To factor a polynomial using greatest common factors (GCF): 1. Find the GCF of the coefficients and of the variables. 2. The GCF is the factored form of the polynomial. 3. Checking the factored form using the distributive property verifies the correct factorization.

2. To factor using a GCF, take the greatest
common factor (GCF), for the numerical
coefficient. When choosing the GCF for the
variables, if all terms have a common variable,
take the ones with the lowest exponent.
Example: 9x4 + 3x3 + 12x2
GCF: Coefficients = 3
Variables (x) = x2

3. GCF = 3x2
Next, you just divide each
monomial by the GCF!
Answer = 3x2(3x2 + x + 4)
Then, check by using the distributive
property!

4. 1)2a + 2b =
2) 5x2 + 5 =
3) 18c – 27d=
4) hb + hc=
5) 6x – 18=
6) 3a2 – 9=
7) 4x2 – 4y2=
8) p + prt=
9) 10x – 15x3=
10) 2x – 4x3=
2(a+b)
5(x2 +1)
9(2c-3d)
h(b+c)
6(x - 3)
3(a2 – 3)
4(x–y) (x+y)
p(1+rt)
5x(2–3x2)
2x(1-2x2)

5. 11) 8x – 12= 4(2x-3)
12) 8 – 4y= 4(2-y)
13) 3ab2 – 6a2b= 3ab(b-2a)
14) 10xy – 15x2y2= 5xy(2-3xy)
15) 21r3s2 – 14r2s= 7r2s(3rs-2)
16) 2x2 + 8x + 4= 2(x2+4x+2)
17) 6c3d – 12c2d2 + 3cd= 3cd(2c2-4cd+1)
18) 3x2 – 6x – 30= 3(x2-2x-10)
19) ay – 4aw – 12a= a(y-4w-12)
20) c3 – c2 + 2c= c(c2 - c+2)

6.  Case I is when there is a coefficient of 1 in
front of your variable2 term (x2).
 You have two hints that will help you:
 When the last sign is addition, both signs are
the same and match the middle term.
 When the last sign is subtraction, both signs
are different and the larger number goes with
the sign of the middle term.

7.  Examples:
Hint #1: Hint #2:
x2 – 5x + 6 x2 + 5x – 36
(x - )(x - ) (x - )(x + )
Find factors of 6, Find factors of
w/ sum of 5. 36 w/ difference
(x – 3)(x – 2) of 5.
CHECK USING FOIL (x – 4)(x + 9)
CHECK USING FOIL

8. 1) a2 + 3a + 2=
2) c2 + 6c + 5=
(a+2)(a+1)
(c+5)(c+1)
3) x2 + 8x + 7= (x+7)(x+1)
4) r2 + 12r + 11= (r+11)(r+1)
5) m2 + 5m + 4= (m+4)(m+1)
6) y2 + 12y + 35=
7) x2 + 11x + 24 =
(y+7)(y+5)
(x+3)(x+8)
8) a2 + 11a + 18 =(a+9)(a+2)
9) 16 + 17c + c2 = (c+16)(c+1)
10)x2 + 2x + 1 = (x+1)(x+1)
11) z2 + 10z + 25 = (z+5)(z+5)

9. 1) 2x2 + 15x + 7 = (2x+1)(x+7)
2) 3x2 – 5x – 12= (3x+4)(x-3)
3) 9x2 + 11x + 2= (9x+2)(x+1)
4) 7x2 – 22x + 3= (7x-1)(x-3)
5) 18x2 – 9x – 2= (6x+1)(3x-2)
6) 4x2 + - 7x – 2= (4x+1)( x-2)

10.  Step 1. Arrange the terms so that the first two terms
have a common factor and the last two terms have a
common factor.
 Step 2. For each pair of terms, use the distributive
property to factor out the pair’s greatest common
factor.
 Step 3. If there is now a common binomial factor,
factor it out.
 Step 4. If there are no common binomial factor in step
3, begin again, rearranging the terms differently. If no
rearrangement leads to a common binomial factor, the
polynomial cannot be factored.

