algebra coefficients

jms11375

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Apr 16, 2013
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This was some test prep for my child and I do not know the correct answer.

In this expression what is the coefficient of y?


(3y2 + 9)5


Thank you for your help.
 
This was some test prep for my child and I do not know the correct answer.

In this expression what is the coefficient of y?


(3y2 + 9)5


Thank you for your help.

jms11375,

is this problem supposed to be \(\displaystyle (3y^2 + 9)5, \ \ or \ \ (3y^2 + 9)^5, \ \ or\ \ something \ \ else?\)
 
All terms will involve powers of 3, powers of 9, and powers of \(\displaystyle y^2\), as well as . In particular, \(\displaystyle (y^2)^n= y^{2n}\) so all powers of \(\displaystyle y\) are even. There are no odd powers of y and, in particular, there is no 1st power. The coefficient of y in this is 0.
 
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Hello, jms11375!

This was some test prep for my child and I do not know the correct answer.
You don't know the correct answer ... no surprise.
But neither of you knows what has to be done?


In this expression what is the coefficient of \(\displaystyle y\)? .\(\displaystyle (3y^2+9)^5\)

HallsofIvy has the best solution.


But we can expand the binomial.

\(\displaystyle (3y^2 + 9)^5 \;=\;(3y^2)^5 + 5(3y^2)^4(9) +10(3y^2)^3(9)^2 + 10(3y^2)^2(9)^3 + 5(3y^2)(9)^4 + (9)^5\)

. . . . . . . . .\(\displaystyle =\;243y^{10} +3645y^8 + 21,\!870y^6 + 65,\!610y^4 + 98,\!415y^2 + 59,\!049\)


There is no \(\displaystyle y\)-term ... Its coefficient is zero.
 
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