Factoring

What are factors?

We are going to rewrite a polynomial as a product of its prime factors. To do this we need to follow a five-step process. Remember to factor completely. Remember your final factored form should equal what you started with. Always start with step one and follow the steps in order. It will make factoring much easier! If a polynomial does not factor at all, it is said to be prime.

I. Greatest Common Factor

Look at each term of the polynomial and see if there is a common factor. If so, factor it out by putting it on the outside of parentheses.

Always factor out the variable to the lowest exponent.

Examples:
6x² + 9x + 3 = 3(2x²+ 3x + 1)
9x⁴y² - 12x³y³ = 3x³y²(3x - 4y)

^{II. Binomials} A. Difference of Squares

         1. Two terms
         2. Both terms must be perfect squares
          A variable must have an even exponent to be a perfect square.
       3.   One term must be positive and one term must be negative

Take note:

All difference of squares will factor.
Put them in the form of perfect square minus perfect square before you try to factor.
The factored format is:

(Square Root plus Square Root) (Square Root minus Square Root)
Remember the commutative property of multiplication.
To get the square root of a variable, divide its exponent by two.
You cannot take the sum of two squares any further: m² + 16 cannot be factored.
Always factor completely:
m⁴ - 1 breaks down to:
(m² + 1) (m² - 1) which breaks down to:
(m² + 1) (m - 1) (m + 1)

Examples:
x² - 36 = (x + 6)(x - 6)
4x² - 1 = (2x + 1)(2x - 1)
x⁴ - 81 = (x² + 9)(x² - 9) = (x² + 9)(x + 3)(x - 3)

B. Difference of Cubes

1. Two terms

2. Both terms must be perfect cubes

A variable must have an exponent divisible by three to be a perfect cube.

3. One term must be positive and one term must be negative

Take note:

All differences of cubes will factor.
Put them in the form of perfect cube minus perfect cube before you try to factor.
The factored format is:

x³- y³= (x - y)(x²+ xy + y²)
Translated into words, this means the difference of two cubes can be factored into the product of a binomial times a trinomial. Begin by taking the cube roots of the perfect cubes. The binomial is the first root minus the second root. Then use that binomial to build the trinomial as follows: Square the first term in the binomial plus the product of the first and second terms in the binomial plus the square of the second term in the binomial.

Example:

C. Sum of Cubes

1. Two terms

2. Both terms must be perfect cubes

A variable must have an exponent divisible by three to be a perfect cube.

3. Both terms must be positive

Take note:

All sums of cubes will factor.
The factored format is:

x³+ y³= (x + y)(x²- xy + y²)

Translated into words, this means the sum of two cubes can be factored into the product of a binomial times a trinomial. Begin by taking the cube roots of the perfect cubes. The binomial is the first root plus the second root. Then use that binomial to build the trinomial as follows: Square the first term in the binomial minus the product of the first and second terms in the binomial plus the square of the second term in the binomial.

Example:

Before you go to the next step:
a. Put the polynomial in descending order.
b. If the leading coefficient is negative, factor out a negative one (-1).

III. Trinomials

    A. Perfect Square Trinomial
          1. Three terms
          2. Two of the terms must be positive perfect squares
         3. The other term must be twice the product of the square roots of the other two terms.

Take note:

All perfect square trinomials factor either as:
(Square Root plus Square Root) (Square Root plus Square Root) OR
(Square Root minus Square Root) (Square Root minus Square Root)

Examples:
x² - 10x + 25 = (x - 5) (x - 5) or (x - 5)²
x² + 10x + 25 = (x + 5) (x + 5) or (x + 5)²
4x² - 12xy + 9y² = (2x - 3y) (2x - 3y) or (2x - 3y)²

All trinomials are either simple or general trinomials. Not all simple trinomials factor and not all general trinomials factor. A simple trinomial has a leading coefficient of one and a general trinomial has a leading coefficient of any number except one.

B. Simple Trinomials that Factor

The simple trinomial will be in the form of x² + bx + c where b and c are nonzero real numbers.

List the factor pairs of c. Make sure you have all the pairs or, if your luck is like mine, the one you don't list will be the one you needed! Then look at the sign in front of the c. A plus tells you to look for the factor pair that gives you a sum of b. A minus tells you to look for the factor pair that gives you a difference of b. If you have the factor pair that gives you what you're looking for, the trinomial will factor. If you don't have the factor pair, the trinomial will not factor. To factor the trinomial:

The first term always factors into its square roots. (Remember FOIL: first times first gives you first.)

