Asked by Becca
Use the distributive property to write (each of) the following expression(s) as the sum of monomials.
M. (t - 1)(t + 1)(t^2 + 1)
I already know that I must multiply across, but I am a bit confused regarding the order to multiply in and which variables to "distribute" or multiply by eachother.
Thank you!
M. (t - 1)(t + 1)(t^2 + 1)
I already know that I must multiply across, but I am a bit confused regarding the order to multiply in and which variables to "distribute" or multiply by eachother.
Thank you!
Answers
Answered by
Steve
The order of distribution does not matter, since the associative property guarantees that the order of grouping does not matter, and the commutative property guarantees that the order of multiplication does not matter.
You can drag it out by brute force:
(t-1)(t+1)(t^2+1)
t(t+1)(t^2+1) - 1(t+1)(t^2+1)
t(t)(t^2+1) + t(1)(t^2+1) - t(t^2+1) - 1(t^2+1)
t^2(t^2)+t^2(1) + t(t^2) + t(1) - t(t^2) - t(1) - 1(t^2)-1(1)
t^4+t^2+t^3+t-t^3-t-t^2-1
t^4-1
or you can remember your difference-of-squares stuff:
(t-1)(t+1)(t^2 + 1)
(t^2-1)(t^2+1)
t^4-1
You can drag it out by brute force:
(t-1)(t+1)(t^2+1)
t(t+1)(t^2+1) - 1(t+1)(t^2+1)
t(t)(t^2+1) + t(1)(t^2+1) - t(t^2+1) - 1(t^2+1)
t^2(t^2)+t^2(1) + t(t^2) + t(1) - t(t^2) - t(1) - 1(t^2)-1(1)
t^4+t^2+t^3+t-t^3-t-t^2-1
t^4-1
or you can remember your difference-of-squares stuff:
(t-1)(t+1)(t^2 + 1)
(t^2-1)(t^2+1)
t^4-1
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