I've never done this kind of problem before and my teacher's never taught it, she expects us to have learned it last year even though we didn't. Can I get a full explanation on how to do it?

Guest Feb 14, 2021

#1**+1 **

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)

= 32x^{5 }+ 80x^{4} + 80x^{3} + 40x^{2} + 10x + 1

--> (2-x)^{6} = (2-x)(2-x)(2-x)(2-x)(2-x)(2-x)

= x^{6} - 12x^{5} + 60x^{4} - 160x^{3} + 240x^{2} - 192x + 64

So if we put this together:

(1+2x)^{5}(2-x)^{6} = (32x^{5 }+ 80x^{4} + 80x^{3} + 40x^{2} + 10x + 1)(x^{6} - 12x^{5} + 60x^{4} - 160x^{3} + 240x^{2} - 192x + 64)

Then when you expand this, you get:

32x^{11} - 304x^{10} + 1040x^{9} - 1240x^{8} - 790x^{7} + 2537x^{6} + 76x^{5} - 2180x^{4} - 320x^{3} + 880x^{2} + 448x + 64

But all you need to pay attention to is 880x^{2} which means the coefficient of x^{2 }, in other words your answer, is 880.

Logarhythm Feb 14, 2021

#2**0 **

Sorry I just realized there is a much quicker way to figure this out.

Going back to this step:

(1+2x)5(2-x)6 = (32x^{5} + 80x^{4} + 80x^{3} + 40x^{2} + 10x + 1)(x^{6} - 12x^{5 }+ 60x^{4} - 160x^{3} + 240x^{2} - 192x + 64)

You can find the coefficient of x^{2} by multiplying the terms that make x^{2} and adding them up:

(10x)(-192x) + (40x^{2})(64) + (1)(240x^{2})

= -1920x^{2} + 2560x^{2} + 240x^{2}

= 880x^{2}

Which means 880 is your answer :)

Guest Feb 14, 2021

#3**+1 **

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

textot Feb 14, 2021