# Math Help - Simple Algebra 1 stuff

1. ## Simple Algebra 1 stuff

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)

Im not mathmatician ; not that great at math. I just wasn't born with good understanding of math probaly why im so clumsy also. Thanks for your help in advance to a math newbie.

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)

Im not mathmatician ; not that great at math. I just wasn't born with good understanding of math probaly why im so clumsy also. Thanks for your help in advance to a math newbie.
The first thing you need to recall is that if $x-\alpha$ is a factor of
$ax^2+bx+c$, then $a\alpha^2+b\alpha+c=0$.

Now for $2t^2+3t-2$ we try putting $t=-2$, then this is:

$2\times (-2)^2+3\times(-2)-2=2\times 4-3\times 2-2=8-6-2=0$

so $t+2$ is a factor of $2t^2+3t-2$, so either by polynomial
long division, inspection or trial and error we have:

$
2t^2+3t-2=(t+2)(2t-1)
$

Now for the second example $x^2+5x - 6$, here we try $x=1$:

$
1^2+5\times 1-6=0
$

so $x-1$ is a factor of $x^2+5x - 6$, and so:

$
x^2+5x - 6 =(x-1)(x+6)
$

RonL

3. 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)

Im not mathmatician ; not that great at math. I just wasn't born with good understanding of math probaly why im so clumsy also. Thanks for your help in advance to a math newbie.
I'm not sure if this is exactly what you are referring to, but it looks the same. I believe it is called the "ac" method.

2x^2+5x-6 isn't a good example because it doesn't factor. I'll show you why using this method.

2*-6=-12
Factors of -12:
1*-12, -1*12
2*-6, -2*6
3*-4, -3,4
None of these pairs adds up to the coefficient of the middle term, 5, so it won't factor.

Let's try 2x^2+7x+6.
2*6=12
12 = 1*12, -1*-12
= 2*6, -2*-6
= 3*4, -3*-4
Which of these pairs adds to 7? 3 + 4 = 7.

So.
2x^2+(3x+4x)+6
(2x^2+3x)+(4x+6)
x(2x+3)+2(2x+3)
(x+2)(2x+3)

Which you can verify by multiplying is 2x^2+7x+6.

The nice thing is this method always works if you can factor the quadratic over the integers.

-Dan

4. ## Different Method

I have never seen this method used before elsewhere, but we were taught it and it works rather well, so here it is:

But the ever-paranoid education system doesn't take any chances, and taught us two other methods, one of which is the method already explained. The other uses algebraic identities, and is for special occasions only:

If the images do not load (they are too big for attachments), go to these direct links:
http://thinkquest.rafflesian.net/others/page1.PNG
http://thinkquest.rafflesian.net/others/page2.PNG
http://thinkquest.rafflesian.net/others/identities.PNG

This post might be too late for your exam, but I hope it will help you in the future.

5. Somehow I managed to pass the exam. If I would have failed I would have had to take the exam over because I had missed 5 days of school and meant I had to take all my exams and would have to retake the class if failed. This was due to pinky surgery. Anyway I Got a 93 percent on my math axam . Yeah Eek. I was really happy everyone else failed or get C and below(i'll be honest all the others without broken bones exempted). Oh I felt like the greatest mathmatician. She said she I should be very proud and no one has ever gotten an A on the 2nd exam as long as she can remember. I managed to figure out how to do every type of problem and many times each. Only struggle I had was wether you shade the top or bottom left or right of the graph for y > or y < . Ahh well. Thanks mhf.