WHAT IS FACTORING?

definition: The factors of a number or expression are the two or more numbers, elements or quantities whose product equals the number or expression.

For example, the factors of 12 are 2 and 6, 3 and 4, 12 and 1.

definition: The prime factors of a given number or expression are all the prime numbers

that are factors of the given number or expression.

So the prime factors of 12 are 1, 2, and 3 since 1 × 2 × 2 × 3 = 12.

For example, 6 and 3 or 9 and 2 are factors of 18. The numbers 7 and 3 are factors of 21 and they are also prime factors of 21 since both 7 and 3 are prime numbers.

Recall that a prime number is one which can be expressed only as the product of itself and one.

So 2, 3, 5, 7, and 11 are all prime numbers since 2 = 2 × 1, 3 = 3 × 1, etc.

Factoring is an extremely important mathematical technique since it allows us to write complicated expressions and equations in a simplified product form, which helps us solve equations and inequalities.

There are a number of different types of factoring, each of which applies to specific situations. It is advisable to learn the names of the different types of factoring in order to help recognize which technique to use at any given time.

The one thing to always remember about all types of factoring is that the expression in factored form must be equal to the original expression.

In other words, if you multiply out the factors, you should get exactly what you started with.

Here are the different types of factoring you should know.

common factor

Factoring out a common factor is doing the opposite of the distributive property.

For example, when we factor ab + ac, which has a common factor of a in both terms,
we get a (b + c). The distributive property says that a (b + c) = ab + ac, so as you can see, we're going backwards when we factor. Instead of distributing or dishing out the multiplier, we collect it or take it out.

It's sort of like when your teacher gives you a test. To start, s/he distributes the test papers and answer booklets -- and then -- after you've suffered for an hour or two -- s/he takes them all back.

so, 5(xy – 3) = 5xy – 15 we distribute the multiplier 5.

5xy – 15 = 5(xy – 3) we " factor " or " take out " the common factor 5.

In common factoring, we write the given expression as the product of two terms:

( common term) ( terms it has to multiply) to produce the original expression.

xyz + xym + xyt = xy ( z + m + t )

Notice that we can divide every term in the original expression by xy.

examples

a) 3x² – 6x + 18 =
3 (x² – 2x + 6)	factor out 3, the common factor
b) 2a³ + 4a² – 2a =
2a (a² + 2a – 1)	the common factor is 2a.
c) ax + ay – bx – by =	there are 2 common factors: a and – b
a(x + y) – b(x + y) =	now (x + y) is the common factor.
(x + y) (a – b)	factor out (x + y)
d) (x – y) + (x – y)(x + y) =	the common factor is (x – y)
(x – y)[1 + (x + y)] =	note the 1 since (x – y) is 1 times (x – y)
(x – y)[1 + x + y]	remove the inner brackets
e) 2ax – 3by – 3ay + 2bx =
2ax + 2bx – 3ay – 3by =	change the order of the terms
2x(a + b) – 3y(a + b) =	now, the common factors are 2x and – 3y
(a + b)(2x – 3y)	the common factor is (a + b)

difference of squares

Think about the words.

What is a difference of squares?

The word difference in math means a minus sign

squares are terms raised to the second power.

So a difference of squares is one square minus another square
such as x² – y ² or a² – 25.

Even a² – 5 can be considered a difference of squares since 5 is the square of .

To factor a difference of squares, we make 2 brackets

First contains (the sum of the roots), second contains (the difference of the roots).

So x² – y² is factored as (x + y)(x – y) and a² – 25 is factored as (a + 5)(a – 5).

The order of the brackets doesn't matter since multiplication is commutative.

That is, we know that a times b equals b times a.

a difference of squares is factored as

(the sum of the roots)(the difference of the roots)

x ² – y ² = (x + y)(x – y)

Note: A sum of squares cannot be factored!

examples

a) 25x² – 36y² =	difference of squares
(5x + 6y)(5x – 6y)	(sum of roots)(difference of roots)
b) 16x4 – 1 =	difference of squares
(4x² + 1)(4x² – 1) =	4x² – 1 is a difference of squares
(4x² + 1)(2x + 1)(2x – 1)
c) 2a³ – 50ab² =	common factor 2a.
2a (a² – 25b² ) =	a² – 25b² difference of squares
2a (a + 5b)(a – 5b)
d) (x + y)² – z² =	difference of squares
[(x + y) + z][(x + y) – z] =
(x + y + z)(x + y – z)
e) 4x² – (3y – 1)² =	difference of squares
[2x + (3y – 1)][2x – (3y – 1)] =	remove the inner brackets
(2x + 3y – 1)(2x – 3y + 1)

sum or difference of cubes

Note: though a sum of squares cannot be factored,

both a sum and difference of cubes can be factored

and they are both factored in the same way.

To factor a sum or difference of cubes make two brackets.

The first contains the cube roots of the 2 terms, with the same sign as the sum or difference.

The second bracket comes from the first.

Square the first term from the first bracket

multiply the two terms together, change the sign,

square the last term.

To check that our factoring is correct, we multiply out the brackets to find that we get what we started with.

Note: it's a good idea to learn the cubes of the first 6 integers.

Note: the last term in the factorization of a sum or difference of cubes is always positive
since it is a perfect square.

examples

a) x³ + y³ =	Sum of cubes. 2 brackets. 1st cube roots, same sign.
(x + y) (x² – xy + y²)	2nd: square 1st term, product of the 2 terms by –1,
	square the last term.
b) 27x³ – 8 =	Difference of cubes.
(3x – 2) (9x² + 6x + 4)	The cube roots of 27x³ and 8 are 3x and 2 so factors as shown.
c) 1 + 125m 6 =	Sum of cubes. cube roots of 1 and 125m6 are 1 and 5m²
(1 + 5m²)(1 – 5m² + 25m4)	factors are as shown.
d) 64x 6 – 1 = (4x² – 1)(16x4 + 4x² + 1) = (2x + 1)(2x – 1)(16x4 + 4x² +1)	Difference of cubes. Now (4x² – 1) is a difference of squares. (sum of roots)( difference of roots)

practice:

Factor Completely if Possible

1) 2x² – 8 ...... 2) ax² + 3ax +2a ...... 3) 3x³ + 24

4) x⁴ – 27x ...... 5) x⁶ – 64 ...... 6) 4x² + 4x + 1 – 16y²

7) 25 – x² – 6x – 9 ...... 8) ax² – ay² + bx + by

solutions

1) 2(x – 2)(x + 2) ......2) a(x + 1)(x + 2) ......3) 3(x + 2)(x² – 2x +4)

4) x(x – 3)(x² + 3x + 9) ......5) (x – 2)(x + 2)(x² + 2x + 4)(x² – 2x + 4)

6) (2x + 1 – 4y)(2x + 1 + 4y) ......7) (2 – x)(8 + x) ......8) (x + y)(ax – ay + b)

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