20p^2-19+3 is this one even possible? AC Method - Factoring. 20*3 = 60 to get two numbers that multiply to sxity but add to -19?
umm two numbers that add up to -19 ,-15 and -4
-15 and -4
Sorry to say this, but this problem is ambiguous as you've stated it. Did you intend this problem statement to present just one problem, or more than one? Mihirb has come up with a solution to the problem "find two numbers that add up to -19 and have the product 60."
write it as: \[20p^2-15p-4p+3\]
I'll let mirhib translate this word problem into the appropriate equations, which must be solved simultaneously to find the values of these two numbers.
@mihirb Ty, I should have looked for more factors instead of just relying on myself. I can solve from there. I just though I went through all the factors =\
now factor by grouping so: \[-5p(-4p+3)+1(-4p+3)\]
so: \[(-5p+1)(-4p+3)\]
To come up with solutions, I chose the letters m and n to represent the two numbers. Then the product of these numbers is m*n=60 and their sum is m+n = -19. Either equation could be used as the constraint equation. Choosing the first n = 60/m. Substituting this expression for n into the other equation, m + (60/m) = -19, or m + 60/m +19 = 0, or (after multiplying through by m), m^2 + 60 + 19m = 0. Going back to the original problem statement, I see that a "p" is missing: I also wonder where that "3" came from.
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