Transcript Factoring Trinomials—with a coefficient of 1 for the squared term

Section 5.4 Factoring
FACTORING
Greatest Common Factor,
Factor by Grouping,
Factoring Trinomials,
Difference of Squares,
Perfect Square Trinomial,
Sum & Difference of Cubes
1
Factoring—define factored form
Factor means to write a quantity as a
multiplication problem
 a product of the factors.
 Factored forms of 18 are:

1 18, 2 9, 3 6
2
Factoring: The Greatest Common
Factor

1.
2.
3.

To find the greatest common factor of a list
of numbers:
Write each number in prime factored form
Choose the least amount of each prime that
occurs in each number
24  2 2 2 3
Multiply them together
Find the GCF of 24 & 36
36  2 2 3 3
2 2 3  12
3
Factoring: The Greatest Common
Factor

1.
2.
3.

To find the greatest common factor of a list
of variable terms:
Choose the variables common to each
term.
Choose the smallest exponent of each
common variable.
Multiply the variables.
Find the GCF of:
x 4 y 2 , x 7 y , x3 y 5 , x 2 y 6 
4
x2 y
Factoring: The Greatest Common
Factor

1.
2.
3.
5
To factor out the greatest common factor
of a polynomial:
Choose the greatest common factor for
the coefficients.
Choose the greatest common factor for
the variable parts.
Multiply the factors.
Factoring: The Greatest Common
Factor
Factor
each of the
following
by factoring
out the
greatest
common
factor:

6
5x + 5 =
4ab + 10a2 =
8p4q3 + 6p3q2 =
2y + 4y2 + 16y3 =
3x(y + 2) -1(y + 2) =
Factoring: The Greatest Common
Factor

The answers are :
5 x  5  5  x  1
4ab  10a 2  2a  2b  5a 
8 p 4 q 3  6 p 3 q 2   2 p 3 q 2   4 pq  3
2 y  4 y 2  16 y 3  2 y 1  2 y  8 y 2 
3x y  2   1 y  2    y  2  3 x  1
7
Factoring: by Grouping

1.
2.
3.

8
Often used when factoring four terms.
Organize the terms in two groups of two
terms.
Factor out the greatest common factor from
each group of two terms.
Factor out the common binomial factor from
the two groups.
Rearranging the terms may be necessary.
Factoring: by Grouping

Factor by grouping:
1.
2 groups of 2 terms
Factor out the GCF
from each group of 2 terms
2n  m  4   3  m  4 
Factor out the
common binomial factor
 m  4  2n  3
2.
3.
9
2mn  8n  3m  12
Factoring: by Grouping
Factor
by
grouping
6 y 2  20w2  15 yw  8 yw


2 3 y 2  10w2  ????????
rearrange the terms and try again
6 y 2  8 yw  15 yw  20w2
2 y  3 y  4 w   5w  3 y  4 w 
3 y  4w 2 y  5w
10
Factoring Trinomials—with a
coefficient of 1 for the squared term

1.
2.
3.
Factor: x  12 y  20
List the factors of 20:
Select the pairs from which 12
may be obtained
Write the two
 x  10  x  2 
binomial factors:
2
x 2  2 x  10 x  20
4.
11
Check using FOIL:
x  12 x  20
2
20
1 20
2 10
45
Factoring Trinomials
TIP
 If the last term of the trinomial is
positive and the middle sign is positive,
both binomials will have the same
“middle” sign as the second term.
x  12 x  20
2
 x  10  x  2 
12
Factoring Trinomials
TIP
 If the last term of the trinomial is
positive and the middle sign is negative,
both binomials will have the same
“middle” sign as the second term.
x  12 x  20
2
 x  10  x  2 
13
Factoring Trinomials—with a
coefficient of 1 for the squared term

1.
2.
3.
4.
14
Factor a  9a  22
22
List the factors of 22
1 22
Select the pair from
which –9 may be obtained
2 11
Write the two
x  11 x  2 

binomial factors:
2
Check using FOIL:
x 2  2 x  11x  22
x 2  9 x  22
Factoring Trinomials TIP
 If the last term of the trinomial is
negative, both binomials will have one
plus and one minus “middle” sign.
x  9 x  22
2
 x  11 x  2
15
Factoring Trinomials—primes
A PRIME POLYNOMIAL cannot be
factored using only integer factors.
2
 Factor : x  2 x  5
 The factors of 5: 1 and 5.
 Since –2 cannot be obtained from 1 and
5, the polynomial is prime.

