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Question 498013: How do you factor this expression?
8c^2-13c+5
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! answer is:
(8c-5) * (c-1)
the logic process was something like this.
the factors had to be:
(ax - b) * (cx - d)
b had to be a 1 and d had to be a 5 because -1 * -5 = +5 and no other combination was possible to get a result of + 5.
don't forget, when you multiply (ax - b) * (cx - d), the result of that multiplication becomes:
a*c*x^2 - a*d*x - b*c*x + b*d
since i made b = 1 and d = 5, then the factors became:
(ax - 1) * (cx - 5)
if you multiply those together, you will get:
a*c*x^2 - a*5*x - 1*c*x - 1*(-5) which becomes:
a*c*x^2 - a*5*x - 1*c*x + 5
now that middle term of - a*5*x - 1*c*x has to add up to -13.
that can only happen if one of the terms is 5 and the other term is 8.
this means that a has to be 1 and c has to be 8 because then we would have:
-1*5*x - 1*8*x which would become:
-5*x - 8*x which would becomes:
-13*x.
so, i assigned 1 to a and 8 to c to get:
(ax - 1) * (cx - 5) becomes:
(x - 1) * (8x - 5)
now, when the factors are multiplied out, we get:
8x^2 - 5x - 8x + 5 which becomes:
8x^2 - 13x + 5 which is the original equation.
i automatically shifted to working with x rather than c.
it doesn't matter.
replace x with c in all the above logic and you get the same answer.
only the name of the variable used has changed.
the logic involved is exactly the same.
you get (c - 1) * (8c - 5) instead of (x - 1) * (8x - 5)
also, note that i showed you the answer as:
(8c - 5) * (c - 1)
that's the same thing as:
(c - 1) * (8c - 5)
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