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# Algebra 1/Algebra 2 Factorization Problems

Hi. I have a hard time with Algebra 1/Algebra 2 Factorization Problems. Can someone explain them in-depth and offer some examples? Clarify the mystery for me. Thanks!

Let me know if any parts of this doesn't make sense to you.

Factorization is the decomposition of an object (almost always a polynomial in algebra I and II) into a product (factors) that can be multiplied together to equal the original object (polynomial).

Ex 1 - simple polynomial

Y=X2-4   can be broken down into    Y=(x-2)(x+2).

These factors when multiplied out equals the originally polynomial. But how did I figure that out you ask. I look at the number in the polynomial that does not have a variable associated with it (so 4 for this example). The numbers in the factorization multiplied together have to equal 4 then (so this limits the options to 2,2 or 1,4). Since the number is a negative, one of the numbers must be a positive, while the other is negative. Furthermore, since there is no variable to the first power (X not sqaured, cubed,etc.) the first power variable must cancel out. Since the coefficient of X2 is 1, the only way to cancel out the X-term is to have a positive and a negative of the same value. This eliminates all other options and the factorization must be 2,-2. To verify this answer, we write out Y=(x+2)(x-2) when worked out completely gives us Y=X2-2x+2x-4. The middle terms cancel out and the factorization is verified correct.

Ex 2: longer polynomial (common example in many books)

Y = X2-5X+6

How do we figure this one out since now a single powered x-value is introduced into the polynomial? It looks complicated, but it really isn't. The 2 numbers chosen have to be multiplied together to yield the non-x number, just like the last one (since it is 6, that means it must be 1/6 or 2/3, and since it is positive, both numbers must be either positive or both negative). To decide which one, we have to figure out that middle X-term. Since the coefficient of X2 is 1, that means the 2 numbers selected have to equal -5 when added together. Options would then be (1/6, -1/-6, 2/3, and -2/-3). When adding these together, only -2/-3 work to equal -5. Therefore the solution must be:

Y=(x-2)(x-3)

Verify your work by working this out to make sure it equals the original polynomial (it is a good practice to get into until you are really comfortable with factorization).

Try some of these and I will put the answers below worked out so you can double check your work.

3.) Y=X2+8x+16
4.) Y=X2-3+2
5.) Y=X2-81

Solutions

3.) non x variable of 16, options of 16/1, 2/8, or 4/4
positive number, so both are positive or both are negative
middle x variable is positive, so both are positive
which options added together = 8, therefore 4/4

Y=(X+4)(X+4) or (X+4)2

4.) non x variable of 2, options 1/2
positive number, so both are positive or both are negative
middle x variable is negative, so both are negative
therefore it is -1/-2

Y=(X-1)(X-2)

5.) non x variable of 81, options 1/81, 3/27, 8/8
negative number, so one is positive and one is negative
no middle x variable, must choose same number to cancel X-term out (because X2 has a coefficient of 1)
therefore it is -8/8

Y=(X+8)(X-8)

3 Quick Examples of X3 polynomials

X3+Y3 = (X+Y)(X2-XY+Y2)
X3-Y3 = (X+Y)(X2+XY+Y2)

^^generally Y will be a number that can be cubed and not a variable for basic algebra

Factor out terms to get two like sets
Y=X3+4X2-X-4   factor out x from first 2
Y=X2(x+4)-X-4   factor out -1 from second 2
Y=X2(X+4)-1(X+4)

(when factoring out a term, you divide it from the set you are removing it from and add parenthesis to indicate that it is being multiplied by the term you divided it from so to factor out X from X3+2X2-X you would divide everything by X because they all have an X-term and add parenthesis around it to show it should be multiplied by those terms to get the original polynomial, so X(X2+2X-1)

back to the example, from this you can write it as

Y=(X2-1)(X+4)    (you can even break down X2-1 with the method used above to get)

Y=(X+1)(X-1)(X+4)

I hope this helps, if you have any more questions, let me know.

sorry - too fast, too complicated for my liking. but thanks anyway.
Factorization is simple...much simpler than you see above, that's for sure. Whew, no wonder so many kids and adults have a problem with understanding the technical babble, as I tend to call it.

Ok, think of pulling something out of a bunch of numbers and letters. Ask yourself what you can pull out of everyone of the numbers. For instance, let's take a polynomial...or a bunch of numbers that are put together, and see if we can factor them out.

Y=X^2+8x+16

We're going to separate all three numbers from each other. Ask yourself what will multiply to get each number. x*x = x^2...so our first letters are separated and we can put them into our parenthesis: (x  )(x  ). Now, let's do the same with the other numbers. Let's look at the last number. It's an important one: 16...what two numbers will multiply to get 16? Let's see...16 times 1, 2 times 8, and 4 times 4 will multiply to get 16. So, I can use any of them. But whichever one I choose, we must add the same two numbers to get the middle number (8) too. So, what two numbers can you 'multiply' together to get 16 that will also add to get 8 (the middle number)?  4 times 4, right? That fills in my parenthesis, except for my signs. (x 4)(x 4) . Now, I just need to look at my signs. Well, my whole polynomial is positive. So, I don't need to change any of my signs. The entire thing is positive. (x+4)(x+4)

But suppose one of my signs is negative? Let's try it and see what happens:

y=X^2-8x+16

Let's repeat our steps above and we're not going to change anything except one small part. We're going to separate all three numbers from each other just like we did before. Now, ask yourself what will multiply to get each number. x*x = x^2...so our first letters are separated and we can put them into our parenthesis again: (x )(x ). Now, let's do the same with the other numbers. Let's look at the last number again: 16...what two numbers will multiply to get 16? Let's see...16 times 1, 2 times 8, and 4 times 4 will multiply to get 16. So, I can use any of them. But whichever one I choose must add to get the middle number (-8) too. So, what two numbers can you 'multiply' together to get positive 16 that will also add to get -8 (the middle number)? -4 times -4, right? That fills in my parenthesis. (x-4)(x-4).

Now, I just need to check to make sure everything adds up correctly. (x-4)(x-4) Use FOIL to check it. The first number times the first number = F. Outside number times outside number, Inside number times inside number and the last number times the last number.

The first number is x in  both parenthesis: So, x times x. = x^2. The inside numbers are -4 times x. If I multiply them, it will total -4x. My outside numbers are -4 and x, also. If I multiply them, I get -4x again, and my last numbers are -4 times -4. Now, let's put them all together. (x*x= x^2) (-4*x = -4x) (-4*x = -4x again) and the last one (-4*-4 = 16). Whew! So, I now have the answers and I see if they match up to what's inside my polynomial. x^2 +(-4x)+(-4x)+16 Yes, it looks like it all works out, because when I put them together, I get: x*2-8x+16 The -4x and -4x added together to make the -8x. Cool, I'm done with that one.

I'll assume that's approximately where you're at, but I'm also going to give you a site that will help explain it in a way that is easily understood. It will give you diagrams and a cool little method, called the box method. Learn it and it will save your life. ;) If you have any questions, feel free to hit me up again.

purplemath.com and do a search for factoring
Rana