Find the binomial raised to the power, whose expansion it came from: 10a^3b^2
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OpenStudy (sleepyjess):
I will tag some users that may be able to help. @Jhannybean @jim_thompson5910 @Callisto
jimthompson5910 (jim_thompson5910):
That's all it asks? Or is there more info?
OpenStudy (anonymous):
thats all it asks.
jimthompson5910 (jim_thompson5910):
what is the sum of the exponents?
OpenStudy (anonymous):
i dont know how to do this , can you explain it to me ?
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jimthompson5910 (jim_thompson5910):
do you see the exponents?
OpenStudy (anonymous):
would it be 6 ?
jimthompson5910 (jim_thompson5910):
3+2 is not 6
OpenStudy (anonymous):
i multiplied them .
OpenStudy (anonymous):
but 5
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jimthompson5910 (jim_thompson5910):
so we know that we have something in the form (x+y)^5
jimthompson5910 (jim_thompson5910):
look at pascal's triangle and look for the row that has 1, 5, etc etc
OpenStudy (anonymous):
1 5 10 10 5 1
jimthompson5910 (jim_thompson5910):
if 'a' is part of the x term in (x+y)^5, then that means we look at the 3rd number in that row of 1, 5, 10, 10, 5, 1
notice the third number is 10
and how there is a 10 as the coefficient of 10a^3b^2
jimthompson5910 (jim_thompson5910):
so that means the two terms x and y must have coefficients of 1 to make 10x^3*y^2 be equal to 10a^3b^2
In this case, x = a and y = b
so we have (a+b)^5
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OpenStudy (anonymous):
so (1+1)^5 ??????
jimthompson5910 (jim_thompson5910):
if you expanded out (a+b)^5, you would get
a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5
and we can see the third term is 10a^3b^2
jimthompson5910 (jim_thompson5910):
no, (a+b)^5
OpenStudy (anonymous):
oh , okay . so where did 'a' come from?
jimthompson5910 (jim_thompson5910):
from 10a^3b^2
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