# Factoring trinomials.

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• Feb 1st 2010, 11:30 PM
Justiceapple
Factoring trinomials.

$\displaystyle 10x^2+3x-18$
$\displaystyle = 10x^2+15x-12x-18$
=... continued

How exactly do you come up with +15x and -12x? The textbook only tells that you multiply 10 and 18=180, and the sum of $\displaystyle 3x$ will be +15x and -12x. What's the point of coming up with 180? And how in the world.... what do you do to get 15x and -12x???

$\displaystyle 15a^2-22ab+8b^2$
=$\displaystyle 15a^2-10ab-12ab+8b^2$
=... continued

This is the same thing. How exactly do you come up with -10ab and 12ab to make -22ab?

Thank you..
• Feb 2nd 2010, 12:20 AM
Sudharaka
Quote:

Originally Posted by Justiceapple

$\displaystyle 10x^2+3x-18$
$\displaystyle = 10x^2+15x-12x-18$
=... continued

How exactly do you come up with +15x and -12x? The textbook only tells that you multiply 10 and 18=180, and the sum of $\displaystyle 3x$ will be +15x and -12x. What's the point of coming up with 180? And how in the world.... what do you do to get 15x and -12x???

$\displaystyle 15a^2-22ab+8b^2$
=$\displaystyle 15a^2-10ab-12ab+8b^2$
=... continued

This is the same thing. How exactly do you come up with -10ab and 12ab to make -22ab?

Thank you..

Dear Justiceapple,

Consider, $\displaystyle 10x^2+3x-18$

$\displaystyle 10\times{18}=180=15\times{12}$

$\displaystyle 15-12=3$

Do you get the point? When a polynomial equation (say $\displaystyle ax^2+bx+c$) of the second degree is given you have to write the first degree term using two terms (say dx and ex) such that their coefficients (d and e) when multiplied together will give you the constant term ($\displaystyle d\times{e}=b$). Then you will be able to factorize them. If you need more help or if you don't understand this please don't hesitate to reply me.

Hope this helps.
• Feb 2nd 2010, 12:20 AM
red_dog
The easiest way for factoring trinomials is to use the roots of trinomials.

$\displaystyle ax^2+bx+c=a(x-x_1)(x-x_2)$, where $\displaystyle x_1,\ x_2$ are the roots of the equation $\displaystyle ax^2+bx+c=0$

and $\displaystyle x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}$, where $\displaystyle \Delta=b^2-4ac$
• Feb 2nd 2010, 12:38 AM
Justiceapple
Sudharaka.
Thank you very much for your help, but what I don't understand is that how do 15 and 12 come to your mind for 180? If you go up on the street and ask someone what 15x12 is, I bet most people can't answer.
Do you understand what I'm trying to say? I try 18x10 is 180, that's the first thing that comes to my mind. And others? I can't think of any other possibilities. (Worried)

red_dog.
Thank you for your help, I haven't learned that concept and for this test, the prof. wants us to solve by this method...
• Feb 2nd 2010, 12:43 AM
Sudharaka
Quote:

Originally Posted by Justiceapple
Sudharaka.
Thank you very much for your help, but what I don't understand is that how do 15 and 12 come to your mind for 180? If you go up on the street and ask someone what 15x12 is, I bet most people can't answer.
Do you understand what I'm trying to say? I try 18x10 is 180, that's the first thing that comes to my mind. And others? I can't think of any other possibilities. (Worried)

red_dog.
Thank you for your help, I haven't learned that concept and for this test, the prof. wants us to solve by this method...

Dear Justiceapple,

As I had told you only 15 and 18 will give you 3 when substracted. It is correct that $\displaystyle 180=18\times{10}$ but can you make 3x in terms of 18x and 10x?
• Feb 2nd 2010, 12:52 AM
Sudharaka
Dear Justiceapple,

Trinomials Factorisation | TutorVista
• Feb 2nd 2010, 12:56 AM
Justiceapple
Sudharaka,

I truly appreciate your help. Yes, I understand that 15 and 12 is the only possibility to get 3 and 180.

