1) 6x^2 + 2xy - 15x - 5y
factor by grouping
2x (3x + y ) - 5 (3x + y) (3x + y) is common to both terms
(3x + y) (2x - 5 )
2) 2x^2 + 7x + 5 we're looking for two terms that multiply to 2 for the first term and to 5 for the last term......trial and error produces
(2x + 5) (x + 1)
Here's another way to do that one....
2x2 + 7x + 5
Multiply the 2 and the 5 together to get 10 .
What two numbers multiply to 10 and add to 7 ? → +2 and +5
So, we can split the middle term like this:
2x2 + 2x + 5x + 5 Notice here how if we combined like terms, we'd get 7x again.
Factor 2x out of the first two terms.
2x(x + 1) + 5x + 5
Factor 5 out of the last two terms.
2x(x + 1) + 5(x + 1)
Factor (x + 1) out of both remaining terms.
(x + 1)(2x + 5)