Use the chosen factor pair to factor the third term. (Remember FOIL: last times last gives you last.)

Again look to the sign in front of c to tell you what signs to use in the parentheses of the factors.

	A plus sign in front of c tells you that you will have like signs in the parentheses - either both pluses or both minuses. The sign in front of b tells you which one.
	A minus sign in front of c tells you that you will have unlike signs - one plus and one minus. You must place them in the parentheses so that when you do the inner and outer products and combine them, you come up with the correct sign for b.

Example 1: Factor x² - 10 x + 24:

Factor pairs of 24 are:
1,24
2,12
3,8
4,6

Because of the plus in front of the 24, we are looking for a sum of 10 (b). 4 and 6 give us that sum so the trinomial factors.

The x² factors into x times x . We chose 4 and 6 for the 24. So we now have:
(x 4)(x 6). We now have to determine the correct signs for the parentheses. The plus in front of the 24 tells us we will have like signs. The minus in front of the 10 tells us they will both be minuses. Thus, we have (x - 4)(x - 6).

Example 2: Factor x² +10 x + 24:

Factor pairs of 24 are:
1,24
2,12
3,8
4,6

Because of the plus in front of the 24, we are looking for a sum of 10 (b). 4 and 6 give us that sum so the trinomial factors.

Example 3: Factor x² + 10 x - 24:

Factor pairs of 24 are:
1,24
2,12
3,8
4,6

Because of the minus in front of the 24, we are looking for a difference of 10 (b). 2 and 12 give us that difference so the trinomial factors.

The x² factors into x times x . We chose 2 and 12 for the 24. So we now have:
(x 2)(x 12). We now have to determine the correct signs for the parentheses. The minus in front of the 24 tells us we will have unlike signs. Because we want to end up with a positive 10x, we will have to place the plus in front of the 12 making the outer product positive 12x and we will place the minus in front of the 2 making the inner product negative 2x. When the positive 12x and the negative 2x are combined, it gives us the positive 10x that we needed. Thus, we have (x - 2)(x + 12).

Example 4: Factor x² - 10 x - 24:

Factor pairs of 24 are:
1,24
2,12
3,8
4,6

Because of the minus in front of the 24, we are looking for a difference of 10 (b). 2 and 12 give us that difference so the trinomial factors.

The x² factors into x times x. We chose 2 and 12 for the 24. So we now have:
(x 2)(x 12). We now have to determine the correct signs for the parentheses. The minus in front of the 24 tells us we will have unlike signs. Because we want to end up with a negative 10x, we will have to place the minus in front of the 12 making the outer product negative 12x and we will place the plus in front of the 2 making the inner product positive 2x. When the negative 12x and the positive 2x are combined, it gives us the negative 10x that we needed. Thus, we have (x + 2)(x - 12).

Another method for factoring simple trinomials:

If we set x² + bx + c equal to the product of x + c₁ and x + c₂, then:
x² + bx + c = (x + c₁) (x + c₂)
= x² + (c₁ + c₂)x + c₁c₂
and thus b = c₁ + c₂ and c = c₁c₂.

Using this line of thinking, we can formulate a procedure for factoring the expression x² + bx + c :

(1) Find two real numbers whose sum is equal to b and whose product is equal to c.
(2) If two real numbers satisfying the above criteria are found, and we denote them as c₁ and c₂ , then (x + c₁) and (x + c₂) are factors of x² + bx + c.

Example 1: Factor x² + 7x + 12

First, list the combinations of integers whose product is equal to 12.   They are
12 and 1 ( 12 · 1 = 12 )
6 and 2 ( 6 · 2 = 12 )
4 and 3 ( 4 · 3 = 12 )

Then find the sum of these combinations:
12 + 1 = 13
6 + 2 = 8
4 + 3 = 7
The integers 4 and 3 have a sum of 7 and a product of 12.
The factors of x² + 7x + 12 are ( x + 3 ) and ( x + 4 ).

Example 2: Factor y² - 3y - 10

The combinations of integers whose product is -10, and their sums are

10 · -1 = -10     10 + ( -1 ) = 9
5 · -2 = -10      5 + ( - 2 ) = 3
-10 · 1 = -10      -10 + 1 = -9
-5 · 2 = -10      -5 + 2 = -3

The integers -5 and 2 have a sum of -3 and a product of -10.
The factors of y² - 3y - 10 are ( y -5 ) and ( y + 2 ).

C. General Trinomials that Factor

The general trinomial will be in the form of ax² + bx + c where a, b, and c are nonzero real numbers.

