Find the coefficient of x2in the expansion of (1 + 2x)^5(2-x)^6

This isn't the best way to do it since it's just brute force but it's still an answer I guess:

I'm sure you've learned how to expand brackets right? If you haven't, you basically expand it like this:


Now, we can expand all of this:

(1+2x)⁵(2-x)⁶  

--> (1+2x)⁵ = (1+2x)(1+2x)(1+2x)(1+2x)(1+2x)
                   = 32x⁵ + 80x⁴ + 80x³ + 40x² + 10x + 1

--> (2-x)⁶ = (2-x)(2-x)(2-x)(2-x)(2-x)(2-x)
                = x⁶ - 12x⁵ + 60x⁴ - 160x³ + 240x² - 192x + 64

So if we put this together:

(1+2x)⁵(2-x)⁶  = (32x⁵ + 80x⁴ + 80x³ + 40x² + 10x + 1)(x⁶ - 12x⁵ + 60x⁴ - 160x³ + 240x² - 192x + 64)

Then when you expand this, you get:

32x¹¹ - 304x¹⁰ + 1040x⁹ - 1240x⁸ - 790x⁷ + 2537x⁶ + 76x⁵ - 2180x⁴ - 320x³ + 880x² + 448x + 64

But all you need to pay attention to is 880x² which means the coefficient of x² , in other words your answer, is 880. 

Nice solution guest and logarhythm!

Here's an alternate solution:

We need to find the coefficients of x^2, x, and constants of both \((1+2x)^5\) and \((2-x)^6\).

To do that, use the binomial theorem (explained here: https://www.mathsisfun.com/algebra/binomial-theorem.html )

Knowing that, the constant, x, and x^2 coefficients of \((1+2x)^5\) are 5 choose 0 = 1 , 5 choose 1 * 2 = 10, and 5 choose 2 * 2^2 = 40, respectively, and the constant, x, and x^2 coefficients of \((2-x)^6\) are 6 choose 0 * 2^6 = 64, 6 choose 1 * -1 * 2^5 = -192, and 6 choose 2 * 2^4 = 240, respectively.

The x^2 term of the product of the 2 expressions will happen when an x term of one expression is multiplied by the x term in the other expression, or when the x^2 term in one expression is multiplied by a constant term in the other expression.

That is equal to \(64\cdot40+-192\cdot10+240\cdot1=\boxed{880}\)\(\)