Factoring, please help me?

1. (x-5)(x+3)

2.

3. (3x+4)(2x+5)

Think of it like this: your equations are (x+a)(x+b)=x^2+(a+b)x+ab)

a and b are constants.

1. x^2-2x-15=x^2+(3-5)x+(3)(-5)

=(x+3)(x-5)

Now when you have a coeffiecint on x^2, you want

(ax+b)(cx+d)=acx^2+bcx+adx+bd

=acx^2+(bc+ad)x+bd

OR, when you just think of it like the first one,

cx^2+

3. 6x^2+23x+20=x^2+23x+120 {ONLY FOR THOUGHT PROCESS, DO NOT WRITE}

if you notice, no factors of 120 will sum up to 23, so this is prime. if it DID exist, and used this thought process, you would keep it as such with a 1 coefficient for x^2. suppose it came out to be (x+18)(x+4). Then, to break it up to the 6, you would find factors of 6 who also factor the a and b in (x+a)(x+b) as well. i.e. (2x+6)(3x+2). If you expand that, you do get 6x^2 and a last term of 12, but the cx term is different because it depends on the sum of factors.

For #2, we use a method called factoring by grouping. This is simple, watch:

2. 2x^3 +6x^2+x+3=2x^2(x+3) + (x+3) [We factored 2x^2 out of the first two terms]

=(x+3)(2x^2 + 1) [since there is a (x+3) in both, we can factor it out to a purely factored form]

Hope that helped!

(x+3)(x-5) the first one is easy because one of the factors of a number ending in 5 is always going to be 5 so it's either 5*3 or 15*1 ... I picked 5*3 and got the right one easy ... and since 3 - 5 = -2 that shows me the order of the numbers

x^2-2x-15

=(x-5)(x+3)

2x^3+6x^2+x+3

=2x^2(x+3)+(x+3)

=(2x^2+1)(x+3)

6x^2+23x+20

=(3x+4)(2x+5)