1. First off, you need to know what the X-axis intersection means: the point of intersection will have y = 0. Now, you know your two points are something like M(m, 0) and N(n, 0). You want to know m and n.
You should also notice that y = (x + a)(x + b) could be written as y = x^2 + x*(a + b) + ab, meaning that if you decide to graph this line, it would be that of a quadratic equation. You are looking for its roots:
x^2 + x*(a + b) + ab = 0
The first statement tells us that a + b is -1. This is insufficient by itself, since in order to solve the above equation, you also need ab.
The second statement tells us that the point (0, -6) can be found on the line. This means that:
0^2 + 0*(a + b) + ab = -6 --- I did nothing else but to replace x with 0 and y with -6 in the highlighted formula.
The second statement thus tells us that ab = -6. This is insufficient as well, since we don't know anything about a + b.
But put both statements together to get that you know both a + b and ab. You can replace them in x^2 + x*(a + b) + ab = 0 and find the roots. Remember: don't waste time with calculation on D-day. Just pick C! I'll just go ahead and calculate for the sake of exercise:
x^2 - x - 6 = 0
(x - 3)(x + 2) = 0
x is 3 or x is -2. The points will be M(3, 0) and N(-2, 0).
2. Number of chairs = c = 5
Number of tables = t = ?
Combinations of 2 chairs + 2 tables = tC2 * cC2 = tC2 * 5C2 = 150.
5C2 = 5!/(2!*3!) = 10.
tC2 = 150/10 = 15.
tC2 = t!/[(t-2)!*2!] = [(t-1)*t]/2 = 15.
This means that t(t - 1) = 30 or that t^2 - t - 30 = 0. Again, a quadratic equation.
t^2 - t - 30 = (t - 6)(t + 5) = 0
Solutions t = 6 and t = -5. t can't be negative (since you can't have a negative number of tables), so t = 6.
I'm kind of weak in permo&combo, so do tell me if I nailed the answer...
