John, I really liked your approach of multiplying the outer terms together and then looking for factors that will result in the middle term when they're added together. I must remember that technique. (I guess the quadratic formula could have been used but it can be a bit complicated). I looked at 15c² -20c -3c + 4 in a different way from the one you took, grouping the 15c² and -3c together and the -20c and the 4 together as follows: (15c² - 3c) - (20c - 4), resulting in 3c(5c - 1) - 4(5c - 1) or (5c - 1)(3c - 4). The same result, of course, so obviously both are feasible..
It is logic that both ways give the same result, because it would be very arbitrary if you had to group 15c^2 - 20c to be able to factor correctly. So it is good that you grouped 15c^2 - 3c, as proof that both ways of grouping do the 'trick'. The conclusion is that you can group in both ways to get the right result. We see that also when we factor the denominator as follows, grouping 10c^2 - 2c and 5c - 1. Then we get: 2c(5c - 1) + 1(5c -1), resulting in (2c + 1)(5c - 1) or (5c - 1)(2c + 1). And of course the same 'end result': 3c - 4/2c + 1). :)
super simple with good old foil. Numerator: 15c² - 23c + 4 we can factor this to: (5c - 1)(3c - 4) Denominator: 10c² + 3c - 1 we can factor this to: (5c - 1)(2c + 1) the numerator and denominator have (5c -1) in common so we can cross them out. Leaving us with (3c-4) / (2c + 1)
Dammit - as usual I tried solving using the thumbnails running down the right side of the screen while I watched other videos. Unfortunately the timestamp mostly covered the -1, which I guessed must be -4. Still working on that one. Reached: ((5c -1) (3c -4)) / (5c + 4) (2c - 1)) Couldn't go any further, so decided to watch! Immediately discovered my imagined -4 was in fact -1, then arrived at your answer - although I tend to avoid the factoring by grouping method.
Factoring by grouping can be a pain, but you just have remember to multiply the two outside terms together to get the number you want to multiply to, then find two factors of that number that add up to the middle coefficient. Once you find these two factors, a useful method is to split the middle term into these two factors. This gives you four terms. Then you factor a GCF out of each pair of terms. If you did it right, you’ll notice that one the terms is repeated. This is your common factor. When you take out this term and multiply it by whats left over, you get your two factors.
Excellent video. Thanks
Amazing approach, thanks for sharing🎉
John, I really liked your approach of multiplying the outer terms together and then looking for factors that will result in the middle term when they're added together. I must remember that technique. (I guess the quadratic formula could have been used but it can be a bit complicated). I looked at 15c² -20c -3c + 4 in a different way from the one you took, grouping the 15c² and -3c together and the -20c and the 4 together as follows: (15c² - 3c) - (20c - 4), resulting in 3c(5c - 1) - 4(5c - 1) or (5c - 1)(3c - 4). The same result, of course, so obviously both are feasible..
It is logic that both ways give the same result, because it would be very arbitrary if you had to group 15c^2 - 20c to be able to factor correctly. So it is good that you grouped 15c^2 - 3c, as proof that both ways of grouping do the 'trick'. The conclusion is that you can group in both ways to get the right result. We see that also when we factor the denominator as follows, grouping 10c^2 - 2c and 5c - 1. Then we get: 2c(5c - 1) + 1(5c -1), resulting in (2c + 1)(5c - 1) or (5c - 1)(2c + 1). And of course the same 'end result': 3c - 4/2c + 1). :)
interesting solution thanks for the fun
super simple with good old foil.
Numerator:
15c² - 23c + 4
we can factor this to:
(5c - 1)(3c - 4)
Denominator:
10c² + 3c - 1
we can factor this to:
(5c - 1)(2c + 1)
the numerator and denominator have (5c -1) in common so we can cross them out. Leaving us with (3c-4) / (2c + 1)
(3c - 4)/(2c + 1)
Dammit - as usual I tried solving using the thumbnails running down the right side of the screen while I watched other videos. Unfortunately the timestamp mostly covered the -1, which I guessed must be -4. Still working on that one.
Reached:
((5c -1) (3c -4)) / (5c + 4) (2c - 1))
Couldn't go any further, so decided to watch!
Immediately discovered my imagined -4 was in fact -1, then arrived at your answer - although I tend to avoid the factoring by grouping method.
Factoring is the key,and following the order of operations.
3c-4/2c+1
My initial assumption was that the 2 quad eq were factorable and they were.
4/3 and1/5
2c + 1
Magic is at 11:30
I don't remeber ever being told the second method before.
I failed on this part of the GED
-3
Wow! What a pain in the patootie...
Factoring by grouping can be a pain, but you just have remember to multiply the two outside terms together to get the number you want to multiply to, then find two factors of that number that add up to the middle coefficient. Once you find these two factors, a useful method is to split the middle term into these two factors. This gives you four terms. Then you factor a GCF out of each pair of terms. If you did it right, you’ll notice that one the terms is repeated. This is your common factor. When you take out this term and multiply it by whats left over, you get your two factors.