4. Factoring
We use the Distributive Property to expand
algebraic expressions.
We sometimes need to reverse this process
(again using the Distributive Property) by:
• Factoring an expression as a product
of simpler ones.
5. Factoring
For example, we can write:
• We say that x – 2 and x + 2 are factors
of x2
– 4.
2
4 ( 2)( 2)
Factoring
Expanding
x x x
→
− = − +
¬
8. E.g. 1—Factoring Out Common Factors
Factor each expression.
(a) 3x2
– 6x
(b) 8x4
y2
+ 6x3
y3
– 2xy4
9. E.g. 1—Common Factors
The greatest common factor of the terms
3x2
and –6x is 3x.
So, we have:
3x2
– 6x = 3x(x – 2)
Example (a)
10. E.g. 1—Common Factors
We note that:
• 8, 6, and –2 have the greatest common
factor 2.
• x4
, x3
, and x have the greatest common
factor x.
• y2
, y3
, and y4
have the greatest common
factor y2
.
Example (b)
11. E.g. 1—Common Factors
So, the greatest common factor of the three
terms in the polynomial is 2xy2
.
Thus, we have:
8x4
y2
+ 6x3
y3
– 2xy4
= (2xy2
)(4x3
) + (2xy2
)(3x2
y)
+ (2xy2
)(–y2
)
= 2xy2
(4x3
+ 3x2
y – y2
)
Example (b)
13. E.g. 2—Factoring Out a Common Factor
The two terms have the common factor x – 3.
(2x + 4)(x – 3) – 5(x – 3)
= [(2x + 4) – 5](x – 3) (Distributive Property)
= (2x – 1)(x – 3) (Simplify)
15. Factoring x2
+ bx + c
In order to factor a trinomial of the form
x2
+ bx + c, we note that:
(x + r)(x + s) = x2
+ (r + s)x + rs
• So, we need to choose numbers r and s
so that r + s = b and rs = c.
16. E.g. 3—Factoring x2
+ bx + c by Trial and Error
Factor: x2
+ 7x + 12
• We need to find two integers whose product is 12
and whose sum is 7.
• By trial and error, we find that they are 3 and 4.
• Thus, the factorization is:
x2
+ 7x + 12 = (x + 3)(x + 4)
17. Factoring ax2
+ bx + c
To factor a trinomial of the form ax2
+ bx + c
with a ≠ 1, we look for factors of the form
px + r and qx + s:
ax2
+ bx + c = (px + r)(qx + s)
= pqx2
+ (ps + qr)x + rs
18. Factoring ax2
+ bx + c
Thus, we try to find numbers p, q, r, and s
such that:
pq = a rs = c ps + qr = b
• If these numbers are all integers, then we will
have a limited number of possibilities to try for
p, q, r, and s.
19. E.g. 4—Factoring ax2
+ bx + c by Trial and Error
Factor: 6x2
+ 7x – 5
• We can factor:
6 as 6 ∙ 1 or 3 ∙ 2
–5 as –25 ∙ 1 or –5 ∙ (–1)
• By trying these possibilities, we arrive at
the factorization
6x2
+ 7x – 5 = (3x + 5)(2x – 1)
20. E.g. 5—Recognizing the Form of an Expression
Factor each expression.
(a) x2
– 2x – 3
(b) (5a + 1)2
– 2(5a + 1) – 3
22. E.g. 5—Recognizing the Form
This expression is of the form
__2
– 2__ – 3
where __ represents 5a + 1.
Example (b)
23. E.g. 5—Recognizing the Form
This is the same form as the expression
in part (a).
So, it will factor as ( __ – 3)( __ + 1).
• (5a + 1)2
– 2(5a + 1) – 3
= [(5a + 1) – 3][(5a + 1) + 1]
= (5a – 2)(5a + 2)
Example (b)
25. Special Factoring Formulas
Some special algebraic expressions can
be factored using the following formulas.
• The first three are simply Special Product
Formulas written backward.
27. E.g. 6—Differences of Squares
Using the Difference of Squares Formula
with A = 2x and B = 5, we have:
4x2
– 25 = (2x)2
– 52
= (2x – 5)(2x + 5)
Example (a)
28. E.g. 6—Differences of Squares
We use the Difference of Squares Formula
with A = x + y and B = z.
