anonymous
  • anonymous
No matter how hard I try I can not figure out how to factor this: 4a^2-32a+63. Any help please? Thanks so much. Please explain the steps!
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
(2 a-9) (2 a-7)
anonymous
  • anonymous
So 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
  • anonymous
you do this with other polynomials but the coefficient on the squared term is usually a 1so most people don't teach it that way.

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anonymous
  • anonymous
You 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
  • anonymous
Your welcome.
anonymous
  • anonymous
I do have one more question...The two numbers I got were 14 and 18.. How did you get 7 and 9 from that?
precal
  • precal
znimon is busy at the moment, have you tried to do factor by grouping
precal
  • precal
|dw:1346462806136:dw|
precal
  • precal
|dw:1346462837874:dw|
precal
  • precal
|dw:1346462861944:dw|
precal
  • precal
|dw:1346462941803:dw|
precal
  • precal
|dw:1346462966127:dw|
precal
  • precal
hope that helps
jim_thompson5910
  • jim_thompson5910
Alternatively, you can set the expression equal to zero to get \[\Large 4a^2-32a+63 = 0\] Now solve for 'a' using the quadratic formula \[\Large a = \frac{-B\pm\sqrt{B^2-4AC}}{2A}\] \[\Large a = \frac{-(-32)\pm\sqrt{(-32)^2-4(4)(63)}}{2(4)}\] \[\Large a = \frac{32\pm\sqrt{1024-1008}}{8}\] \[\Large a = \frac{32\pm\sqrt{16}}{8}\] \[\Large a = \frac{32+\sqrt{16}}{8} \ \text{or} \ a = \frac{32-\sqrt{16}}{8}\] \[\Large a = \frac{32+4}{8} \ \text{or} \ a = \frac{32-4}{8}\] \[\Large a = \frac{36}{8} \ \text{or} \ a = \frac{28}{8}\] \[\Large a = \frac{9}{2} \ \text{or} \ a = \frac{7}{2}\] 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 \[\Large 4a^2-32a+63 = 0\] factors to \[\Large (2a - 9)(2a - 7)\]
anonymous
  • anonymous
Wow, extraordinarily helpful. Thank you!
precal
  • precal
yw

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