+506 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 \(64\cdot40+-192\cdot10+240\cdot1=\boxed{880}\)\(\)
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