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if the coefficient of x^3 is 4320 and the coefficient of x^4 is -1620 in the expansion...

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xdsmith | (Level 2) eNoter

Posted April 8, 2013 at 10:13 PM via web

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if the coefficient of x^3 is 4320 and the coefficient of x^4 is -1620 in the expansion (ax+b)^5, find a and b

using binomial theorem

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted April 9, 2013 at 12:54 AM (Answer #1)

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Using binomial thereom we can say;

`(ax+b)^5` = `sum_(r=0)^(r=5)` `^5C_r(ax)^rb^(5-r)`

The coefficient of x^3 is given when r = 3

coefficient of `x^3 = ^5C_3xxa^3xxb^(5-3) = 10a^3b^2`

`4320 = 10a^3b^2` ----(1)

 

The coefficient of `x^4` is given when r = 4

coefficient of `x^4 = ^5C_4xxa^4xxb^(5-4) = 5a^4b^1`

`-1620 = 5a^4b ` ---(2)

 

`(2)^2/(1)`

`(-1620^2)/(4320) = (5a^4b)^2/(10a^3b^2)`

`243 = a^5`

`a = (243)^(1/5)`

`a = 3`

 

From (2)

`b = (-1620)/(5a^4)`

`b = -5`

 

So the answers are a = 3 and b = -5

 

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