3 Hard Factoring Problems when a is not 1 

I have a test tmr and your video suddenly appeared thank you so much!❤

Nice explain sir

You could factor the first one by grouping, splitting 16x into 10x and 6x, 2x(2x+3)+5(2x+3), which becomes (2x+5)(2x+3).

The best - you are

Number 1 must be (2n + 15) (2n + 1)

yesss

Tysm! ;-;