# Thread: Factoring Trinomials ....

1. ## Factoring Trinomials ....

Sorry These were the first forums I found and I thought maybe since you all seem high in math to be able to help me. And people post on these regulary unlike the other empty math forums.

Im having a really hard time Factoring by grouping...

Can someone explain to me how it works I understand Foil somewhat of how the order goes so no need to clarify on that part.
2t^2+3t-2 (for those of you not following the way my teacher taught as to start was to do this )
2 * -2 = -4
4 * -1 = -4
4 + -1 = 3 I don't even understand this first step. I kno there muliplying the outnumber and then finding out what multiplies to get those numbers combined and what ADDS up to get the Middle number , well not that complicated. Now my teacher said I write it like this (2t^2-t) + (4+-2) next line is the 3rd step t(2t-1) + 2(2t-1) . Now that the parenthese are the same theres one more step but I don't kno if I did it right 2t^2-1 + 4t-1. See thats totally wrong I have no idea what im doing here PLEASE any help is greatly appreciated for I have a test on it tommorow.

If that problem was too difficult or strange can you help me with this one
x^2+5x - 6 . (factor the expression)

2. Originally Posted by D@nny
Sorry These were the first forums I found and I thought maybe since you all seem high in math to be able to help me. And people post on these regulary unlike the other empty math forums.

Im having a really hard time Factoring by grouping...

Can someone explain to me how it works I understand Foil somewhat of how the order goes so no need to clarify on that part.
2t^2+3t-2 (for those of you not following the way my teacher taught as to start was to do this )
2 * -2 = -4
4 * -1 = -4
4 + -1 = 3 I don't even understand this first step. I kno there muliplying the outnumber and then finding out what multiplies to get those numbers combined and what ADDS up to get the Middle number , well not that complicated. Now my teacher said I write it like this (2t^2-t) + (4+-2) next line is the 3rd step t(2t-1) + 2(2t-1) . Now that the parenthese are the same theres one more step but I don't kno if I did it right 2t^2-1 + 4t-1. See thats totally wrong I have no idea what im doing here PLEASE any help is greatly appreciated for I have a test on it tommorow.

If that problem was too difficult or strange can you help me with this one
x^2+5x - 6 . (factor the expression)
You need to factor,
$\displaystyle x^2+5x-6$
1)Find all the numbers such as when multiplied give, -6.
Those are,
$\displaystyle \left\{ \begin{array}{c} 1,-6\\-1,6\\2,-3,\\-2,3$
Then you add them up,
$\displaystyle \left\{ \begin{array}{c}1+(-6)=-5\\-1+6=5\\2+(-3)=-1\\-2+3=1$
Since the middle number in this problem is 5, the correct pair is, $\displaystyle -1,6$
That means it factors as
$\displaystyle (x-1)(x+6)$.

Note you always need to express your trinomial with the squared term first then the x terms and finally the number. And this method only works for a trinomial having a 1 in front of the x squared terms like here,
$\displaystyle 1x^2+5x-6$ so it works.

3. Not the way my teacher does it but it defnitly does work. Can I only do it this way when the number thats squared is just a varible or would this method work for
6x^2 + 5x + 1
How would I do that one?
I have the anser which is (3x+1) ( 2x+1) I just don't know how to show the work.
Also can you verify if the method I mentioned for the first problem the same thing as what you did by finding all the factors? That method I hate and yours I understand but it will also be difficult when you have to factor larger numbers no?
Like persay 100b^2 + 180b + 81 My teacher would tell me to multiply 100b^2 * 81 then find out what 2 numbers multiply to get 1800b^2 LOL man that's totally wrong and equal 180. Nm im getting all the methods confused on how to do each one damnit. Can you help me out some more plz... I do understand factoring things like 16k^2 - 49 I'd make it (8k - 7 )^2 or 25w^2 - 196 factoring would be (5w - 14 )^2 . Thats just like square rooting everything. factoring by grouping the book gives me an example y^3 + 3y^2 + 4y + 12 = y^2(y+3) + 4(y+3) = (y^2+4)(y+3) How did it get that? Im not following it. Please explain further . thanks tons

4. Originally Posted by D@nny
Not the way my teacher does it but it defnitly does work. Can I only do it this way when the number thats squared is just a varible or would this method work for
6x^2 + 5x + 1
How would I do that one?
I have the anser which is (3x+1) ( 2x+1) I just don't know how to show the work.
Also can you verify if the method I mentioned for the first problem the same thing as what you did by finding all the factors? That method I hate and yours I understand but it will also be difficult when you have to factor larger numbers no?
Like persay 100b^2 + 180b + 81 My teacher would tell me to multiply 100b^2 * 81 then find out what 2 numbers multiply to get 1800b^2 LOL man that's totally wrong and equal 180. Nm im getting all the methods confused on how to do each one damnit. Can you help me out some more plz... I do understand factoring things like 16k^2 - 49 I'd make it (8k - 7 )^2 or 25w^2 - 196 factoring would be (5w - 14 )^2 . Thats just like square rooting everything. factoring by grouping the book gives me an example y^3 + 3y^2 + 4y + 12 = y^2(y+3) + 4(y+3) = (y^2+4)(y+3) How did it get that? Im not following it. Please explain further . thanks tons
You want to factor,
$\displaystyle 6x^2+5x+1$
This one is tricker, the thing is to exhaust all the possibilities.
You are going to have,
(_X+_)(_X+_)
You need to fill in the blanks
You need them to be factors of six, the ones in front and the ones in the back need to be multiply to one.
Thus, the possibilities are,
(_X+1)(_X+1)
or
(_X-1)(_X-1)
Let us pick one, say the first one.
The blanks multiply to six thus, try all the possibilites,
(6X+1)(X+1)
(2X+1)(3X+1)

Note:For the front ones you do not need to try the negative ones also.

If you open them up you see that the second factor is good thus,
$\displaystyle 6x^2+5x+1=(2x+1)(3x+1)$