# factorization with binomials

• Nov 28th 2009, 04:19 PM
masterinex
factorization with binomials
Hi everyone,

I know that the common method is like this :
x^2 + 5x + 4

Sum = 4
product = 5

so a = 1 and b = 4

=(x+1)(x+4)

If I know the product is 36 (9*4) , how does knowing a= -6 b=-6 help me to find the factorized form of (3x-2)(3x-2) ?

9x^2 -12x +4

Sum = -12
product = 36

a= -6 b=-6

= 9x^2 -6x-6x+4
(3x-2)(3x-2)
• Nov 28th 2009, 11:35 PM
Hello masterinex
Quote:

Originally Posted by masterinex
...
If I know the product is 36 (9*4) , how does knowing a= -6 b=-6 help me to find the factorized form of (3x-2)(3x-2) ?

9x^2 -12x +4

Sum = -12
product = 36

a= -6 b=-6

= 9x^2 -6x-6x+4
(3x-2)(3x-2)

Take out the highest common factor from the first pair of terms, and then from the second pair:
$9x^2-6x-6x+4=3x\color{red}(3x-2)\color{black}-2\color{red}(3x-2)$
Then note that the two brackets are the same. So you can now complete the factorisation:
$=(3x-2)\color{red}(3x-2)$
• Nov 29th 2009, 10:38 AM
masterinex
if i know that a = -6 and b =-6 , is there a connection between the greatest common factor 3x and 2 ?
• Nov 29th 2009, 11:03 AM
Jameson
Quote:

Originally Posted by masterinex
Hi everyone,

I know that the common method is like this :
x^2 + 5x + 4

Sum = 4
product = 5

so a = 1 and b = 4

=(x+1)(x+4)

If I know the product is 36 (9*4) , how does knowing a= -6 b=-6 help me to find the factorized form of (3x-2)(3x-2) ?

9x^2 -12x +4

Sum = -12
product = 36

a= -6 b=-6

= 9x^2 -6x-6x+4
(3x-2)(3x-2)

I think you are talking about relating the roots of f(x) by coefficients a,b and c of $f(x)=ax^2+bx+c$. I believe you are making errors doing this as:

$x_1+x_2= -\frac{b}{a}$
$x_1*x_2= \frac{c}{a}$

So if you plug in a, b and c for any particular quadratic equation, you can solve for x_1 and x_2.
• Nov 30th 2009, 12:40 PM
masterinex
I guess the idea is like this :

9x^2 -12x +4

Sum = -12
product = 36

a and b are just two numbers which added together give -12 and multiplied get 36 . Then the linear term -12x can be rewritten in terms of ax + bx
a= -6 b=-6

so 9x^2 -12x +4 becomes
= 9x^2 -6x-6x+4
(3x-2)(3x-2)

for this example the method worked fine . But what about if I have something like this :

2x^2 + 11x + 5 for which the factored form should be (2x+1)(x+5)

Then by the method
product = 10 and sum = 11
the only two numbers I can think of are 2 and 5 , but the sum wouldnt be 11. Any suggestions ?
• Nov 30th 2009, 12:47 PM
Jameson
Quote:

Originally Posted by masterinex
I guess the idea is like this :

9x^2 -12x +4

Sum = -12
product = 36

a and b are just two numbers which added together give -12 and multiplied get 36 . Then the linear term -12x can be rewritten in terms of ax + bx
a= -6 b=-6

so 9x^2 -12x +4 becomes
= 9x^2 -6x-6x+4
(3x-2)(3x-2)

for this example the method worked fine . But what about if I have something like this :

2x^2 + 11x + 5 for which the factored form should be (2x+1)(x+5)

Then by the method
product = 10 and sum = 11
the only two numbers I can think of are 2 and 5 , but the sum wouldnt be 11. Any suggestions ?

You are right that when factoring you basically do this method without writing it out, meaning you think of two numbers that add up to something and multiply to something else as well. I don't know of an easy way to use the formulas for sum and product of roots to automatically lead to a factored polynomial, sorry. It might very well exist, but I don't know it.

Notice that x^2 has a coefficient other than 1. This always complicates the process of factoring. That 2 in front of x^2 is going to multiply with the two terms in the other parentheses but not the term in the same one. So now you aren't trying to find a and b such that a+b is the coefficient of x and ab is the constant term. I know this doesn't really help, but maybe there is something you didn't know buried in this post.
• Nov 30th 2009, 01:18 PM
Hello masterinex
Quote:

Originally Posted by masterinex
I guess the idea is like this :

9x^2 -12x +4

Sum = -12
product = 36

a and b are just two numbers which added together give -12 and multiplied get 36 . Then the linear term -12x can be rewritten in terms of ax + bx
a= -6 b=-6

so 9x^2 -12x +4 becomes
= 9x^2 -6x-6x+4
(3x-2)(3x-2)

for this example the method worked fine . But what about if I have something like this :

2x^2 + 11x + 5 for which the factored form should be (2x+1)(x+5)

Then by the method
product = 10 and sum = 11
the only two numbers I can think of are 2 and 5 , but the sum wouldnt be 11. Any suggestions ?

You'll kick yourself - what about $10$ and $1$?

Then you can say (using the same method as before):
$2x^2 + 11x + 5 = 2x^2+10x+x+5$
$=2x(x+5)+1(x+5)$
$=(2x+1)(x+5)$
Any quadratic that can be factorised will work in the same way, even with numbers with a large number of factors. For example:
$45x^2+126x-80$
Product $= 45 \times (-80) = -3600$. So we want two factors of $3600$ whose difference is $126$. A certain amount of trial and error is necessary, but $150$ and $24$ are the factors required. So:
$45x^2+126x-80$
$=45x^2 +150x-24x-80$

$=15x(3x+10)-8(3x+10)$

$=(15x-8)(3x+10)$