How do you find the product #(9t+6)(9t-6)#?

2 Answers
Apr 29, 2017

Answer:

See the solution process below:

Explanation:

This has a rule we can use to multiply the two terms:

#(a + b)(a - b) = a^2 - b^2#

Substituting ##9t# for #a# and #6# for #b# gives:

#(9t + 6)(9t - 6) = (9t)^2 - 6^2 = 81t^2 - 36#

We can also use the long way to multiply these two terms and get the same result. To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

#(color(red)(9t) + color(red)(6))(color(blue)(9t) - color(blue)(6))# becomes:

#(color(red)(9t) xx color(blue)(9t)) - (color(red)(9t) xx color(blue)(6)) + (color(red)(6) xx color(blue)(9t)) - (color(red)(6) xx color(blue)(6))#

#81t^2 - 54t + 54t - 36#

We can now combine like terms:

#81t^2 + (-54 + 54)t - 36#

#81t^2 + 0t - 36#

#81t^2 - 36#

Apr 29, 2017

Answer:

#81t^2-36#

Explanation:

FOIL it! First, multiply #9t# by #9t#. Then multiply #9t# by #-6#. Then multiply #6# by #9t#. And lastly, multiply #6# by #-6#.

This is what it should look like:

#81t^2-54t+54t-36#

Which can then be simplified into:

#81t^2-36#