So, what I am guessing at this point is that you just go over whatever numbers to get to those? Like, you try 13x10, "no, it's 130, not 180. Then you try, 14x11, "Oh it's 154, not it". Maybe you skip one and try 16x13, "Oh it's 208, not it". It must be the one in between. 15x12!!! Bingo.........
You just try like that until you get 180?
It seems very time consuming,,, not?
• Feb 2nd 2010, 12:57 AM
Justiceapple
Oh, thank you for the website link. I'll take a look :)
• Feb 2nd 2010, 01:07 AM
Justiceapple
Hmm I'm still confused. The websites doesn't answer my question either. The example 2 of Factoring a trinomial type ... by splitting the middle term, where it shows that 90 is (9x10) or (5x18) or (15x6). To get the difference of (-9)
I can quickly come up with 9x10, but I don't think I will be able to come up with 5x18 and 15x6 in the given time for the test. It seems to me that the possibilities I'd have to try from are just vast.
• Feb 2nd 2010, 01:25 AM
Sudharaka
Quote:

Originally Posted by Justiceapple
Sudharaka,

I truly appreciate your help. Yes, I understand that 15 and 12 is the only possibility to get 3 and 180.

So, what I am guessing at this point is that you just go over whatever numbers to get to those? Like, you try 13x10, "no, it's 130, not 180. Then you try, 14x11, "Oh it's 154, not it". Maybe you skip one and try 16x13, "Oh it's 208, not it". It must be the one in between. 15x12!!! Bingo.........
You just try like that until you get 180?
It seems very time consuming,,, not?

Dear Justiceapple,

Of course by looking at a number you can predict it's divisors. In this example 180 is divided by 10,15,18,.... but not 13,14 or 16 so you would never be able to get 180 by multiplying 13,14 or 16. Then you can rule out the obvious. For example 2 divides 180 but since 180=2x90 you can neglect it. Same goes for 3,4,5..... etc. First you have to think about the divisors of 180 and then the most appropiate. I know this method of factorization needs a bit of practice but by doing it several times you will be able to solve the problems very easily.
• Feb 2nd 2010, 02:01 AM
Sudharaka
Dear Justiceapple,

When I learnt solving trinomial equations I refered some books. And a good one I want you to refer is Elementary Algebra for School by H.S.Hall. In this book CHAPTER XVII has a good introduction to factorization and also many problems that you could try yourself.

Elementary Algebra for Schools

Chapter XVII starts at page 125.

Hope this helps.
• Feb 2nd 2010, 02:14 AM
Gusbob
Hopefully this can help supplement Sudharaka's point.

Consider a polynomial:

$\displaystyle (ax+b)(dx+e)$

Expanding this we see that

$\displaystyle (ad)x^2 + (ae)x + (bd)x + be$

Notice that the coefficient (number in front) of the x² term multiplied by the constant term is the same as the product of the coefficients two x terms.

$\displaystyle ad \times be = ae \times bd = abde$

So when attempting to factorise trinomials like this, you will be trying to find two integers which has the same product as the coefficient of x² multiplied by the constant term.

If you list down all the pairs you can clearly see which two can be added or subtracted to make the x term you have (when you only have one).

As for your point about people not knowing how to find these pairs, you can always try prime factorisation. For example:

180 = 2 x 90 = 2 x 2 x 45 = 2 x 2 x 3 x 15 = 2 x 2 x 3 x 3 x 5.

Of course you don't have to check every possible factor. The coefficient of your lone x term should give a clue. If the coefficient is large, it is likely the difference between the two factors is quite large. If the coefficient is small, the factors should be numerically close to each other.
• Feb 2nd 2010, 02:15 AM
Justiceapple
You've been a great help and you answered my question, Sudharaka.
(Wink) Thank youuuu.
• Feb 2nd 2010, 02:18 AM
Justiceapple
Gusbob. Yes, I tried doing the prime factorization, but I didn't know what to do with the numbers I got. I guess you multiply the first three numbers to get 12, and 3x5 for 15, but I couldn't apply the same rule to other problems of this kind of factoring.. so, I wasn't sure. Thank you.
• Feb 2nd 2010, 02:38 AM
Gusbob
Quote:

Originally Posted by Justiceapple
Gusbob. Yes, I tried doing the prime factorization, but I didn't know what to do with the numbers I got. I guess you multiply the first three numbers to get 12, and 3x5 for 15, but I couldn't apply the same rule to other problems of this kind of factoring.. so, I wasn't sure. Thank you.

Ok. Given a prime factorisation of 2 x 2 x 3 x 3 x 5, you can mix and match all the numbers. Anything from:

(2 x 2) x (3 x 3 x 5) to (5 x 3 x 2) x (2 x 3) to (5 x 2) x (2 x 3 x 3)
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