G33k
  • G33k
i needed to edit this question :p
Mathematics
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katieb
  • katieb
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anonymous
  • anonymous
i am lousy at factoring how about you?
anonymous
  • anonymous
you kind have a choice for the first part could be \[(2x+a)(2x+b)\] or \[(4x+a)(x+b)\]
anonymous
  • anonymous
there are people who tell you there is a method for doing it, but i have never seen one that did not involve knowing the answer to begin with i usually grind it till you find it we can try \[(4x+2)(x+3)\] and see if it works, or at least see why it does not work

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jim_thompson5910
  • jim_thompson5910
4x^2+11x+6 The first coefficient is 4 The last term is 6 Multiply them: 4*6 = 24 Now find two numbers that multiply to 24 and add to 11 (middle coefficient) at the same time These two numbers are 3 and 8 so we can break up the 11x into 8x+3x ---------------------------- 4x^2+11x+6 turns into 4x^2+8x+3x+6 Now factor by grouping
jim_thompson5910
  • jim_thompson5910
Alternatively, you can use the quadratic formula to find the roots. Then you use the roots to find the factorization
anonymous
  • anonymous
you will definitely get \(4x^2\) and also get the \(6\) but you will not get \(11x\) because \(2x+12x\neq 11x\)
anonymous
  • anonymous
try \[(4x+3)(x+2)\]
anonymous
  • anonymous
@jim_thompson5910 yes, if we know this Now find two numbers that multiply to 24 and add to 11 (middle coefficient) at the same time then we know it, but that seems to me the crux of the matter
jim_thompson5910
  • jim_thompson5910
yeah the guess and check can be a bit tedious sometimes. Which is why the quadratic formula method is probably more efficient
anonymous
  • anonymous
true. but i have NEVER seen it taught that way. always factoring comes first, then solving by factoring, then completing the square, then the quadratic formula
anonymous
  • anonymous
with the third and most useful step usually skipped
anonymous
  • anonymous
@G33k did you try multiplying \[(4x+3)(x+2)\] to see if it works?

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