Download Factoring protocols and reference sheet, by K. Davis. Factoring is a

Factoring protocols and reference sheet, by K. Davis.
Factoring is a means of rewriting expressions in terms of factors, or simpler terms being multiplied together.
The process is the exact opposite of the distributive property.
When factoring, the first thing you should look for is the greatest common factor, or GCF.
GCF factoring: If every term in a group shares a common factor, that factor may be set apart from the group as a
shared factor of the sum of the remnants of the remaining terms.
Factor: AB + AC - ADE
A(B + C - DE)
//There are 3 terms, each term shares the common factor, A.
//A is set outside a pair of parenthesis, the remaining parts of the other
terms are rewritten inside the parenthesis.
How to proceed after checking for common factors depends on the number of terms left.
If there are 2 terms left, check to see if one of the special factoring formulas work.
Difference of squares: 𝐴2 − 𝐵2 = (𝐴 + 𝐵)(𝐴 − 𝐵)
Difference of cubes: 𝐴3 − 𝐵3 = (𝐴 − 𝐵)(𝐴2 + 𝐴𝐵 + 𝐵2 )
Sum of cubes: 𝐴3 + 𝐵3 = (𝐴 + 𝐵)(𝐴2 − 𝐴𝐵 + 𝐵2 )
Factor: 𝑥 6 − 27
//There are 2 terms, so we think special factoring formula, they are being subtracted so
difference of squares or cubes. 𝑥 6 can be rewritten as (𝑥 2 )3 𝑜𝑟 (𝑥 3 )2 and so
either formula would work. However, 27 can only be rewritten as 33 so
difference of cubes is most appropriate.
2
3
3
= (𝑥 ) − (3)
//The original expression is now rewritten to look like a difference of things being
cubed, the things being cubed are 𝑥 2 and 3. So use the formula but replace A
with 𝑥 2 and B with 3.
= (𝑥 2 )3 − (3)3 = (𝑥 2 − 3)((𝑥 2 )2 + 3𝑥 2 + 32 )
//Use basic algebra to simplify the answer.
2
4
2
= (𝑥 − 3)(𝑥 + 3𝑥 + 9)
Factor: 𝑥 4 𝑦 2 − 100
= (𝑥 2 𝑦)2 − (10)2
//There are two terms, so we think special factoring formula, the
exponents on the first term are divisible by 2, and 100 is a perfect
square, so difference of squares is appropriate.
//Rewrite as the difference of two things being squared. Replace A with
𝑥 2 𝑦 and B with 10 in the difference of squares formula.
= (𝑥 2 𝑦 + 10)(𝑥 2 𝑦 − 10)
If there are 3 terms left, and the expression is of the form 𝑥 2 + 𝑏𝑥 + 𝑐 use simple trinomial factoring
𝑥 2 + 𝑏𝑥 + 𝑐 = (𝑥 + 𝑗)(𝑥 + 𝑘)if you can find two numbers, j and k, such that j*k = c, j+k = b.
If no such numbers can be found, it is called non-factorable.
Factor: 𝑥 2 + 10𝑥 − 24
//This has three terms and is of the form 𝑥 2 + 𝑏𝑥 + 𝑐 where b = 10, c = -24
//If jk = -24, one of the two must be negative. J+k = 10 means the bigger one
must be positive (since the sum is positive).
//Factors of -24 that fit that criteria are -1, 24 ; -2, 12 ; -3, 8 ; -4, 6
// -2, 12 add up to ten, those are values for j and k that we need.
= (x+12)(x-2)
//Write the solution by using the values for j, k.
*If c is negative, j and k have different signs. if c is positive, they have the same sign.
*If b is negative, the bigger value is also negative. If b is positive, the bigger value is also positive.
If there are 3 terms left, and the expression is of the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, use less simple trinomial factoring.
Step 1. Multiply a * c.
Step 2. Find two numbers, j and k, such that jk = ac, and j + k = b.
Step 3. Rewrite equation as 𝑎𝑥 2 + 𝑗𝑥 + 𝑘𝑥 + 𝑐
Step 4. Factor by grouping-- if you can find j, k problems from step 3, this will work every time.
Factor: 4𝑥 2 + 12𝑥 + 9
4𝑥 2 + 6𝑥 + 6𝑥 + 9
2x(2x+3) + 3(x+3)
(2x+3)(2x+3)
//This is a trinomial of the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, a = 4, b = 12, c = 9.
//ac = 4(9) = 36. We need factors of 36 that add up to 12.
//Factors are: 1, 36 ; 2, 18 ; 3, 12 ; 4, 9 ; 6, 6. 6+6 = 12, those are our j, k values.
//Step 3...bx gets split up into jx + kx
//factor by grouping
//factor by GCF
If there are 4 terms left, factor by grouping.
To factor by grouping, create two groups with the same number of terms in each group. Let one group contain
those terms with the highest degrees and the other contain those terms with the lowest degrees.
Try GCF factoring with each group, and then see if the resulting terms have any common factors. If they do, try
GCF factoring again, else, it is not factorable.
Factor: 𝑥 5 + 5𝑥 2 + 𝑥 3 + 5
(𝑥 5 + 𝑥 3 ) + (5𝑥 2 + 5)
𝑥 3 (𝑥 2 + 1) + 5(𝑥 2 + 1)
(𝑥 3 + 5)(𝑥 2 + 1)
//There are 4 terms, the highest degree terms are 𝑥 5 𝑎𝑛𝑑 𝑥 3 group them.
//Each group now has common factors, attempt GCF factoring.
//The new terms now share the common term, 𝑥 2 + 1
//Since 5 is not a perfect cube and there is no sum of squares formula, this
cannot be factored any further.
If you have to solve an equation and the variable appears 1 time, solve using inverse operations.
If you have to solve an equation and the variable appears more than 1 time, or if that variable is raised to a
power:
Factor the expression
Set each factor equal to zero.
Solve each separate equation using inverse operations.
In a polynomial, the number of zeros is equal to the degree of the polynomial, though not all zeros must be real
or unique.
*Remember, learning math takes practice...not pieces of paper with the answers on them. Even with this
wonderfully crafted handout, if you don't practice all the different types of factoring and encounter all the
different tricks and problems associated with them, then you're going to spend more time staring at pages of
notes and scratching your head than actually solving problems when it comes time to take a test, solve modeling
problems, or make sense of even trickier topics to come in this and other math classes.