Question 33232
Check this out in case I do it wrong myself!

Your equation {{{x^2-11x-60}}} is very simple to factor because {{{x^2}}} is not multiplied by anything.  I'll tell you how to solve those at the end.

I'll use the substitutes A, B, and C, defined {{{Ax^2+Bx+C}}}.   Your A=1 B=-11 C=-60.

Basically, the only hard part is finding the two numbers that ADD to B and MULTIPLY to A*C.  [Let the numbers be X and Y: {{{XY=AC}}}, and {{{X+Y=B}}}].  In your equation, X+Y=-11 and XY=-60.  We know one number is positive, and one negative.

How do we know that?  Since it's product (XY) is negative, we know that one is negative and one positive (because (+ times +) and (- times -) both have negative answers.  We also know that the bigger number is negative because the sum is negative, it makes sense.

Now, what numbers add up to -11 and multiply to -60?  First, we find the numbers that add up to 11 and multiply to 60.  It's simpler.  {{{10 +- 1=11 or 9}}}, but 10*1=10, not 60.  10*6=60, but {{{10 +- 6=16 or 4}}}.  How about 15? No, 15*4=60, but 15+4, and 15-4=11. Ha! It equals 11!  We know from earlier that we only have one negative and its the larger number, now you know why. -15+4=4-15=-11, and as we concluded 4*-15=*60!

Now that we've found the numbers, we put them in the equation, previously {{{X^2-11x-60}}}.  We change the -11x to +4x-15x (it's the same).  The equation now yields {{{X^2+4x-15x-60}}}.  Now, you see that {{{X^2}}} and -15x both have something in common with 4x and -60.  {{{X^2}}} and 4X are both divisible by X, so {{{X^2+4X}}} is replaced by x(x+4).  -15x and -60 have a negative 15 in common (-60= -15*4).  So we replace -15x-60 by -15(x+4).  Our entire equation is now x(x+4)-15(x+4).

Put the x+4's together and what's leftover? x-15. there's your equation: 
(X+4)(X-15)

Any questions IM me.