11. Example 1) Factor 3x2 + 4xy – 3x – 4y
by grouping.
3x2 + 4xy – 3x – 4y =
x(3x + 4y) –(3x + 4y) =
(3x + 4y)(x – 1)
Reminder: When factoring a
polynomial, make sure the polynomial
is written as a product. Do not write
the final answer as a sum or
difference of terms like: x(3x+4y) -
(3x+4y) This form is not a factored
form of the original polynomial. The
factored form is the product:
(3x + 4y)(x – 1)

12. Example 2) Factor ax – ab – 2bx + 2b2
by grouping.
ax – ab – 2bx + 2b2 =
a(x – b) – 2b(x – b) =
(x – b)(a -2b)

13. A. Simplifying Rational Expressions
 Recall that a rational number is a quotient of
integers. A rational expression is a quotient of
polynomials, such as:
𝑃
𝑄
where Q  0 and P and Q are polynomial
 A rational expression can be evaluated just as
any polynomial, except that a rational expression
can be undefined when the denominator is equal
to zero.

14. Recall that
−𝑎
𝑏
=
𝑎
−𝑏
= −
𝑎
𝑏
when 𝑏 ≠ 0
This is also true of polynomials. My
suggestion is to simply pull any negative signs
out and consider the entire expression as
either a positive or negative expression.
Example:
−(𝑥+3)
𝑥2−3𝑥+2
= -
𝑥+3
𝑥2−3𝑥+2

15. Just as when we were dealing with
fractions, if you multiply the
numerator and denominator by the
same thing the resulting expression
is equivalent. This is called the
Fundamental Principle of Rational
Expressions, when we are discussing
a fraction of polynomials.
𝑃𝑅
𝑄𝑅
=
𝑃
𝑄
if P, Q and R are polynomials
and Q and R ≠ 0

16.  Concept Example:
15
35
=
35
75
=
3
7
 In order to simplify rational expressions we
will use the Fundamental Principle of Rational
Expressions just as we used the Fundamental
Principle of Fractions to simplify fractions.
 Steps to Simplifying a Rational Expression
1) Factor the numerator and the denominator
completely
2) Cancel common factors

17.  Example: Simplify
15𝑥2𝑦
35𝑥𝑦2 =
3𝑥
7𝑦

18. Recall that another way of
using the principle is division!
Example 1: Simplify
𝑥2
𝑥2+2𝑥
=
𝑥2
𝑥2 + 2𝑥
=
𝑥2
𝑥(𝑥 + 2)
=
𝑥
𝑥 + 2

19. Example 2:
Simplify
𝑥+5
𝑥2+2𝑥−15
=
𝑥 + 5
𝑥2 + 2𝑥 − 15
=
𝑥 + 5
(𝑥 + 5)(𝑥 − 3)
=
1
𝑥 − 3

20. Example 3:
Simplify
𝑥2−𝑥−2
𝑥2+5𝑥−14
=
𝑥2
− 𝑥 − 2
𝑥2 + 5𝑥 − 14
=
(𝑥 − 2)(𝑥 + 1)
(𝑥 + 7)(𝑥 − 2)
=
𝑥 + 1
𝑥 + 7

21.  Example 4:
 Simplify
−𝑥+𝑦
𝑥−𝑦
=
=
−𝑥 + 𝑦
𝑥 − 𝑦
=
−(𝑥 − 𝑦)
𝑥 − 𝑦
= −1

22. Steps for Multiplying Rational Expressions
1) Factor numerators and denominators
completely
2) Cancel if possible
3) Multiply numerators and denominators
Concept Example:
2
3

3
8
=
1
4

23. EXAMPLE 1:
𝑥 + 2
𝑥2 + 7𝑥 + 6

𝑥 + 1
𝑥3 + 8
=
𝑥 + 2
𝑥2 + 7𝑥 + 6

𝑥 + 1
𝑥3 + 8
=
𝑥 + 2
𝑥 + 6 𝑥 + 1
𝑥 + 1
𝑥 + 2 𝑥2 − 2𝑥 + 4
=
1
(𝑥 + 6)(𝑥2 − 2𝑥 + 4)

