Question

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

Answer 1

Hey noone helping you anymore?

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

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

Answer 6

@mxolisi3903 can you help me?

Answer 7

I hate math...

Answer 8

Im checking it. But I am not sure yet. If you find numbers that add to 11 but multiply to 3 tell me.

Answer 9

none that I can think of

Answer 10

Let me look

Answer 11

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

Answer 12

But thats not factorising by grouping Daniel

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

The a and the b are the unkown terms @jewels89

Answer 18

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.

Answer 19

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

Answer 20

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.

Answer 21

Takes some typing to make all that.

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

:-(

Answer 29

no

Answer 30

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.

Answer 31

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

Answer 32

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