Free points for the people who thinks that u shouldnt report or delete free point posts like fr they just deleted my post all im doing is helping ppl out. ):<<< lol

Enter the number of complex zeros for the polynomial function f(x) = x 4+ 5x² +6

Prove (a) cosh2(x) − sinh2(x) = 1 and (b) 1 − tanh 2(x) = sech 2(x). solution (a) cosh2(x) − sinh2(x) = ex + e−x 2 2 − 2 = e2x + 2 + e−2x 4 − = 4 = . (b) we start with the identity proved in part (a): cosh2(x) − sinh2(x) = 1. if we divide both sides by cosh2(x), we get 1 − sinh2(x) cosh2(x) = 1 or 1 − tanh 2(x) = .

3c2 + 2d)(–5c2 + d) select all of the partial products for the multiplication problem above.