24. Example 2:
7𝑥2
15𝑦3 
−3𝑦
14𝑥3
=
7𝑥2
15𝑦3

−3𝑦
14𝑥3
=
7𝑥2
15𝑦3

−3𝑦
14𝑥3
=
−1
10𝑥𝑦2

25. 1) Take the reciprocal of the divisor.
2) Multiply the dividend and the reciprocal of
the divisor
Concept Example:
1
2
÷
2
3
=
1
2

3
2

26. Example 1:
𝑥 + 1
2𝑥2 − 𝑥 + 1
÷
3𝑥2
− 3
𝑥2 − 3𝑥 + 2
=
𝑥 + 1
2𝑥2 − 𝑥 + 1

𝑥2 − 3𝑥 + 2
3𝑥2 − 3
=
𝑥 + 1
2𝑥 + 1 𝑥 − 1

𝑥 − 2 𝑥 − 1
3 𝑥2 − 1
=
𝑥 + 1
(2𝑥 + 1)(𝑥 − 1)

(𝑥 − 2)(𝑥 − 1)
3(𝑥 − 1)(𝑥 + 1)
=
𝑥 − 2
3(2𝑥 + 1)(𝑥 − 1)

27. Example 2:
4𝑥2
− 16
𝑥 + 1
÷
3x2  3
x2 3 x + 2
=
4(𝑥2−4)
𝑥+1
÷
3(𝑥2−1)
(𝑥−2)(𝑥−1)
=
4(𝑥−2)(𝑥+2)
𝑥+1
÷
3(𝑥−1)(𝑥+1)
(𝑥−2)(𝑥−1)
=
4(𝑥−2)(𝑥+2)
𝑥+1
∗
(𝑥−2)(𝑥−1)
3 𝑥−1 𝑥+1
=
4 𝑥+2 𝑥−2 2
3 𝑥+1 2

28.  When we have a common denominator, as
with fractions, we simply add or subtract the
numerators. We do have to be cautious
because when subtracting it is the entire
numerator that's subtracted, so we must use
the distributive property to subtract.
 Concept Example:
7
15
−
3
15
=
4
15

29. Example 1:
7
15
-
3
15
=
4
15
Example 2:
2𝑥+5
𝑦+2
-
𝑥−5
𝑦+2
=
(2𝑥+5)−(𝑥−5)
𝑦+2
=
2𝑥+5−𝑥+5
𝑦+2
=
𝑥+10
𝑦+2

30. Concept Example:
2
15
+
4
15
=
6
15
=
2
5
Example:
𝑦
𝑦−2
−
2
𝑦−2
=
𝑦−2
𝑦−2
= 1
This leads us to the point that we always
need to factor the numerator and
denominator in order to reduce and also
for our next application of finding a
common denominator.

31. Finding the LCD of a rational expression is the
same as finding the LCD of fractions. We just
need to remember that a polynomial must
first be factored. Each factor that isn't a
constant is considered like a prime number.
Concept Example:
Find the LCD of
2
15
,
7
36
Prime Factors of 15: 5, 3
Prime Factors of 36: 2, 2, 3, 3
Common Factors: 3
Uncommon Factors: 5, 2, 2, 3
LCD(Combination of ALL factors):
3* 5* 2* 2*3 = 180

32. Equivalent Fractions:

2
15
=
24
180
 Using the LCD to find its
equivalent fraction, for its
numerator, divide the LCD (180) by
the denominator (of original
fraction-15), then multiply it by the
numerator (2).

7
36
=
35
180
 Using the LCD to find its
equivalent fraction, for its
numerator, divide the LCD (180) by
the denominator (of original
fraction-36), then multiply it by the
numerator (7).

