2t(cubed) + 32t(squared) + 128 t I need to factor this equation. Just struggling with how to begin.I know that 128 / 32 is 4...its the t variables that are giving me trouble.
You need to factor out `2t` such that:
`2t(t^2 + 16t + 64)`
You need to write the factored form of quadratic `t^2 + 16t + 64` such that:
`t^2 + 16t + 64 = (t - t_1)(t - t_2)`
Notice that t_1 and t_2 represent the solutions to the quadratic `t^2 + 16t + 64 = 0` such that:
`t_(1,2) = (-16+-sqrt(16^2 - 4*64))/2`
`t_(1,2) = (-16+-sqrt(256-256))/2 => t_1=t_2 = -16/2 = -8`
`t^2 + 16t + 64 = (t+8)(t+8)`
`t^2 + 16t + 64 = (t+8)^2`
Hence, the quadratic `t^2 + 16t + 64` represents the expansion of binomial `(t+8)^2` .
Hence, reducing to the factored form the given expression yields `2t^3 + 32t^2 + 128t = 2t(t+8)^2` .