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anonymous
 3 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!
anonymous
 3 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!

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you do this with other polynomials but the coefficient on the squared term is usually a 1so most people don't teach it that way.

anonymous
 3 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!

anonymous
 3 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
 3 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
 3 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 \[\Large (2a  9)(2a  7)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wow, extraordinarily helpful. Thank you!
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