Alan

This should help:

As follows:

Copy and paste it into the calculator on the Home page here and press enter.

A general term in the expansion is \(\begin{pmatrix} 9\\ n \end{pmatrix} (2z)^nz^{-(\frac{9-n}{2})}(-1)^{9-n}\)

The constant term means the z terms don't appear, which means we must have \(n-\frac{9-n}{2}=0\)  or n = 3

So the constant is given by \(\begin{pmatrix} 9\\ 3 \end{pmatrix}2^3(-1)^6=672\)

You have not incorporated the condition "... at least three of the squares are red" Guestonwebsite.

Go to https://www.desmos.com/calculator and simply type in your equation.

I guess this should look like:  \(\frac{2x-2}{x^2-x-12}\)

If so, it can be written as \(\frac{2x-2}{(x-4)(x+3)}\)

Express as partial fractions:  \(\frac{2x-2}{(x-4)(x+3)} = \frac{A}{x-4}+\frac{B}{x+3}\)

from which we find    

\(A + B = 2\\\\3A-4B=-2\)

i'll leave you to find A and B.

This is all I can think of (though the result is rather ugly!):

1.  As follows:

Set y = 56, so  56 = -16*t^2 + 60*t

Divide by 16 and rearrange this:   t^2 - (15/4)*t + (7/2) = 0

This factorises as (t - 7/4)*(t - 2) = 0

So, there are two times.  Which do you think represents the answer to the question?