Simplify the rational expression by factoring the polynomial – MUST KNOW ALGEBRA!

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[johnmarchington]

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..

[nmin]

Amazing approach, thanks for sharing🎉

[russelllomando]

interesting solution thanks for the fun

[bigdog]

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)

[edpottinger]

Factoring is the key, and following the order of operations.

[vespa]

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.

[terry_willis]

My initial assumption was that the 2 quad eq were factorable and they were.

[stephena]

I don't remeber ever being told the second method before.