factor by grouping
\[6y ^{2}+11y+3\]

- anonymous

factor by grouping
\[6y ^{2}+11y+3\]

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- anonymous

Hey noone helping you anymore?

- anonymous

a+b=11/6
a x b=3/6

- 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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## More answers

- 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

so it would be prime. because there isn't anything that can go into all the numbers. Idk.. I'm confused

- anonymous

@mxolisi3903 can you help me?

- anonymous

I hate math...

- 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

none that I can think of

- anonymous

Let me look

- anonymous

\[6(y^2+ \frac{11}{6} y +\frac{3}{6})\]

- anonymous

But thats not factorising by grouping Daniel

- anonymous

It is not. I think I got it now, kinda
\[(2y + a)(3y+b)\]

- anonymous

\[2b+3a=11 \\
a\times b=3
\]

- anonymous

Idk either way.. I'm completely lost because there is no a or b in the actual equation.

- anonymous

But Daniel when you multiply that out you dont get back to what we had before

- anonymous

The a and the b are the unkown terms @jewels89

- 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

Yes that is absolutely correct ! I am gonna explain in a moment

- 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

Takes some typing to make all that.

- anonymous

|dw:1370227560562:dw|

- 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

I'm not understand how we go from 3y(2y+3) + (2y+3) to (3y+1)(2y+3)

- anonymous

You just take out (2y+3) out as a common factor

- anonymous

ok so then it's 3y + (2y+3) so where does the +1 come from

- anonymous

and you can see that if you do that inside you will have (3y+1)

- 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

no

- 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

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

Ok. Now I think I am beginning to understand.
Thank you!
:)

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