For general trinomials you will need to list the factor pairs of a and c. The sign in front of the c still tells you whether you are looking for a sum or difference. This time, however, it will not be a single factor pair that you're looking for but the sum or difference of the products of factors from a factor pair of the a and a factor pair of the c. If you have what you're looking for, the trinomial will factor. If you don't, the trinomial will not factor. To factor the trinomial:

Use the factors of the factor pair you chose for a. The variable always factors into its square roots. Put them in the first positions of both parentheses. (Remember FOIL: first times first gives you first.)

Use the factors of the factor pair you chose for c. Put them in the last positions of both parentheses but make sure that you place them so that the inner and outer products combine to give you the middle term. (Remember FOIL: last times last gives you last and a combination of the inner and outer products give you the middle term.)

Again look to the sign in front of c to tell you what signs to use in the parentheses of the factors.

	A plus sign in front of c tells you that you will have like signs in the parentheses - either both pluses or both minuses. The sign in front of b tells you which one.
	A minus sign in front of c tells you that you will have unlike signs - one plus and one minus. You must place them in the parentheses so that when you do the inner and outer products and combine them, you come up with the correct sign for b.

Example 1: Factor 3p² - 4p - 4

Factor pairs of 3 are
1,3

Factor pairs of 4 are
1,4
2,2

We are looking for a difference of 4 (b).

Using the 1,3 and the 1,4
1 · 1 = 1 and 3 · 4 = 12 1 and 12 do not give us a difference of 4.
1 · 4 = 4 and 3 · 1 = 3 4 and 3 do not give us a difference of 4.

Using the 1,3 and the 2,2
1 · 2 = 2 and 3 · 2 = 6 2 and 6 give us the difference of 4 we were looking for. Thus, this trinomial factors into (1p - 2)(3p + 2). We had to use different signs since the sign of the third term is minus. We placed the signs so that the inner and outer products combine to give us a negative 4p.

Example 2: Factor 20 u² + 19uv + 3v²

Factor pairs of 20 are
1,20
2,10
4,5

Factor pairs of 3 are
1,3

We are looking for a sum of 19.

Using the 1,20 and the 1,3
1 · 1 = 1 and 20 · 3 = 60 1 and 60 do not give us a sum of 19.
1 · 3 = 3 and 20 · 1 = 20 3 and 20 do not give us sum of 19.

Using the 2,10 and 1,3
2 · 1 = 2 and 10 · 3 = 30 2 and 30 do not give us a sum of 19.
2 · 3 = 6 and 10 · 1 = 10 6 and 10 do not give us a sum of 19.

Using the 4,5 and the 1,3
4 · 1 = 4 and 5 · 3 = 15 4 and 15 give us the sum of 19 we are looking for.

Thus this trinomial factors into (4u + 3v)(5u + 1v). We used like signs because the sign of the third term is plus and they both had to be pluses because the sign of the second term is plus. Also, notice we placed them so that the 4 and 1 are paired and the 5 and 3 are paired.

IV. Four or More Terms

You must factor by grouping. Group terms to form one or more of the above factorable polynomials.

Example 1: Factor x³ + 2x² + 8x + 16

Solution: There are no common factors to all four terms. Try grouping the first two terms and the last two terms and factoring out the common factor with each grouping as follows:

(x³ + 2x²) + (8x + 16)

x²(x + 2) + 8(x + 2)

(x + 2) (x² + 8)

Example 2: Factor xy - 4y + 3x - 12

Solution: Again, there are no common factors to all four terms. Try grouping the first two terms and the last two terms and factoring out the common factor with each grouping as follows:

(xy - 4y) + (3x - 12)

y(x - 4) + 3(x - 4)

(x - 4) (y + 3)

Example 3: Factor xy - 4y - 3x + 12

Solution: Group the first two terms and the last two terms again.

(xy - 4y) + (-3x + 12)

y(x - 4) + 3(-x + 4) This time there is no common terms and we can't stop here because it is not all factors. What if I take out a -3 rather than a +3 in the second grouping?

y(x - 4) - 3(x - 4)

(x - 4) (y - 3)

Now it's all factors!

Example 4: Factor xy - 4y + 3x + 12

Solution: (xy - 4y) + (3x + 12)

y(x - 4) + 3(x + 4)

At this point it is important to realize that no common factor resulted and there is nothing I can do to get a common factor. As a matter of fact, there is nothing I can do to factor this polynomial. This polynomial can't be factored. Remember, not all polynomials can be factored. What do we call those that can't be factored?

PRIME!