Posted by lynne on Monday, November 1, 2010 at 10:11pm.
I use a method that nobody seems to either teach or learn anymore, yet it is the fastest way to find the factors if they exist
I make a pair of two columns, one for the factors of 40, another for the factors of 30
(they line up vertically)
2 4 5 -------2 3 5 15 10 6
20 10 8 ----15 10 6 2 3 5
I then form cross-products with a column pair from the first and a column pair from the second
In this case I am looking for a difference of 23, since the last term is -30
Had it been +30 I would look for a sum of 23
I notice my pairs
5 --- 6
8 --- 5
give the cross-product difference of 5x5 - 6x8 = -23
so my factors must be
(5x .....6)(8x .... 5)
A quick mental check gives me
(5x+6)(8x-5)
Notice I repeat the factors in one of the groups by flipping the order. Most of the time I don't list all the combinations, often skipping the extreme ones like
1
40
In actual practise it only took me about 30 seconds to find the above
so the correct answer is (5x+6)(8x-5) ?
can be easily checked by expanding it.
i did and came up with the same. however when i entered it they said it was wrong, s/b X= 5/8, -6/5...not sure how
Since the hard part was to factor it, I assumed that you would finish it.
so
(5x+6)(8x-5) = 0
x = -6/5 or x = 5/8
i totally dropped the ball in this one...i should have definitely know to find x!!!
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