Expand (1-6x)^4 (1+2x)^7 - Solve Binomial Expansion

[Fly_High]

Homework Statement

Expand (1-6x)^4 (1+2x)^7 in ascending powers of x up to and including the terms in x^3

Homework Equations

(1-6x)^4 (1+2x)^7

The Attempt at a Solution

Firstly, I expand (1-6x)^4
= (1)^4 + 4C1 (-6x) + 4C2 (-6x)^2 + 4C3 (-6x)^3 + ...
= 1 + 4(-6) + 6 (36x^2) + 4(-216^3) + ...
= 1 -24x + 216^2 - 864x^3 + ...


Then after that, I expand (1+2x)^7
= (1)^7 + 7C1 (2x) + 7C2 (2x)^2 + 7C3 (2x)^3 +...
= 1 + 7 (2x) + 21 (4x^2) + 35 (8x^3) + ...
= 1 + 14x + 84x^2 + 280x^3 + ...

Finally, I expand (1-6x)^4 (1+2x)^7 together...
= (1-24x+216x^2=864x^3+...) (1+14x+84x^2+280x^3+...)
= 1+14x+84x^2+280x^3-24x-336x^2-2016x^3+3024x^3-864x^3+...
= 1 -10x-252x^2+2240x^3+...

My final answer is wrong. It should be 1-10x-36x^2+424x^3+...

Can someone help me solve this? Thanks.

 

[AlephZero]

= (1-24x+216x^2=864x^3+...) (1+14x+84x^2+280x^3+...)

That's OK

= 1+14x+84x^2+280x^3 OK
-24x-336x^2-2016x^3 OK
+3024x^3-864x^3+... That's where you went wrong. There's a term missing, and probably you made another mistake adding up the terms because I can't see how you got 2240x^3 from your (wrong) numbers.

 

[Architeuthis Dux]

Your final answer is incorrect because when multiplying the two expressions together, you have not properly accounted for the terms with the same power of x. For example, the term with x^2 should be (6x)^2 from the first expression multiplied by (2x)^2 from the second expression, which gives a coefficient of 36. Similarly, for the term with x^3, it should be (6x)^3 from the first expression multiplied by (2x)^0 from the second expression, which gives a coefficient of 216.

A correct solution would be:

(1-6x)^4 (1+2x)^7 = (1+4(-6x)+6(36x^2)+4(-216x^3)+...) (1+7(2x)+21(4x^2)+35(8x^3)+...)
= 1 + 28x - 108x^2 + 424x^3 + ...