16
Factoring Trinomials—2 variables
y 2  6 yz  8z 2
Factor:
 The factors of 8 are: 1,8 & 2,4, & -1,-8 & -2, -4
 Choose the pairs from which
–6 can be obtained: 2 & 4
 Use y in the first
y  4z y  2z
position and z in the
second position
2
2
y  2 yz  4 yz  8 z
 Write the two binomial
factors and
2
2
y

6
yz

8
z
check your answer


17


Factoring Trinomials—with a GCF


If there is a greatest common factor?
If yes, factor it out first.
Factor: 3z 4  15 z 3  18 z 2
3z 2  z 2  5 z  6  factors of 6: 1 6 & 2 3
3z
2
 z   z  
Choose 2 3
3z 2  z  2  z  3
18
Factoring Trinomials—always
check your factored form

Always
check your
answer with
multiplication
of the factors.

The check:
3z
2
3z
2
3z 2
 z  2  z  3
z
z
2
 3z  2 z  6 
2
 5z  6
3 z  15 z  18 z
4
19
3
3
Factoring Trinomials—
when the coefficient is not 1 on the
squared term

Factor: 3 x 2  4 x  1
1. Multiply 3 1  3
2. List the factors of 3: 1 3
3. Rewrite the middle term of the trinomial using
the factors of 3.
3x 2  x  3x  1
4. Factor by grouping.
x  3x  1  1 3 x  1 
20
 x  1 3x  1
Factoring Trinomials---use
grouping

Factor: 6x 2  x  1
1. Multiply 6 1  6
2. List the factors of 6: 1 6 & 2 3
3. Rewrite the middle term of the trinomial using the
factors of 6 which add to be 1 ie. 2 3.
6 x 2  3x  2 x  1
4. Factor by grouping.
3x  2 x  1  1 2 x  1 
21
 2 x  1 3x  1
Factoring Trinomials---use
grouping

Factor: 12x 2  5 x  2
1. Multiply 12 2  24
2. List the factors of 24: 1 24; 2 12; 3 8; & 4 6
3. Rewrite the middle term of the trinomial using
the factors of 24 which add to be 5 ie. 3 8.
12 x 2  8 x  3 x  2
4. Factor by grouping.
4 x  3 x  2   1 3 x  2  
22
 3x  2  4 x  1
Factoring Trinomials---use FOIL
and Trial and Error

Factor: 6 x 2  19 x  10
There will be two binomial factors.

Both middle signs will be positive.
The factors of 6 are 1  6 and 2  3.
The factors of 10 are 1  10 and 2  5.
(Continued on next screen.)
23







Factoring Trinomials---use FOIL
and Trial and Error
 Think
FOIL and focus on the outers and inners
while using trial and error to position the
factors correctly.
 2x  5  3x  2 
inners: 15x
outers: 4x
Always check your answer.
24
2
2
x

5
3
x

2

6
x
 19 x  10



Factoring Trinomials---use FOIL
and Trial and Error

Factor: 10x 2  23 x  12
There will be two binomial factors.
Both middle signs will be negative.




The factors of 10 are 1  10 and 2  5.
The factors of 12 are 1  12, 2  6, and 3  4.
(Continued on next screen.)
25




Factoring Trinomials---use FOIL
and Trial and Error
 Think
FOIL and focus on the outers and inners
while using trial and error to position the
 2x  3 5 x  4 
factors correctly.
inners:  15x
outers:  8 x
Always check your answer.
26
 2x  3 5 x  4   10 x
2
 23 x  12
Factoring Trinomials---use FOIL
and Trial and Error
2
Factor:
5
x
 13x  6

There will be two binomial factors.
One middle sign will be negative.
One middle sign will be positive.
The factors of 5 are 1  5.
The factors of 6 are 1  6, and 2  3.
(Continued on next screen.)
27








Factoring Trinomials---use FOIL
and Trial and Error

Think FOIL and focus on the outers and inners
while using trial and error to position the
factors correctly.
 5 x  2  x  3
inners:  2 x
outers: 15x
Always check your answer.
28
2
5
x

2
x

3

5
x
 13x  6



Factoring Trinomials---with a
negative GCF



Is the squared term negative?
If yes, factor our a negative GCF.
Factor:  4a 2  2a  30
 2a  2a 2  a  15 
 2a  2a  5  a  3
The sum of inners: 5a and outers:  6a is  a
29
Special Factoring—difference of 2
squares

The following must be true:
1.
There must be only two terms in the
polynomial.
Both terms must be perfect squares.
There must be a “minus” sign between
the two terms.
2.
3.
30
Special Factoring—difference of 2
squares
 The
following pattern holds true
for the difference of 2 squares:

31
x  y   x  y  x  y 
2
2
Special Factoring—difference of 2
squares


The pattern: x2  y 2   x  y  x  y 
Factor: x 2  25
x 2 is a perfect square since
x2  x
25 is a perfect square since 25  5
Use the pattern letting y  5.
x   5    x  5  x  5 
2
32
2
Special Factoring—difference of 2
squares


The pattern: x2  y 2   x  y  x  y 
Factor: a  b
2
2
a 2 is a perfect square since a 2  a
b 2 is a perfect square since b 2  b
Use the pattern letting x  a & y  b.
a  b   a  b  a  b 
2
33
2
Special Factoring—difference of 2
squares


