Solve the following equation: 2t (t^2 -1) - 4(t^2-1) = 0?

Question

Please help!

And thank YOU! :)

Answer 1

(2t - 4)(t^2-1) = 0

(2t-4)(t-1)(t+1)=0

Split it into three different equations

2t - 4 = 0

2t = 4

t = 2

t-1 = 0

t=1

t+1 = 0

t=-1

All three are solutions.

Answer 2

I Cannot remember the name of the rule just remember this is what you do it.

First of all, since the info within both of the parens are the same, set that to zero

t^2 - 1 = 0

t^2 = 1

t = 1 and -1 since (1)(1) and (-1)(-1) both = 1

then take the information outside of the parens and group together and set to zero.

2t -4 = 0

2t = 4

t = 2

Now if you plug in "2" and solve, you should get zero

and if you plug in "1" and solve, you should get zero

HOPE THIS HELPS

Answer 3

(t^2-1)(2t-4)=0

(t+1)(t-1)2(t-2)=0

t=1, -1, 2