Question 1205514
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I've often wondered about problems like this:

L = S+5
L*S = 84

(S+5)*S = 84
S<sup>2</sup>+5S = 84
S<sup>2</sup>+5S-84 = 0

To factor that, you have to think of two integers that have
product 84 and difference 5, which is exactly what the problem
asks for in the first place!! So you have to essentially solve 
the problem in your head in order to solve it by factoring!!  
That seems so silly!

I guess to keep it from being silly, you'd have to complete the
square or use the quadratic formula, but that's much harder than
thinking up 7 x 12 = 84 and 12 - 7 = 5.  

When I was teaching, I never assigned this type of problem, 
especially when teaching factoring binomials, but every textbook
would always include them!

Edwin</pre>