(x + y)2
– z2
= (x + y – z)(x + y + z)
Example (b)
30. E.g. 7—Difference of Cubes
Using the Difference of Cubes Formula
with A = 3x and B = 1, we get:
27x3
– 1 = (3x)3
– 13
= (3x – 1)[ (3x)2
+ (3x)(1) + 12
]
= (3x – 1)(9x2
+ 3x + 1)
Example (a)
31. E.g. 7—Sum of Cubes
Using the Sum of Cubes Formula with
A = x2
and B = 2, we have:
x6
+ 8 = (x2
)3
+ 23
= (x2
+ 2)(x4
– 2x2
+ 4)
Example (b)
32. Perfect Square
A trinomial is a perfect square if it is
of the form
A2
+ 2AB + B2
or A2
– 2AB + B2
• So, we recognize a perfect square if the middle
term (2AB or –2AB) is plus or minus twice the
product of the square roots of the outer two terms.
34. E.g. 8—Perfect Squares
Here, A = x and B = 3.
So, 2AB = 2 .
x .
3 = 6x.
• Since the middle term is 6x, the trinomial is
a perfect square.
• By the Perfect Square Formula, we have:
x2
+ 6x + 9 = (x + 3)2
Example (a)
35. E.g. 8—Perfect Squares
Here, A = 2x and B = y.
So, 2AB = 2 .
2x .
y = 4xy.
• Since the middle term is –4xy, the trinomial is
a perfect square.
• By the Perfect Square Formula, we have:
4x2
– 4xy + y2
= (2x – y)2
Example (b)
37. Factoring an Expression Completely
When we factor an expression, the result
can sometimes be factored further.
In general,
• We first factor out common factors.
• Then, we inspect the result to see if it can be factored
by any of the other methods of this section.
• We repeat this process until we have factored
the expression completely.
38. E.g. 9—Factoring an Expression Completely
Factor each expression completely.
(a) 2x4
– 8x2
(b) x5
y2
– xy6
39. E.g. 9—Factoring Completely
We first factor out the power of x with
the smallest exponent.
2x4
– 8x2
= 2x2
(x2
– 4) (Common factor is 2x2
)
= 2x2
(x – 2)(x + 2) (Factor x2
– 4 as a difference
of squares)
Example (a)
40. E.g. 9—Factoring Completely
We first factor out the powers of x and y
with the smallest exponents.
x5
y2
– xy6
= xy2
(x4
– y4
) (Common factor is xy2
)
= xy2
(x2
+ y2
)(x2
– y2
) (Factor x4
– y4
as
a difference of squares)
= xy2
(x2
+ y2
)(x + y)(x – y) (Factor x2
– y2
as
a difference of squares)
Example (b)
41. Factoring Expressions with Fractional Exponents
In the next example, we factor out
variables with fractional exponents.
• This type of factoring occurs in calculus.
42. E.g. 10—Factoring with Fractional Exponents
Factor each expression.
(a)3x3/2
– 9x1/2
+ 6x–1/2
(b)(2 + x)–2/3
x + (2 + x)1/3
43. E.g. 10—Fractional Exponents
Factor out the power of x with the smallest
exponent—that is, x–1/2
.
3x3/2
– 9x1/2
+ 6x–1/2
= 3x–1/2
(x2
– 3x + 2) (Factor out 3x–1/2
)
= 3x–1/2
(x – 1)(x – 2) (Factor the quadratic
x2
– 3x + 2)
Example (a)
44. E.g. 10—Fractional Exponents
Factor out the power of 2 + x with
the smallest exponent—that is, (2 + x)–2/3
.
(2 + x)–2/3
x + (2 + x)1/3
= (2 + x)–2/3
[x + (2 + x)] (Factor out (2 + x)–2/3
)
= (2 + x)–2/3
(2 + 2x) (Simplify)
= 2(2 + x)–2/3
(1 + x) (Factor out 2)
Example (b)
46. Factoring by Grouping
Polynomials with at least four terms
can sometimes be factored by grouping
terms.
• The following example illustrates the idea.
47. E.g. 11—Factoring by Grouping
Factor each polynomial.
(a) x3
+ x2
+4x + 4
(b) x3
– 2x2
– 3x + 6
48. E.g. 11—Factoring by Grouping
x3
+ x2
+4x + 4
= (x3
+ x2
) + (4x + 4) (Group terms)
= x2
(x + 1) + 4(x + 1)(Factor out common
factors)
= (x2
+ 4)(x + 1) (Factor out x + 1 from
each term)
Example (a)
49. E.g. 11—Factoring by Grouping
x3
– 2x2
– 3x + 6
= (x3
– 2x2
) – (3x – 6) (Group terms)
= x2
(x – 2) – 3(x – 2)(Factor out common
factors)
= (x2
– 3)(x – 2) (Factor out x – 2 from
each term)
Example (b)
