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=X^{2}-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 X^{2} 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=X^{2}-2x+2x-4. The middle terms cancel out and the factorization is verified correct.

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

Y = X^{2}-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 X^{2
}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=X^{2}+8x+16

4.) Y=X^{2}-3+2

5.) Y=X^{2}-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 X^{2} has a coefficient of 1)

therefore it is -8/8

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

3 Quick Examples of X^{3} polynomials

X^{3}+Y^{3} = (X+Y)(X^{2}-XY+Y^{2})

X^{3}-Y^{3} = (X+Y)(X^{2}+XY+Y^{2})

^^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=X^{3}+4X^{2}-X-4 factor out x from first 2

Y=X^{2}(x+4)-X-4 factor out -1 from second 2

Y=X^{2}(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 X^{3}+2X^{2}-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(X^{2}+2X-1)

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

Y=(X^{2}-1)(X+4) (you can even break down X^{2}-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.

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