Factoring With Respect to the Integers

[Mackenzie25]

I'm a little bit stuck on a few of my homework problems, and can't really find out how to do them in my text book. I don't really need help on both of them unless there're different ways to find the answer, I just need to learn how to solve them. Any help would be fantastic.
Thanks in advance,
Mackenzie

"Completely factor the polynomial with respect to the integers"
1. x[sup:htd7v9lz]2[/sup:htd7v9lz]-9x-22
2. 2x[sup:htd7v9lz]2[/sup:htd7v9lz]-5x+2

 

[Denis]

http://www.google.ca/search?hl=en&sourc ... =&aq=f&oq=

 

[galactus]

"Completely factor the polynomial with respect to the integers"

1. \(\displaystyle x^{2}-9x-22\)

What two numbers when multiplied equal -22 and when added equal -9?.

How about -11 and 2?.

\(\displaystyle x^{2}-11x+2x-22\)

\(\displaystyle x(x-11)+2(x-11)\)

\(\displaystyle (x+2)(x-11)\)

See?. That's all there is to it. Now, you try the other one.

 

[Mackenzie25]

For some reason, I understand where all of the numbers came from and how you did it towards the middle, but I can't really figure out how where to put the numbers after you find the common factors.

I have a quick question, couldn't you also solve this problem using factor boxes and working backwards?
I attatched my example and hopefully it worked out okay....

In the first box, I took the first and last term and put them in their places in the box
Next, I found the factors of the product of 1and-22
The only pair that will have a sum or difference of -9 was 11and2 (-11+2=-9)
Therefor the two empty boxes were filled with -11x and 2x
From this point, I figured out which numbers would fit around the box to make the equation true.

The bottom half is (hopefully) the solution to two

 

[BigGlenntheHeavy]

Mackenzie25, when factoring a quadratic, if everything else fails, one can always used the quadratic formula, to wit:

\(\displaystyle x \ - \ \frac{-b \pm \sqrt(b^{2}-4ac)}{2a}\)

All the other forms are fine when dealing with only integers, but how would you solve for x (or factor for that matter),

\(\displaystyle 3.7x^{2}-66.80755x+10.79753085?\)

 

[fasteddie65]

For some reason, I understand where all of the numbers came from and how you did it towards the middle, but I can't really figure out how where to put the numbers after you find the common factors.

I have a quick question, couldn't you also solve this problem using factor boxes and working backwards?
I attatched my example and hopefully it worked out okay....

In the first box, I took the first and last term and put them in their places in the box
Next, I found the factors of the product of 1and-22
The only pair that will have a sum or difference of -9 was 11and2 (-11+2=-9)
Therefor the two empty boxes were filled with -11x and 2x
From this point, I figured out which numbers would fit around the box to make the equation true.

The bottom half is (hopefully) the solution to two

This is often the way that factoring is taught in some high schools.

I used to use the Product Sum Method, which goes like this:

2x^2 - 5x + 2

Multiply 2 and 2: 4
Find factors of 4 whose sum is -5: -4 and -1

Rewrite the trinomial as a 4-term polynomial: 2x^2 - 4x - 1x + 2
Regroup: (2x^2 - 4x) - (x - 2)
Factor each binomial: 2x(x - 2) - 1(x - 2)
x - 2 is the common factor; (x - 2)(2x - 1) are the factors.

Basically, it is the same as all the boxes, but much easier to use, imho.