Answer:
(3g-5)(2g+7)
Explanation:
Compare
6g^2+11g-35 to
ag^2+bg+c.
We should see that a=6, b=11,c=-35.
It these is factoable over the rationals we should be able to find two numbers that multiply to be ac and add up to be b.
ac=6(-35)
b=11
Now I really don't want to actually find the product of 6(-35). I'm just going to play with the factors until I see a pair that adds up to 11.
6(-35)
30(-7) Moved a factor of 5 around.
10(-21) Moved a factor of 3 around.
10 and -21 is almost it. We just need to switch where the negative is because we want a sum of 11 when we add the numbers (not -11).
So b=-10+21 and ac=-10*21.
We are going to replace b in
6g^2+11g-35
with -10+21.
We can do this because 11 is -10+21.
Let's do it.
6g^2+(-10+21)g-35
6g^2+-10g+21g-35
Now we are going to factor the first two terms together and the second two terms together.
Like so:
(6g^2-10g)+(21g-35)
We are going to factor what we can from each pair.
2g(3g-5)+7(3g-5)
There are two terms both of these terms have a common factor of (3g-5) so we can factor it out:
(3g-5)(2g+7)