Question

9b^2 + 3b -20
factor completely

Answer 1

Can be done by factoring. You will not have to resort to the general solution for quadratics. Try both (9b )(b ) and (3b )(3b ). Any more help than that and I'm just handing you an answer.

Answer 2

Still there? Need help with how to factor?

Answer 3

yes here, i dont think that (9b ) (b ) is right

Answer 4

If you don't think that one works (it may or may not be the right method), then try the other. If you can't get that one to work either, neither one, then we need to walk through how to factor, and I can help you.

Answer 5

i was doing the method where m + n = b and m*n=b

Answer 6

m + n=c

Answer 7

that's right. Very good logic.

Answer 8

but im stuck i can not figure it out.

Answer 9

No problem, I can help you. Some quick words first though.

Answer 10

Your logic is great and is to be commended. However, the method that that gives just leads to another quadratic, so factoring has to be a little bit intuitive and less analytical. The form (3b )(3b ) is the correct one and we just have to find 2 numbers that when multiplied give -20 but when added and then multiplied by 3 give 3. I know you know that, but the question is how? Believe it or not, it's trial and error.

Answer 11

thank you, so is this prime?

Answer 12

No, not prime. Follow along here. This will help you. So, we know the two numbers are of opposite signs because we have a negative 20. This gives (3b + )(3b - ). So we have the signs, which is a big improvement.

Answer 13

wait (3b+5) (3b-4)

Answer 14

Yes. and you have had to trial-and-error that with intuitively looking for the bigger # to be + because of the "3".

Answer 15

Review this. It's a good problem and you will learn how to combine analytic thinking with intuitive which is powerful.

Answer 16

thank you.. I still do not understand how that -4 and 5, relate to the 3...

Answer 17

I just did foiled and got the original

Answer 18

It's because we are "trial-and-error"ing factors of -20. We pick +5 and -4 instead of -5 and +4, not because of the value of "3", but because it is positive and 5 is bigger in absolute value than the -4.

Answer 19

so (3b - 4)(3b + 5) is your answer and you got it yourself basically. I just filled in clarity in thinking.

Answer 20

haha yeah i just don't know how to begin sometimes, and sometimes get lost while doing it and just have always been horrible at factoring. I have a new problem, 15x^3 - 9x -6x

Answer 21

ok, good, first step is to take out a 3 from all terms. That's always the first thing to look at, to see if you can factor the coefficients.

Answer 22

Are you sure that the second term is correct? You have 2 terms with the power of 1 on x. I'm thinking that maybe a power of 2 is supposed to go on one of them?

Answer 23

9x^2 Sorry

Answer 24

np.

Answer 25

so, a 3 can factor out and so also can an "x". So, you will have 3x(5x^2 - 3x - 2) to start with which is a BIG improvement since the coefficients are smaller and we have a quadratic instead a cubic.

Answer 26

So you still have to factor out (5x^2 - 3x - 2) which can also be done by factoring. It will of course have the form (5x +- )(x +- ).

Answer 27

To get the rest, you only have 1 x 2 = 2, so you have to figure out (again, trial-and-error) where to put the signs and where then to put either 1 then 2 or 2 and then 1. That's all there is to it.

Answer 28

Making sense to you?

Answer 29

Hint: In (5x +- )(x +- ), one sign will be - and the other will be + because in (5x^2 - 3x - 2) we have a negative 2. But here, unlike the last problem where (3b )(3b ) where we could arbitrarily put + then -, we have 5x and x, so we have to try both ways. You still there?