SOLUTION: I seem to be having a small problem understanding Binomial Theorems as far as substitution and Expansion. Your assistancewith this problem would be greatly appreciated: It stat

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Question 61330: I seem to be having a small problem understanding Binomial Theorems as far as substitution and Expansion. Your assistancewith this problem would be greatly appreciated:
It states: Use substitution to find the expansion of (c-2i)^5. Simplify your final result.
Please help!
Thanks in advance!
Answer by Cintchr(481) About Me  (Show Source):
You can put this solution on YOUR website!
+%28c-2i%29%5E5+
let's expand this out
+%28c-2i%29%28c-2i%29%28c-2i%29%28c-2i%29%28c-2i%29+
Looking at just the first 2 quantities ... let's FOIL them
F: +c+%2A+c+=+c%5E2+
O: +c+%2A+-2i+=+-2ci+
I: +-2i+%2A+c+=+-2ci+
L: +-2i+%2A+-2i+=+4i%5E2+=4%28-1%29+=+-4+
remember i^2 = -1
Combine like terms
+c%5E2+-4ci+-4+
Now the 3rd and 4th terms when FOILed, give the same response.
You now have the following
+%28c%5E2+-4ci+-4%29%28c%5E2+-4ci+-4%29%28c-2i%29+
now multiply out the two trinomials

simplify
+c%5E4+-4c%5E3i+-4c%5E2+-4c%5E3i+-16c%5E2i%5E2+%2B16ci+-4c%5E2+%2B16ci+%2B16+
multiply, working the i^2 as well
+c%5E4+-8c%5E3i+-8c%5E2+-16c%5E2%28-1%29+%2B16ci+%2B16+
combine like terms
+c%5E4+-+8c%5E3i+-8c%5E2+%2B16c%5E2+%2B16ci+%2B16+
+c%5E4+-+8c%5E3i+%2B8c%5E2+%2B16ci+%2B16+
Now using this quantity ... multiply it by the 5th (c-2i)
+%28c-2i%29%28c%5E4+-8c%5E3i+%2B8c%5E2+%2B16ci+%2B16%29+

Multiply, working the i^2 as well

combine like terms
+c%5E5+-10c%5E4i+-+8c%5E3i+%2B+48c+-+32i+
the c^2 cancel out