T(t + 4)
-13
(4x - t2)= [(2√x)2 - t2] (This is now the difference of squares)= (2√x - t)(2√x + t)
All that is factorable here is the common factor t.t3 - 8tt(t2 - 8)======
If you mean t^2 +11t +10 then it is (t+10)(t+1) when factorized
T(t + 4)
(t - 7)(t - 11)
4(t + 3)(t2 - 3t + 9)
t2 - 59t + 54 - 82t2 + 60t = - 81t2 + t + 54
-13
(t-8)(t + 3)
(4x - t2)= [(2√x)2 - t2] (This is now the difference of squares)= (2√x - t)(2√x + t)
If you mean t(t+7) = 4(3+2t), then it is factored. If you meant solve for 't', then the solution is t = t2 - 12.
All that is factorable here is the common factor t.t3 - 8tt(t2 - 8)======
(t - 15)(t + 3)
12t(t2-4) 12t(t-2)(t+2)
If you mean t^2 +11t +10 then it is (t+10)(t+1) when factorized