A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
No matter how hard I try I can not figure out how to factor this: 4a^232a+63. Any help please? Thanks so much. Please explain the steps!
 2 years ago
No matter how hard I try I can not figure out how to factor this: 4a^232a+63. Any help please? Thanks so much. Please explain the steps!

This Question is Closed

znimon
 2 years ago
Best ResponseYou've already chosen the best response.1So take the coefficient on the a squared term and multiply that to 63 you need to find two number (we will call them A and B) such that A * B = 4 * 63 and A + B = 32

znimon
 2 years ago
Best ResponseYou've already chosen the best response.1you do this with other polynomials but the coefficient on the squared term is usually a 1so most people don't teach it that way.

lildancngurl
 2 years ago
Best ResponseYou've already chosen the best response.0You are fabulous. Thank you so much! No matter how hard I tried, I could not understand it. You've made it so much simpler!

lildancngurl
 2 years ago
Best ResponseYou've already chosen the best response.0I do have one more question...The two numbers I got were 14 and 18.. How did you get 7 and 9 from that?

precal
 2 years ago
Best ResponseYou've already chosen the best response.0znimon is busy at the moment, have you tried to do factor by grouping

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1Alternatively, you can set the expression equal to zero to get \[\Large 4a^232a+63 = 0\] Now solve for 'a' using the quadratic formula \[\Large a = \frac{B\pm\sqrt{B^24AC}}{2A}\] \[\Large a = \frac{(32)\pm\sqrt{(32)^24(4)(63)}}{2(4)}\] \[\Large a = \frac{32\pm\sqrt{10241008}}{8}\] \[\Large a = \frac{32\pm\sqrt{16}}{8}\] \[\Large a = \frac{32+\sqrt{16}}{8} \ \text{or} \ a = \frac{32\sqrt{16}}{8}\] \[\Large a = \frac{32+4}{8} \ \text{or} \ a = \frac{324}{8}\] \[\Large a = \frac{36}{8} \ \text{or} \ a = \frac{28}{8}\] \[\Large a = \frac{9}{2} \ \text{or} \ a = \frac{7}{2}\] Now rearrange to get the right sides equal to zero \[\Large a = \frac{9}{2} \ \text{or} \ a = \frac{7}{2}\] \[\Large 2a = 9 \ \text{or} \ 2a = 7\] \[\Large 2a  9 = 0 \ \text{or} \ 2a  7 = 0\] Finally, use the zero product property to get \[\Large (2a  9)(2a  7) = 0\] So this shows us that \[\Large 4a^232a+63 = 0\] factors to \[\Large (2a  9)(2a  7)\]

lildancngurl
 2 years ago
Best ResponseYou've already chosen the best response.0Wow, extraordinarily helpful. Thank you!
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.