Ask your own question, for FREE!
Mathematics 16 Online
OpenStudy (shiree19):

Please Help I don't understand!!!! Medal! Factor the express completely. 9r^2 + 3r - 30

OpenStudy (shiree19):

\[9r^{2} + 3r - 30\]

OpenStudy (shiree19):

@Loser66 @freckles @Nnesha

OpenStudy (shiree19):

I have a few problems similar to this. I know the answer I just don't know how to solve or explain it.

jimthompson5910 (jim_thompson5910):

First factor out the GCF 3 9r^2 + 3r - 30 = 3(3r^2 + r - 10) Now we factor 3r^2 + r - 10. One way to factor 3r^2 + r - 10 quickly is to use the quadratic formula. Use the quadratic formula to solve 3r^2 + r - 10 = 0 for r to get r = 5/3 or r = -2 You can then use those roots to construct the factorization r = 5/3 or r = -2 3r = 5 or r = -2 3r - 5 = 0 or r + 2 = 0 (3r - 5)(r + 2) = 0 So 3r^2 + r - 10 factors to (3r - 5)(r + 2) Therefore, 9r^2 + 3r - 30 factors to 3(3r - 5)(r + 2)

OpenStudy (shiree19):

sorry I typed the wrong thing. wrong problem's answer.

OpenStudy (shiree19):

Okay I understand this one thanks. Can you help me with a few more?

jimthompson5910 (jim_thompson5910):

sure, 2 more

OpenStudy (shiree19):

\[27w ^{2} - 12\]

OpenStudy (shiree19):

and the other is \[12y ^{3} + 4y ^{2} - 9y - 3\]

OpenStudy (shiree19):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

the first one, we can factor out the GCF 3 27w^2 - 12 = 3(9w^2 - 4) then we can factor 9w^2 - 4 further using the difference of squares rule a^2 - b^2 = (a-b)(a+b) 9w^2 - 4 = (3w)^2 - (2)^2 = (3w - 2)(3w + 2) making 27w^2 - 12 completely factor into 3(3w - 2)(3w + 2)

jimthompson5910 (jim_thompson5910):

alternatively, you can think of 9w^2 - 4 as 9w^2 + 0w - 4 and either use the quadratic formula (similar to shown above) or think to yourself "what two numbers multiply to -4*9 = -36 and also add to 0?". The answer to that question is "-6 and +6". Those two values will help you factor 9w^2 + 0w - 4 by grouping

jimthompson5910 (jim_thompson5910):

making sense?

OpenStudy (shiree19):

Yes thank you!

OpenStudy (shiree19):

And the last one?

jimthompson5910 (jim_thompson5910):

Use factor by grouping to factor \[\Large 12y^3 + 4y^2 - 9y - 3\] \[\Large (12y^3 + 4y^2) + (-9y - 3)\] \[\Large 4y^2(3y + 1) + (-9y - 3)\] \[\Large 4y^2(3y + 1) -3(3y + 1)\] \[\Large (4y^2 -3)(3y + 1)\]

OpenStudy (shiree19):

Thank you Thank you Thank you Thank you Thank you Thank you Thank you

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!