33. Example: Find the LCD of
1
𝑥
,
2𝑥+5
𝑥(𝑥+5)
Prime Factors of x: x
Prime Factors of x: x, (x+5)
Common Factors: x
Uncommon Factors: x+5
LCD(Combination of ALL factors):
x(x+5)
Equivalent Fractions:
1
𝑥
=
(𝑥+5)
𝑥(𝑥+5)
2𝑥+5
𝑥(𝑥+5)
=
2𝑥+5
𝑥(𝑥+5)
 Since the
denominator and LCD are equal, we
have the same fraction

34.  Example: Find the LCD of
𝑥+2
𝑥2+3𝑥+2
,
𝑥+1
𝑥+2
Prime Factors of
(x2 + 3x + 2): (x+2)(x+1)
Prime Factors of (x+2): (x+2)
Common Factors: (x+2)
Uncommon Factors: (x+1)
LCD(Combination of ALL factors):
(x+2)(x+1)
Equivalent Fractions:
𝑥+2
𝑥2+3𝑥+2
=
𝑥+2
(𝑥+2)(𝑥+1)

Since the denominator and LCD are equal,
we have the same fraction
𝑥 + 1
𝑥 + 2
=
(𝑥 + 1)(𝑥 + 1)
(𝑥 + 2)(𝑥 + 1)
=
𝑥2 + 2𝑥 + 1
(𝑥 + 2)(𝑥 + 1)

35.  Example: Find LCD of
2
4𝑥2+13𝑥+3
,
𝑥+1
𝑥2+6𝑥+9
1) Factor 4x2 + 13x + 3: (4x+1)(x+3)
2) Factor x2 + 6x + 9: (x+3)(x+3)
3) LCD: (4x+1)(x+3)(x+3)

36. Next, we need to build a higher term so
that we can add and subtract rational
expressions with unlike denominators.
Building A Higher Term (another
solution to get the equivalent fraction)
1) Find LCD
2) Divide LCD by denominator
3) Multiply numerator by quotient

37. Concept Example: Build the higher term
2
15
,
7
36
1) LCD was 180
2) 180  15 = 12 180  36 = 5
3)
2
15

12
12
=
24
180
7
36

5
5
=
35
180

38. Example:
Find the LCD of
1
𝑥
,
2𝑥+5
𝑥(𝑥+5)
1) LCD was x(x + 5)
2)
𝑥(𝑥+5)
𝑥
= 𝑥 + 5
𝑥(𝑥+5)
𝑥(𝑥+5)
= 1
3)
1
𝑥

𝑥+5
𝑥+5
=
𝑥+5
𝑥(𝑥+5)
2𝑥+5
𝑥(𝑥+5)

39. Addition and Subtraction of
Rational Expressions w/ Unlike
Denominators
 1) Find LCD
 2) Build Higher Term
 3) Add or Subtract as normal

40. Example:
15𝑎
𝑏
+
6𝑏
5
LCD : 5b
1) Build Higher Term
2)15a  5 = 75a 6b  b = 6b2
b 5 5b 5 b 5b
3) Add:
75𝑎
5𝑏
+
6𝑏2
5𝑏
=
75𝑎+6𝑏2
5𝑏

41.  Example:
15
2𝑥−4
+
𝑥
𝑥2−4
1) Factors: 2x – 4 = 2(x – 2)
Factors: x2 – 4 = (x – 2)(x + 2)
LCD: 2(x – 2)(x + 2)
2) Build Higher Term
15
2 𝑥 − 2
𝑥 + 2
𝑥 + 2
=
15 𝑥 + 2
2 𝑥 − 2 𝑥 + 2
𝑥
(𝑥 − 2)(𝑥 + 2)
2
2
=
2𝑥
2(𝑥 − 2)(𝑥 + 2)
3) Add:
15 𝑥+2
2 𝑥−2 𝑥+2
+
2𝑥
2(𝑥−2)(𝑥+2)
=
15𝑥 + 30 + 2𝑥
2(𝑥 − 2)(𝑥 + 2)
=
17𝑥 + 30
2(𝑥 − 2)(𝑥 + 2)

42. Example:
𝑥
𝑥2−4
-
2
4−𝑥2
𝑥
𝑥2 − 4
−
2
4 − 𝑥2
=
𝑥
𝑥2 − 4
−
2
− 𝑥2 − 4
=
𝑥
𝑥2 − 4
+
2
𝑥2 − 4
=
𝑥 + 2
𝑥 − 2 𝑥 + 2
=
1
𝑥 − 2

