How do you find the product of #t^2/((t-4)(t+4))*(t-4)/(6t)#?

Question

Abigail Allen · Accepted Answer

See a solution process below:

First, cancel common terms in the numerator and denominator: #t^color(red)(cancel(color(black)(2)))/(color(blue)(cancel(color(black)((y - 4))))(t + 4)) * color(blue)(cancel(color(black)(t - 4)))/(6color(red)(cancel(color(black)(t)))) =># #t/(t + 4) * 1/6 =># #(t * 1)/((t + 4)6) =># #t/((6 * t) + (6 * 4)) =># #t/(6t + 24)# However, because we cannot divide by #0# in the original equation we must exclude some values for #t#: #t - 4 = 0# means #t != 4# #t + 4 = 0# means #t != -4# #6t = 0# means #t != 0# Therefore: #t/(6t + 24)# where #t != 4# and #t != -4# and #t != 0#

Eli Assouline · Answer

undefined To find the product of the given expression, we can simplify it step by step:

1. Start by canceling out common factors between the numerator and denominator:
   t^2 / ((t-4)(t+4)) * (t-4) / (6t)
   = t^2 / (t^2 - 16) * (t-4) / (6t)

2. Next, simplify the expression by multiplying the numerators and denominators:
   (t^2 * (t-4)) / ((t^2 - 16) * (6t))

3. Expand the numerator:
   (t^3 - 4t^2) / ((t^2 - 16) * (6t))

Therefore, the product of the given expression is (t^3 - 4t^2) / ((t^2 - 16) * (6t)).