Ask your own question, for FREE!
Algebra 20 Online
OpenStudy (sand-lock53):

How do you do this algebra? Question in the comments...

OpenStudy (sand-lock53):

OpenStudy (3mar):

What you picked?

OpenStudy (sand-lock53):

1

OpenStudy (emobunny2200):

On the first worksheet the answer would be D

OpenStudy (emobunny2200):

Answer to second worksheet is B

OpenStudy (sand-lock53):

I don't want the answers, I want to know how to do it

OpenStudy (3mar):

So I will

OpenStudy (3mar):

For 1st pic, just make two parentheses. \[(..............)(.................)\] ok?

OpenStudy (emobunny2200):

Oh. Well. I'm just using a site for these.

OpenStudy (emobunny2200):

GoodLuck!

OpenStudy (sand-lock53):

k, what goes in the parenthese?

OpenStudy (3mar):

Nice question! in the 1st and 2nd, type y \[(y..........)(y............)\]

OpenStudy (sand-lock53):

(y+?) (y-?) is this correct?

OpenStudy (phi):

in this problem, they give you the possible factors, so we can simplify by using this info: look at the sign of the last number i.e. -60 if the sign is *negative* then the two number in the factors will have *different signs* that rule (which you should memorize) helps to eliminate lots of choices also, look at the sign of the middle number i.e. +4 that means the bigger factor will be positive. (and because the signs are different due to rule 1 up above), the other factor will be minus

OpenStudy (3mar):

Correct!

OpenStudy (3mar):

so what is the next ? we are looking for two numbers that their sum is 4 and their product is -60 what are these two numbers?

OpenStudy (sand-lock53):

10 & -6

OpenStudy (3mar):

type the parentheses fully now, please!

OpenStudy (sand-lock53):

(y+10) (y-6)

OpenStudy (3mar):

Great! Then it will be ...# choice?

OpenStudy (phi):

one more rule look at the sign of the last number is *positive* then the two number in the factors will have *same signs* the other rule for the middle number is the same: the sign of the middle number is the sign of the bigger factor. (and because both have the same sign, it is also the sign of the smaller factor)

OpenStudy (sand-lock53):

That would give me D

OpenStudy (3mar):

Congratulations! It is D.

OpenStudy (sand-lock53):

I'll try to do question 2 by myself...

OpenStudy (phi):

The last one is trickier. The key idea is you can factor 4 from each term in other words, you can write the problem as 4(h^2 + 6h +5) but we now have to factor h^2+6h+5

OpenStudy (3mar):

I am happy for that. and I am here for any obstacles.

OpenStudy (3mar):

Let me know when you have finished, and take care of the last one!

OpenStudy (sand-lock53):

okay, I got (x-15) (x-15), which would give me D

OpenStudy (3mar):

Great! Correct! and the last one?

OpenStudy (3mar):

phi said "The last one is trickier. The key idea is you can factor 4 from each term in other words, you can write the problem as 4(h^2 + 6h +5) but we now have to factor h^2+6h+5"

OpenStudy (3mar):

Time for pray Salam! 15 min

OpenStudy (sand-lock53):

???

OpenStudy (3mar):

Thank you for the medal!

OpenStudy (sand-lock53):

You're welcome!

OpenStudy (sand-lock53):

I am a bit confused on the last one

OpenStudy (3mar):

which one?

OpenStudy (3mar):

@sand-lock53

OpenStudy (3mar):

any idea?

OpenStudy (3mar):

Where are you stuck?

OpenStudy (3mar):

ha?

OpenStudy (sand-lock53):

I don't recall learning it in the lesson, but I think the answer is C

OpenStudy (3mar):

Let me explain, please There are two ways to proceed. 1. take 4 as a common factor from the three terms or 2.resolve the three terms directly I prefer 1st one

OpenStudy (sand-lock53):

Would the numbers be 5 & 6?

OpenStudy (3mar):

\[4h^2+24h+20=4(h^2+6h+5)=4(h+1)(h+5)\] try this , test it, and let me know when you are pretty sure

OpenStudy (3mar):

5 and 1

OpenStudy (3mar):

got it?

OpenStudy (sand-lock53):

yes

OpenStudy (sand-lock53):

So, i have to divide all three numbers by 4?

OpenStudy (sand-lock53):

@3mar

OpenStudy (3mar):

Yes you have, then you can easily factor them!

OpenStudy (3mar):

Sorry, I was not online!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!