anonymous
  • anonymous
f(x) = 6x2 - x - 12 and g(x) = 2x - 3 find the function (f/g)(x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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UsukiDoll
  • UsukiDoll
\[f(x) = 6x^2-x-12\] \[g(x) = 2x-3 \] If your question is just asking to find the function \[\frac{f(x)}{g(x)}\] then we just place the f(x) here ------------------ place the g(x) here the notation I'm using is the same as (f/g)(x)
anonymous
  • anonymous
But I can't simplify 6x^2 - x -12. So there is no real answer for this equation?
UsukiDoll
  • UsukiDoll
we could try to simplify 6^x-x-12 using b^2-4ac which is the discriminant.

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UsukiDoll
  • UsukiDoll
(-1)^2-4(6)(-12) for a = 6, b = -1, and c = -12
UsukiDoll
  • UsukiDoll
if we have a perfect square number like 25 16 9 then we can factor if we have a non-perfect square number like 3 -32... then we can't factor and have to use the quadratic equation (or just leave the problem alone in this case)
UsukiDoll
  • UsukiDoll
so from the discriminant (-1)^2-4(6)(-12) = 1-24(-12) =1+288 =289 we have a perfect square .. because the square root of 289 is 17
UsukiDoll
  • UsukiDoll
so we can factor 6x^2-x-12... it becomes (2x-3)(3x+4)
anonymous
  • anonymous
Oh okay. so then the 2x -3 crosses each other out which leaves 3x+4?
UsukiDoll
  • UsukiDoll
yes! that's right! the 2x-3 terms cancels out
UsukiDoll
  • UsukiDoll
\[\frac{(2x-3)(3x+4)}{2x-3} \rightarrow 3x+4 \]
anonymous
  • anonymous
Okay so if the question is Given f(x) = x + 2 and g(x) = x2 - 4, find the function (fg)(x) then the answer would be x^3 -2x^2- 4x - 8
anonymous
  • anonymous
Or did I do that wrong? Sorry. Still trying to get the hang of this
UsukiDoll
  • UsukiDoll
so (fg(x) or f(x)(g(x)) means that we multiply the f(x) function with the g(x) function
UsukiDoll
  • UsukiDoll
f(x)(g(x) = \[(x+2)(x^2-4) \] then we use foil method
UsukiDoll
  • UsukiDoll
\[x^3-4x+2x^2-8\]
UsukiDoll
  • UsukiDoll
you've done it right....it's just that my terms have been rearranged. It means the same thing
anonymous
  • anonymous
Okay got it! Thank you so much

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