-12}{2}\ \vee\ t_2=\frac42\\t_1=-6\ \vee\ t_2=2\)\(2t(t^2+4t-12)=0\\2t(t+6)(t-2)=0\)
Answer:
2t (t-2) (t+6)
Step-by-step explanation:
