anonymous
  • anonymous
Factor Completely: 18x^2-35x+12 I know how to factor, but I am not sure of the rule if products of 12 don't equal 35.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
x(18x-35)+12 (2x-3)(9x-4)
anonymous
  • anonymous
So you factored out the x, left the 12 by itself, but then how did you get that?
anonymous
  • anonymous
Ah, but that \[18x^{2} \] changes everything.

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anonymous
  • anonymous
Okay, I see what you did there. I think, anyway. You pulled out x(18x-35) + 12. Well, actually, I still don't get how you got (2x-3)(9x-4)
anonymous
  • anonymous
Since it's not a monic quadratic, you're trying to find two numbers that multiply to be 216 rather than 35, and add to be -35. [[216 comes from 18*12]] -27 and -8 work, so split 35x into -27x and -8x 18x² - 35x + 12 = 18x² - 27x - 8x + 12 = 9x(2x - 3) - 4(2x - 3) = (9x - 4) (2x-3)
anonymous
  • anonymous
So you can multiple the coefficient of a by c, then find something in there that equals 35?
anonymous
  • anonymous
*multiply
anonymous
  • anonymous
Just pick some representation of 18 as the product of two factors: 18 & 1 9 & 2 6 & 3 Since the middle term is negative and the first and last terms are positive you know you have something in the form (ax - c)(bx - d) And you know that a and b must be 2 numbers on that list. So you have c*d = 12 ad + bc = 35 And you have 3 different sets you can test for a/b to solve for the c and d using the system above.
anonymous
  • anonymous
To factorize the polynomial ax² + bx + c, you need to find 2 numbers (let them be m and n) that satisfy these two conditions: 1) m*n = a*c 2) m + n = b Then you split bx into mx and nx and factorize by grouping in pairs
anonymous
  • anonymous
Okay, I get that, ty both vm :D
anonymous
  • anonymous
no problem =)

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