Factoring polynomials involves finding common factors that can be divided out of terms, similar to factoring numbers but with variables; this is done by looking for a single variable or number that is a common factor of all terms that can be pulled out in front of parentheses. The document provides examples of different types of factoring polynomials including using the greatest common factor, difference of squares, grouping, and perfect squares and cubes.

1. -- Factoring polynomial expressions is not quite the same
as factoring numbers, but the concept is very similar.
When factoring numbers or factoring polynomials, you are
finding numbers or polynomials that divide out evenly from
the original numbers or polynomials. But in the case of
polynomials, you are dividing numbers and variables out
of expressions, not just dividing numbers out of numbers.

2. 1: One common factor. a x + a y = a (x + y)
2: Sevearl grouped common factor. a x + a y + b x + b y = a(x + y) + b(x + y) = (a +
b ) (x + y)
3: Difference of two squares (1). x 2 - y 2 = (x + y)(x - y)
4: Difference of two squares (2). (x + y) 2 - z 2 = (x + y + z)(x + y - z)
5: Sum of two cubes. x 3 + y 3 = (x + y)(x 2 - x y + y 2)
6: Difference of two cubes. x 3 - y 3 = (x - y)(x 2 + x y + y 2)
7: Difference of fourth powers. x 4 - y 4 = (x 2 - y 2)(x 2 + y 2) = (x + y)(x - y)(x 2 + y 2)
8: Perfet square x 2 + 2xy + y 2 = (x + y) 2
9: Perfet square x 2 - 2xy + y 2 = (x - y) 2
10: Perfect cube x 3 + 3x 2y + 3xy 2 + y 3 = (x + y) 3
11: Perfect cube x 3 - 3x 2y + 3xy 2 - y 3 = (x - y) 3

3.  Previously, you have simplified expressions by
distributing through parentheses, such as:
Example :
2(x + 3) = 2(x) + 2(3) = 2x + 6
The trick is to see what can be factored out of every
term in the expression. Warning: Don't make the
mistake of thinking that "factoring" means
"dividing something off and making it magically
disappear". Remember that "factoring" means
"dividing out and putting in front of the
parentheses". Nothing "disappears" when you
factor; things merely get rearranged.

4.  Any number or variable that is a factor of
both terms in a binomial can be factored
out. For example, the binomial 3x^3 + 6x
can be factored to (3x)(x^2 + 2) because
3x is a factor of 3x^2 (which is 3x * x^2)
and of 6 (which is 3x * 2).

5.  Find the greatest common factor (GCF) of
both terms. The greatest common factor is
the largest value that can be factored out
of both terms. In the expression 6y^2 - 24,
3 is a common factor of both terms, but it
is not the GCF. Six is the GCF, since both
numbers can be divided by 6. Factoring it
out, we get 6(y^2 - 4).

6.  Find out if you have a difference of
squares. A difference of squares is a
variable squared minus a constant, like
y^2 - 4. If you have factored out the GCF
and don't have a minus sign in your
binomial, you are done.

7.  Solve the difference of squares. First
make sure the numbers are arranged in
the proper order, with the positive term
before the negative term, then find the
square root of each term. In the example
above, the square root of y^2 is y, and the
square root of 4 is 2.

8.  Set up two sets of parentheses. Each will
have the first square root followed by the
second square root. In the first, they will
be separated by a addition sign, and in
the second, a subtraction sign. To take the
example from steps 3 and 4, we get y^2 -
4 = (y + 2)(y - 2). Looking at the whole
problem for step 3, we get 6y^2 - 24 =
6(y^2 - 4) = 6(y + 2)(y - 2).

9. Example : 7x -7
A "7" can come out of each term, so I'll factor
this out front:
7x – 7 = 7( )
Dividing the 7 out of "7x" leaves just an "x":
7x – 7 = 7(x )
What am I left with when I divide the 7 out of
the second term? Well, if "nothing" is
left, then "1" is left. (Remember: 7 ÷ 7 = 1.)
So I get:
7x – 7 = 7(x – 1)
 Take careful note: When "nothing" is left after
factoring, a "1" is left behind in the
parentheses.

10. Example : x2y3 + xy
I can factor an "x" and a "y" out of each
term: x2y3 = xy(x1y2)
= xy(xy2) and xy = xy(1).
x2y3 + xy = xy( )
= xy(xy2 )
= xy(xy2 + 1)

11. Example : 3x3 + 6x2 – 15x.
I can factor a "3" and an "x" out of each
term: 3x3 = 3x(x2), 6x2 = 3x(2x), and
–15x = 3x(–5). Being careful of my signs, I
get:
3x3 + 6x2 – 15x = 3x( )
= 3x(x2 )
= 3x(x2 + 2x )
= 3x(x2 + 2x – 5)

12. If the Polynomial has 4 terms or more,
Factor by Grouping
example : x^3 + x^2 + 2x + 2
x^3 + x^2 | + 2x + 2
= x^2(x +1) | + 2 (x +1)
=(x + 1 ) ( x^2 +2 )

13.  1. x2 + 4x – x – 4.
 2. x2 – 4x + 6x – 24.
 3.3x – 12.
 4. 12y2 – 5y.
 5. 9 - 4x 2.

14.  1. x2 + 4x – x – 4
= x(x + 4) – 1(x + 4)
= (x + 4)(x – 1)
 2. x2 – 4x + 6x – 24
= x(x – 4) + 6(x – 4)
= (x – 4)(x + 6)
 3. 3x – 12 = 3( )
3x – 12 = 3(x )
3x – 12 = 3(x – 4)

15.  4. 12y2 – 5y = y( )
12y2 – 5y = y(12y )
12y2 – 5y = y(12y – 5)
 5. 9 - 4x 2
= 3 2 - (2x) 2
= (3 - 2x)(3 + 2x)

16.  a) 3x² + 8x + 5
 b) 3x² + 16x + 5
 c) 2x² + 9x + 7
 d) 2x² + 15x + 7
 e) 5x² + 8x + 3
 f) 5x² + 16x + 3
 g) 2x² + 5x − 3
 h) 2x² − 5x − 3
 i) 2x² + x − 3
 j) 2x² − 13x + 21

17.  A. (3x + 5)(x + 1)
 B. (3x + 1)(x + 5)
 C. (2x + 7)(x + 1)
 D. (2x + 1)(x + 7)
 E. (5x + 3)(x + 1)
 F. (5x + 1)(x + 3)
 G. (2x − 1)(x + 3)
 H. (2x + 1)(x − 3)
 I. (2x + 3)(x − 1)
 J. (2x − 7 )(x −3)

18. Leader : Ghia Adrielle Alpuerto
Members :
Marco Ruelos
Xina Sularte
Justine tan
Michelle Mendoza