# Factorization Question 2

• Sep 24th 2011, 03:36 PM
sparky
Factorization Question 2
For the question below, does anyone have a quick and easy method for finding the two numbers that when added together gives 11x, and when multiplied together gives -180? I know the numbers are 20 and -9 but it takes me forever to find them? Is there is quick way?

$\text{Factorize } 12x^2+11x-15$
• Sep 24th 2011, 04:19 PM
Furyan
Re: Factorization Question 2
Hello

I don't now if this will help, but I do it this way. It doesn't take ages.
You know you have either:

1,15 or 3,5

and your looking for a difference of 11

Write out:

(12x )(x )
(6x )(2x )
(4x )(3x )

It doesn't tke long to work out the difference of the possible products and see that 4x5 - 3x3 = 11

So you have

(4x - 3)(3x + 5)
• Sep 24th 2011, 09:30 PM
Siron
Re: Factorization Question 2
If you have a polynomial $f(x)$ and the zero's $x_1,x_2,...x_n$ are known then the polynomial can be factored as:
$f(x)=a(x-x_1)(x-x_2)....(x-x_n)$ (where $a$ is the coefficient of the highest degree term).
Offcourse in this case you have to find the zero's first (by using the quadratic formula) ...
• Sep 25th 2011, 05:18 AM
Soroban
Re: Factorization Question 2
Hello, sparky!

Quote:

For the question below, does anyone have a quick and easy method for
finding the two numbers whose difference is 11 and whose product 180?
I know the numbers are 20 and -9, but it takes me forever to find them?
Is there is quick way?

$\text{Factor: }\:12x^2+11x-15$

I use a primitive approach.
It takes a bit of time, but it is direct.

Factor 180 into pairs of factors, and note the difference.

. . $\begin{array}{cc}\text{Factors} & \text{Diff.} \\ \hline 1,180 & 179 \\ 2,90 & 88 \\ 3,60 & 57 \\ 4,45 & 41 \\ 5,36 & 31 \\ 6,30 & 24 \\ 7,? & - \\ 8,? & - \\ 9,20 & 11 \\ 10,18 & 8 \\ 11,? & - \\ 12,15 & 3 \\ 13,? & - \\ \hline \end{array} \begin{array}{c} \\ \\ \\ \\ \\ \leftarrow\text{ Here!} \end{array}$

How do we find these pairs?
Divide 180 by 1, 2, 3, . . .
Some of the divisions do not "come out even", of course.

When can we stop?
We can stop at the integer part of $\sqrt{180} \,\rightarrow\, 13$

We see that the pair $(9,20)$ has a difference of 11.
(Of course, we can stop listing the moment we find it.)

We have: . $12x^2 \quad 9x\quad 20x - 15$

We want the middle term to be +11x, so we will use -9x and +20x.

So we have: . $12x^2 - 9x + 20x - 15$

Factor "by grouping": . $3x(4x-3) + 5(4x-3) \;=\;(4x-3)(3x+5)$

• Sep 25th 2011, 04:05 PM
sparky
Re: Factorization Question 2
Quote:

Originally Posted by Soroban
Hello, sparky!

I use a primitive approach.
It takes a bit of time, but it is direct.

Factor 180 into pairs of factors, and note the difference.

. . $\begin{array}{cc}\text{Factors} & \text{Diff.} \\ \hline 1,180 & 179 \\ 2,90 & 88 \\ 3,60 & 57 \\ 4,45 & 41 \\ 5,36 & 31 \\ 6,30 & 24 \\ 7,? & - \\ 8,? & - \\ 9,20 & 11 \\ 10,18 & 8 \\ 11,? & - \\ 12,15 & 3 \\ 13,? & - \\ \hline \end{array} \begin{array}{c} \\ \\ \\ \\ \\ \leftarrow\text{ Here!} \end{array}$

How do we find these pairs?
Divide 180 by 1, 2, 3, . . .
Some of the divisions do not "come out even", of course.

When can we stop?
We can stop at the integer part of $\sqrt{180} \,\rightarrow\, 13$

We see that the pair $(9,20)$ has a difference of 11.
(Of course, we can stop listing the moment we find it.)

We have: . $12x^2 \quad 9x\quad 20x - 15$

We want the middle term to be +11x, so we will use -9x and +20x.

So we have: . $12x^2 - 9x + 20x - 15$

Factor "by grouping": . $3x(4x-3) + 5(4x-3) \;=\;(4x-3)(3x+5)$

Wow, I love this! Why didn't I think of this? This was so obvious! Thanks
• Sep 25th 2011, 05:36 PM
Deveno
Re: Factorization Question 2
Quote:

Originally Posted by Soroban
Hello, sparky!

I use a primitive approach.
It takes a bit of time, but it is direct.

Factor 180 into pairs of factors, and note the difference.

. . $\begin{array}{cc}\text{Factors} & \text{Diff.} \\ \hline 1,180 & 179 \\ 2,90 & 88 \\ 3,60 & 57 \\ 4,45 & 41 \\ 5,36 & 31 \\ 6,30 & 24 \\ 7,? & - \\ 8,? & - \\ 9,20 & 11 \\ 10,18 & 8 \\ 11,? & - \\ 12,15 & 3 \\ 13,? & - \\ \hline \end{array} \begin{array}{c} \\ \\ \\ \\ \\ \leftarrow\text{ Here!} \end{array}$

How do we find these pairs?
Divide 180 by 1, 2, 3, . . .
Some of the divisions do not "come out even", of course.

When can we stop?
We can stop at the integer part of $\sqrt{180} \,\rightarrow\, 13$

We see that the pair $(9,20)$ has a difference of 11.
(Of course, we can stop listing the moment we find it.)

We have: . $12x^2 \quad 9x\quad 20x - 15$

We want the middle term to be +11x, so we will use -9x and +20x.

So we have: . $12x^2 - 9x + 20x - 15$

Factor "by grouping": . $3x(4x-3) + 5(4x-3) \;=\;(4x-3)(3x+5)$

^this. is beautiful.