43. Example:
𝑦+2
𝑦+3
− 2
𝑦 + 2
𝑦 + 3
− 2 =
𝑦 + 2
𝑦 + 3
−
2 𝑦 + 3
𝑦 + 3
=
𝑦 + 2 − 2 𝑦 + 3
𝑦 + 3
=
𝑦 + 2 − 2𝑦 − 6
𝑦 + 3
=
−𝑦 − 4
𝑦 + 3
𝑜𝑟
−(𝑦 + 4)
𝑦 + 3

44. Example:
𝑥
𝑥2−1
−
2
𝑥2−2𝑥+1
=
𝑥
𝑥 − 1 𝑥 + 1
−
2
𝑥 − 1 𝑥 − 1
=
𝑥 𝑥 − 1 − 2 𝑥 + 1
𝑥 − 1 𝑥 − 1 𝑥 + 1
=
𝑥2
− 𝑥 − 2𝑥 − 2
𝑥 − 1 𝑥 − 1 𝑥 + 1
=
𝑥2
− 3𝑥 − 2
𝑥 − 1 𝑥 − 1 𝑥 + 1

45. A complex fraction is a fraction with an
expression in the numerator and an expression
in the denominator.
 Simplifying Complex Fractions Method #1
1) Solve or simplify the problem in the
numerator
2) Solve or simplify the problem in the
denominator
3) Divide the numerator by the denominator
4) Reduce

46. Example:
3
4
2
3
=
3
4
÷
2
3
=
3
4

3
2
=
9
8

47. Example:
6𝑥 − 3
5𝑥2
2𝑥 − 1
10𝑥
=
6𝑥 − 3
5𝑥2
÷
2𝑥 − 1
10𝑥
=
6𝑥 − 3
5𝑥2
10𝑥
2𝑥 − 1
=
3(2𝑥 − 1)
5𝑥2
10𝑥
2𝑥 − 1
=
6
𝑥

48. Example:
1
2
+
2
3
5
9
−
5
6
=
1
2
+
2
3
÷
5
9
−
5
6
=
3 + 4
6
÷
10 − 15
18
=
7
6
÷
−5
18
=
7
6
−
18
5
= −
21
5

49. Example:
1
5
−
1
𝑥
7
10
−
1
𝑥2
=
1
5
−
1
𝑥
÷
7
10
−
1
𝑥2
=
𝑥 − 5
5𝑥
÷
7𝑥2
− 10
10𝑥2
=
𝑥 − 5
5𝑥
10𝑥2
7𝑥2 − 10
=
2𝑥(𝑥 − 5)
7𝑥2 − 10

50.  Simplifying Complex
Fractions Method #2
1) Find the LCD of the
numerator and the
denominator fractions
2) Multiply numerator and
denominator by LCD
3) Simplify resulting
fraction

51.  Example:
3
4
2
3
=
3
4
2
3

12
12
=
3
4
(12)
2
3
(12)
=
9
8

52.  Example:
6𝑥 − 3
5𝑥2
2𝑥 − 1
10𝑥
=
6𝑥 − 3
5𝑥2
2𝑥 − 1
10𝑥

10𝑥2
10𝑥2
=
6𝑥 − 3
5𝑥2 (10𝑥2
)
2𝑥 − 1
10𝑥
(10𝑥2)
=
2(6𝑥 − 3)
𝑥(2𝑥 − 1)
=
(2)(3)(2𝑥 − 1)
𝑥(2𝑥 − 1)
=
6
𝑥

53. Example:
1
2
+
2
3
5
9
−
5
6
=
1
2
+
2
3
5
9
−
5
6

18
18
=
1
2
(18) +
2
3
(18)
5
9
(18) −
5
6
(18)
=
9 + 12
10 − 15
= −
21
5

54. Example:
1
5
−
1
𝑥
7
10
−
1
𝑥2
=
1
5
−
1
𝑥
7
10
−
1
𝑥2
10𝑥2
10𝑥2
=
1
5
(10𝑥2) −
1
𝑥
(10𝑥2)
7
10
(10𝑥2) −
1
𝑥2 (10𝑥2)
=
2𝑥2
− 10𝑥
7𝑥2 − 10
=
2𝑥(𝑥 − 5)
7𝑥2 − 10

55. THANK
YOU 