Answer :
[tex]15(2z+1)^3+10(2z+1)^2-25(2z+1)\\\\subtitute\ x=2z+1\\\\15t^3+10t^2-25t=5t\cdot3t^2+5t\cdot2t-5t\cdot5=5t(3t^2+2t-5)\\\\=5t(3t^2-3t+5t-5)=5t[3t(t-1)+5(t-1)]\\\\=5t(t-1)(3t+5)\\\\=5(2z+1)(2z+1-1)[3(2z+1)+5]\\\\=5(2z+1)(2z)(6z+3+5)=5(2z)(2z+1)(6z+8)\\\\=10z(2z+1)(6z+8)=10z(2z+1)(2)(3z+4)\\\\=\boxed{20z(2z+1)(3z+4)}[/tex]