The pattern: x2  y 2   x  y  x  y 
Factor: 9 x  4 y
2
2
9 x 2 is a perfect square since 9 x 2  3 x
4 y 2 is a perfect square since 4 y 2  2 y
Use the pattern letting x  3x & y  2 y.
 3x    2 y 
2
34
2
  3x  2 y  3 x  2 y 
Special Factoring—difference of 2
squares


The pattern: x2  y 2   x  y  x  y 
Factor: 9 x  4 y
Although both terms are perfect squares,
2
2
the pattern does not apply because the
problem is a sum and not a difference.
35
Special Factoring—perfect square
trinomial

A perfect square trinomial is a trinomial
that is the square of a binomial.

4 x 2  4 xy  y 2 is a perfect square trinomial
because:
 2 x  y    2 x  y  2 x  y  
2
2
2
2
 2 x   2 xy  2 xy  y  4 x  4 xy  y
2
36
Special Factoring—perfect square
trinomial
 4 x2




37
 4 xy  y 2 is a perfect square trinomial
because:
The first and third terms are perfect squares.
AND the middle term is twice the product of the
square roots of the first and third terms
TEST THE MIDDLE TERM:
4x  2x
2
y  y 2  2 x  y   4 xy
2
Special Factoring—perfect square
trinomial

The patterns for a perfect square
trinomial are:
x  2 xy  y   x  y  x  y    x  y 
2
2
x  2 xy  y   x  y  x  y    x  y 
2
38
2
2
2
Special Factoring—perfect square
trinomial

Factor the following using the perfect square
trinomial pattern:
16a 2  56a  49   4a  7 
2
The first and third terms are perfect squares.
16a 2  4a
49  7
The middle term is twice the product of the
square root of the first and third terms.
2  4a  7   56a
39
Special Factoring—perfect square
trinomial

Factor the following using the perfect square
trinomial pattern:
x  22 xz  121z   x  11z 
2
2
2
The first and third terms are perfect squares.
x2  x
121z 2  11z
The middle term is twice the product of the
square root of the first and third terms.
2  x 11z   22 xz
40
Special Factoring—difference of
3
3
2
2
x

y

x

y
x

xy

y


two cubes


Factor using the pattern.
27 x 3  8 y 3
Both terms must be perfect cubes.
3
27x 3  3 x;
3
8y 3  2 y
 3x    2 y  Let x  3x, Let y  2 y
2
2
 3x  2 y    3x    3x  2 y    2 y  
3
3
2
2
3
x

2
y
9
x

6
xy

4
y



41
Special Factoring—sum of two
cubes x3  y 3   x  y   x 2  xy  y 2 

Factor using the pattern.
1000 x3  125 y 3
Both terms must be perfect cubes.
3
1000 x3  10 x;
3
125 y 3  5 y
10 x    5 y  Let x  10 x, Let y  5 y
2
2
10
x

5
y
10
x

10
x
5
y

5
y

         
3
3
10 x  5 y  100 x 2  50 xy  25 y 2 
42
Solving quadratic equation with
factoring
 A quadratic
equation has a
“squared” term.
ax  bx  c  0
2
43
ZERO FACTOR PROPERTY
To Factor a Quadratic,
Apply the Zero-Factor Property.
 If
a and b are real numbers and
if ab = 0, then a = 0 or b = 0.
44
Solving quadratic equations with
factoring—Zero-Factor Property

Solve the equation:
(x + 2)(x - 8) = 0.

Apply the zero-factor property:
(x + 2) = 0
x = -2
45
or
or
(x – 8) = 0
x=8
Solving quadratic equations with
factoring—Zero-Factor Property
There are two answers for x:
-2 and 8.
 Check by substituting the values
calculated for x into the original equation.
(x + 2)(x - 8) = 0.
(-2 + 2)(-2 – 8) = 0 (8 + 2)(8 – 8) = 0
0=0
0=0

46
Solving quadratic equations with
factoring—Standard Form

1.
To solve a quadratic equation,
Write the equation in standard form.
 (Solve the equation for 0.)
x  6 x  8  x  6 x  8  0
2
47
2
Solving quadratic equations with
factoring

2.
To solve a quadratic equation,
Factor the quadratic expression.
x  6x  8  0
2
 x  2  x  4   0
48
Solving quadratic equations with
factoring

3.
To solve a quadratic equation,
Apply the Zero-Factor Property
x  6x  8  0
2
 x  2  x  4   0
x20 x40
x  2
49
x  4
Solving quadratic equations with
factoring

4.
To solve a quadratic equation,
Check your answers
x2  6 x  8  0
x  2, or x  4
 2
 4
2
 6  2  8  0
4  12  8  0
8  8  0
00
50
2
 6  4   8  0
16  24  8  0
88  0
00