SOLUTION: Expand completely (2x3+1)5

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Question 1109181: Expand completely (2x3+1)5
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

%282x%5E3%2B1%29%5E5...I guess you have that
to expand, multiply base five times
%282x%5E3%2B1%29%282x%5E3%2B1%29%282x%5E3%2B1%29%282x%5E3%2B1%29%282x%5E3%2B1%29
%284x%5E6+%2B+4x%5E3+%2B+1%29%284x%5E6+%2B+4x%5E3+%2B+1%29%282x%5E3%2B1%29
%2816x%5E12+%2B+32+x%5E9+%2B+24x%5E6+%2B+8x%5E3+%2B+1%29%282x%5E3%2B1%29
32x%5E15+%2B+80x%5E12+%2B+80x%5E9+%2B+40x%5E6+%2B+10x%5E3+%2B+1

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Multiplying the expression times itself 5 times is inefficient. Use the binomial theorem.

(2x^3+1)^5
= C(5,5)*(2x^3)^5*(1)^0
+ C(5,4)*(2x^3)^4*(1)^1
+ C(5,3)*(2x^3)^3*(1)^2
+ C(5,2)*(2x^3)^2*(1)^3
+ C(5,1)*(2x^3)^1*(1)^4)
+ C(5,0)*(2x^3)^0*(1)^5

= 1(2x^3)^5 + 5(2x^3)^4 + 10(2x^3)^3 + 10(2x^3)^2 + 5(2x^3)^1 + 1(2x^3)^0

= 32x^15 + 5*16x^12 + 10*8x^9 + 10*4x^6 + 5*2x^3 + 1

= 32x^15 + 80x^12 + 80x^9 + 40x^6 + 10x^3 + 1