Question 1185961
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(1) Expand (1/2-2x)^5 up to the term in x^3<br>
{{{(1/2)^5 + (5)((1/2)^4)(-2x)^1 + (10)((1/2)^3)(-2x)^2 + 10((1/2)^2)(-2x)^3}}} + ...<br>
ANSWER: {{{(1/32) - (5/8)x + 5x^2 - 20x^3}}} + ...<br>
(2) Find a, given that the coefficient of x^2 in {{{(1+ax+3x^2)(1/2-2x)^5}}} is 13/2<br>
{{{(1+ax+3x^2)((1/32) - (5/8)x + 5x^2 - 20x^3))}}}<br>
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:<br>
{{{(1/32)(3)+(-5/8)(a)+(5)(1) = 13/2}}}
{{{3/32-(5/8)a+5 = 13/2}}}
{{{3-20a+160 = 208}}}
{{{20a=-45}}}
{{{a=-9/4}}}<br>
ANSWER: a=-9/4<br>
(3) Find the coefficient of x^3 in (1+ax+3x^2)(1/2-2x)^5<br>
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:<br>
{{{(-20)(1)+(5)(-9/4)+(-5/8)(3) = (-160-90-15)/8 = -265/8}}}<br>
ANSWER: -265/8<br>
The answers are confirmed with this completely expanded polynomial as calculated on wolframalpha.com:<br>
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<br>