Factor the expression 6g^2+11g-35

(3g-5) (2g+7)

Step-by-step 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)