Question 90546
Factor 8x^2+28x-60
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Your first step was to take out a factor of 4. This is the correct thing to do. After that
you are left with:
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4*(2x^2 + 7x - 15)
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The 2x^2 can only factor to 2x * x so the next step (if 2x^2 + 7x -15 factors) involves:
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4*(2x ....)*(x ....)
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the 15 can only factor to 15*1 or 5*3 (ignoring the signs on the two numbers). The potential
factors of 2x^2 + 7x - 15 are therefore:
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(2x 15)*(x 1)
(2x 1)*(x 15)
(2x 5)*(x 3)
(2x 3)*(x 5)
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If it factors, there must be some combination with signs between the x terms and the numbers
that adds to +7. Becomes pretty apparent that the factors involving 15 are not likely.
So the most likely factors are (2x 5)*(x 3) or (2x 3)*(x 5).
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We could have:
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(2x -5)*(x + 3) but this doesn't work because 2x*3 - 5x = 6x - 5x = +x not +7x
or
(2x + 5)*(x - 3) but this doesn't work because 2x*(-3) + 5x = -6x + 5x = -x not +7x
so if it factors it must be either (2x + 3)*(x - 5) or (2x - 3)*(x + 5). Try:
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(2x + 3)*(x - 5). The cross products are 2x*(-5) and 3*x = -10x + 3x = -7x. Close, but
the sign is wrong.
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We are down to trying (2x - 3)*(x + 5). The cross products are 2x*5 and -3*x which lead
to 10x - 3x = +7x. That's what we are looking for.  So multiplying out the factors:
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(2x - 3)*(x + 5) gives 2x*x + 2x*5 -3*x -3*5 = 2x^2 + 6x - 5x -15 = 2x^2 +7x - 15.
Hooray! Now we can say that 2x^2 + 7x - 15 factors to (2x  - 3)*(x + 5) and as a result the
whole factoring process results in:
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4*(2x - 3)*(x + 5)
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That's the answer and if you multiply this out the product will be the original polynomial 
you were given to factor. 
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Factoring like this involves a lot of judgment, guessing, and trial-and-error as you can
see from the above analysis. The more you work these kind of problems the better you will 
become at getting to the answer quicker because you will develop a "feel" for what 
combinations  are more likely to work and what combinations are likely to lead to a 
wrong answer. 
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Hope you can follow the above process ... especially with regard to using the cross
products (multiplying a term containing x by a term that is just a number) to look for
a factor that works.
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And hope that the answer above points out where you made a few simple mistakes in your 
logic that led you to an answer that wouldn't check.