anonymous
  • anonymous
Simplifying Algebraic Terms
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\frac{ 4x+6 }{ 4x ^{2}-9 }\]
DebbieG
  • DebbieG
Both the num'r and den'r can be factored. I'd start there.
DebbieG
  • DebbieG
there is a GCF in the num'r, so pull that out. The den'r is a "special" kind of binomial, that factors into the product of two binomials....

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anonymous
  • anonymous
IT will be something like this \[\[\frac{ 4x+6 }{ 4x ^{2} -9} = \frac{ 2\left( x+3 \right) }{ 4x ^{2}-9 }\]\]
DebbieG
  • DebbieG
Double-check that num'r (did you CHECK FACTORING WITH MULTIPLICATION?? I THINK NOT! ;) but that's almost it. Now the den'r....? do you know where to go with that?
anonymous
  • anonymous
Kinda lost :(
DebbieG
  • DebbieG
ok, let's start with the num'r. You have: 4x + 6 = 2(x + 3) But if you CHECK that factoring with multiplication, you'll see that there's something wrong: 2(x + 3)=2x + 6 So you need to fix that factoring (but more importantly, you need to always, ALWAYS, CHECK FACTORING WITH MULTIPLICATION! :)
anonymous
  • anonymous
So it will just stay the same?
DebbieG
  • DebbieG
I'm not sure what you mean. You need to factor it, but you need to do so correctly. You are correct that the GCF is 2, you just didn't factor it out correctly. 4x + 6 = 2(? + 3) What is 4x/2?
DebbieG
  • DebbieG
For the den'r, you'll need the "special product" rule, in reverse: product of a sum and a difference gives you a difference of squares: \(\Large (a +b)(a-b)=a^2-b^2\) Right? So the "reverse" direction means, that you can FACTOR a difference of squares into the product of a sum and a difference: \(\Large a^2-b^2=(a +b)(a-b)\) Now, does any of that look helpful with your den'r?

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