anonymous
  • anonymous
factor by grouping \[6y ^{2}+11y+3\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Hey noone helping you anymore?
anonymous
  • anonymous
a+b=11/6 a x b=3/6
anonymous
  • anonymous
no. And I'm google searching trying to find out how to do this. what do you mean by the a+b=11/6 axb=3/6

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anonymous
  • anonymous
you have to find a and b, one times the other is 3/6 and the sum is -11/6, I am not sure the easiest way to solve this one.
anonymous
  • anonymous
so it would be prime. because there isn't anything that can go into all the numbers. Idk.. I'm confused
anonymous
  • anonymous
@mxolisi3903 can you help me?
anonymous
  • anonymous
I hate math...
anonymous
  • anonymous
Im checking it. But I am not sure yet. If you find numbers that add to 11 but multiply to 3 tell me.
anonymous
  • anonymous
none that I can think of
anonymous
  • anonymous
Let me look
anonymous
  • anonymous
\[6(y^2+ \frac{11}{6} y +\frac{3}{6})\]
anonymous
  • anonymous
But thats not factorising by grouping Daniel
anonymous
  • anonymous
It is not. I think I got it now, kinda \[(2y + a)(3y+b)\]
anonymous
  • anonymous
\[2b+3a=11 \\ a\times b=3 \]
anonymous
  • anonymous
Idk either way.. I'm completely lost because there is no a or b in the actual equation.
anonymous
  • anonymous
But Daniel when you multiply that out you dont get back to what we had before
anonymous
  • anonymous
The a and the b are the unkown terms @jewels89
anonymous
  • anonymous
6y^2+11y+3 --> 6y^2+9y+2y+3 --> 3y(2y+3) + (2y+3) --> (3y+1)(2y+3) Would this be right and if so can someone explain it to me? My fiance doesn't know how to explain much of any of this to my level.
anonymous
  • anonymous
Yes that is absolutely correct ! I am gonna explain in a moment
anonymous
  • anonymous
I think that is the right answer. To check \[(3y+1)(2y+3)\] You multiply the first item for both in the second parenthesis: \[3y \times 2y\ + 3y \times 3\] And sum with the second term multiplied by the second parenthesis. I think it matches the original trinomial.
anonymous
  • anonymous
Takes some typing to make all that.
anonymous
  • anonymous
|dw:1370227560562:dw|
anonymous
  • anonymous
Because when we grouping and we only have three terms we have to think of an intuitive way of writing the polynomial in four terms as you did say \[6y ^{2} +9y +2y + 3\] noticing that 9 and 2 add up to the middle term we had before which is 11 and so we are correct surely and yeah we can see that just for the first two terms there comes out a common factor \[3y(2y +3) +(2y+3)\] and hey we can see that \[(2y+3)\] appears twice in the expression and so it is indeed the common factor as well so we just take it out \[(2y+3){3y+1} = (2y+3)(3y+1)\] as expected and you can check by finding the product that this is indeed what we had before
anonymous
  • anonymous
I'm not understand how we go from 3y(2y+3) + (2y+3) to (3y+1)(2y+3)
anonymous
  • anonymous
You just take out (2y+3) out as a common factor
anonymous
  • anonymous
ok so then it's 3y + (2y+3) so where does the +1 come from
anonymous
  • anonymous
and you can see that if you do that inside you will have (3y+1)
anonymous
  • anonymous
Oh my friend when you factor out say \[x ^{2} +x = x(x+1)\] do you understand how I got that 1 in this expression please feel free to say no if you don't so I can explain
anonymous
  • anonymous
:-(
anonymous
  • anonymous
no
anonymous
  • anonymous
I'm math illiterate and if it weren't for me going to into Nursing, I wouldn't be taking this right now... even though nursing doesn't really use mathematics of this sort. Only measurements.
anonymous
  • anonymous
Good ...I took out the common factor \['x'\] and then I divided each term by \['x'\] like this \[\frac{ x^2 }{ x } +\frac{ x }{ x } = x +1 \] but remembering that I had a common factor which I took out and so I should multiply the result I got by the commmon factor which should be of course \['x'\] and so we have \[x(x+1)\]....so in general whenever you take out a common factor you have to take it out and divide each term of the original expression by the common factor you took out and when you write your final answer dont forget to multiply the result by the common factor...I hope tha helped
anonymous
  • anonymous
Ok. Now I think I am beginning to understand. Thank you! :)

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