SOLUTION: Expand (1/2-2x)^5 up to the term in x^3. If the coefficient of x^2 in the expansion of (1+ax+3x^2)(1/2-2x)^5 is 13/2, find the coefficient of x^3.

Algebra ->  Test -> SOLUTION: Expand (1/2-2x)^5 up to the term in x^3. If the coefficient of x^2 in the expansion of (1+ax+3x^2)(1/2-2x)^5 is 13/2, find the coefficient of x^3.      Log On


   

Question 1185961: Expand (1/2-2x)^5 up to the term in x^3. If the coefficient of x^2 in the expansion of (1+ax+3x^2)(1/2-2x)^5 is 13/2, find the coefficient of x^3.
Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


(1) Expand (1/2-2x)^5 up to the term in x^3

+ ...

ANSWER: %281%2F32%29+-+%285%2F8%29x+%2B+5x%5E2+-+20x%5E3 + ...

(2) Find a, given that the coefficient of x^2 in %281%2Bax%2B3x%5E2%29%281%2F2-2x%29%5E5 is 13/2

%281%2Bax%2B3x%5E2%29%28%281%2F32%29+-+%285%2F8%29x+%2B+5x%5E2+-+20x%5E3%29%29

The contributions to the x^2 coefficient come from the constant term in the first polynomial times the coefficient of x^2 in the second, the coefficient of the x term in the first polynomial times the coefficient of the x term in the second, and the coefficient of the x^2 term in the first polynomial times the constant term in the second:

%281%2F32%29%283%29%2B%28-5%2F8%29%28a%29%2B%285%29%281%29+=+13%2F2
3%2F32-%285%2F8%29a%2B5+=+13%2F2
3-20a%2B160+=+208
20a=-45
a=-9%2F4

ANSWER: a=-9/4

(3) Find the coefficient of x^3 in (1+ax+3x^2)(1/2-2x)^5

The contributions to the x^3 coefficient come from the constant term in the first polynomial times the coefficient of the x^3 term in the second, the coefficient of the x term in the first polynomial times the coefficient of the x^2 term in the second, and the coefficient of the x^2 term in the first polynomial times the coefficient of the x term in the second:



ANSWER: -265/8

The answers are confirmed with this completely expanded polynomial as calculated on wolframalpha.com:

1/32 - (89 x)/128 + (13 x^2)/2 - (265 x^3)/8 + 100 x^4 - 182 x^5 + 192 x^6 - 96 x^7