Download how to factor_1

Always work in the following
order… Square
2) Perfect Trinomial
3) Quadratic functions
4) Difference of two Squares…
5) By grouping
Look for…
 A constant,
 A variable,
 Or a combination of both.
Ex 1: Factor 7x+7 ……
7( x  1)
Ex 2: Factor 3x+x² …..
x(3  x)
Ex 3: Factor 2x² - 8x…
2 x ( x  4)



Is the first term a perfect square?
Is the last term a perfect square?
Is the middle term…
2·
1st term • 3rd term
=2 ·x ·4 ?
If so, it factors as…
Ex 1: x² +8x+ 16 = (x + 4)2
Use the middle sign
Ex 2: x² -8x + 16 = (x - 4)2
Use the middle sign

Ex: x 2 + 10x + 16
Possible Products
16 = 1 x 16
=2 x 8
=4x4
But Middle term must add up to +10!
2 +8=10
Factors as : (x + 2)(x + 8)

Ex: x 2 - 10x + 16
Possible Products
16 = 1 x 16
=2x8
=4x4
But Middle term must add up to -10!
-2 + -8= -10
Factors as : (x - 2)(x - 8)

Ex: 2x 2 - 5x – 12
Multiply (2)(-12) = -24
Possible Factors:
-24
= -1 x 24
= -2 x 12
= 3x 8
= -4 x 6
Yet… the Middle term is (- 5x)
so, we need more negatives!
3x + - 8x and…




2x 2 - 5x - 12
2x 2 + 3x + -8x -12
(2x 2 + 3x) + (-8x-12)
x(2x+3) – 4(2x+3)
(2x+3) (x-4)
Original
Bust Middle.
Group.
Factor GCF.
(Write what they have in common), then (write
what’s left).


Ex: 6x 2 - 8x – 8
Multiply (6)(-8) = -48
Possible Factors:
-48
= -1 x 48
= -2 x 24
= -3 x16
= 4 x 12
= -6 x 8
Yet…the middle term is(- 8x) ,so we need
more negatives!
-12x + 4x so…




6x 2 - 8x - 8
6x 2 - 12x + 4x -8
(6x 2 -12x) + (4x-8)
6x(x-2) + 4(x-2)
(x-2) (6x+4)
Original
Bust Middle.
Group.
Factor GCF.
(Write what they have in common), then (write
what’s left).




Is the first term a perfect square?
Is the second term a perfect square?
Is there a minus between the two terms?
Are there only two terms?...if yes
Factor as:
( 1st term  2nd term)( 1st term  2nd term)
( x 2 + 16 )(
Ex: x 2 - 16 =
x
– 16 )
(x+4)(x-4)
2
a 3- b
Factor:
3
3
= (a - b)(a 2+ ab + b 2)
64x
3
- 27
3
64 x  4 x
3
27  3
Factors as…
(4 x  3)(16 x  12 x  9)
2
a 3+ b
Factor:
3
3
= (a + b)(a 2 - ab + b 2)
64x
3
+ 27
3
64 x  4 x
3
27  3
Factors as…
(4 x  3)(16 x - 12 x  9)
2
a
4-
b
4
= (a
Factor…
4
)  (b ) =(a 2 - b 2)(a 2+ b 2)
2 2
2 2
= (a - b)(a + b)(a 2+ b 2)
x  16
4
x x
4
4
Factors as…
16  2
( x  2)( x  2)( x  4)
2

Used when there are 4 or more terms.
Ex: 2ab + 14a + b + 7
(2ab +14a) + (b + 7)
Group
2a(b+7) + 1(b+7)
(b+7) (2a+1)
factor, also 1
…
(Write what they have in common), then
(write what’s left).

Ex: xy + 3x – y²- 3y
(xy + 3x) + (-y²-3y)… Always use a plus sign
between parentheses,

x(y+3) - y(y+3)
(y+3) (x-y)
then factor GCF
(Write what they have in common), then (write
what’s left).
x 2  x  30  0
Try to factor…
His implies
that either
expression
must be 0
( x  6)( x  5)  0
x 6  0  x  6
x  5  0  x  5
There are TWO solutions:
x=6; x=-5

Solve by substitution.
4 x  13x  3  0
4
2
Factoring
Using Synthetic Division
Synthetic Division
We can use this method as long as you know one of the zero’s,
x  4 x  5x  2, one of the zero' s is x  2
3
2
x  4 x  5x  2
Place the zero’s value
3
2
of x here.
2
1
-4
5 -2
Add these
upbelow
Bring first
these
line
up these up
2 number
- 4 downAdd
2 Add
Multiply
Multiply
these and
these and
put answer
This is the remainder
put answer
above line in
above line in
next column
next column
2
List all coefficients (numbers in front of x's) and the constant along
the top. If a term is missing, put in a 0.
Put variables back in (one x was divided out in the process, so
2
the first term is x²).
Is the factored form
1
-2
1
( x  2)( x  2 x  1)
( x  2)( x  1)
0
You Try to factor!
4 x 3  8 x 2  25 x  50
Given one of its factors is x  2
set the factor = 0 and
solve for x, this will be
your divisor.
-2
4
8
-25 -50
Add these
upbelow
Bring
these
line
up these up
- 8first number
0downAdd
50Add
Multiply
Multiply
these and
these and
No remainder so x + 2 IS a
put answer
2
put answer
factor because it divided in
above line in
above line in
evenly
next column
next column
2
So the
Listanswer
all coefficients
is the divisor
(numbers
timesinthe
front
quotient:
of x's) and the constant along
back in (one
x was
the top.Put
If variables
a term is missing,
put in
a 0. divided out in the process, so
the first term is x²).
You could factor even further…
4x +0 x - 25
0
x  24 x
 25
x  22x  5(2x  5)

You can use synthetic division to divide
polynomials
Set divisor = 0 and
solve. Put answer here.
1 x  6 x  8x  2
3
2
x3
-3
1
x + 3 = 0 so x = - 3
6
8
-2
Add these
upbelow
Bring
these
line
up these up
- 3first number
- 9 downAdd
3 Add
Multiply
Multiply
these and
these and
put answer
2
This is the remainder
put answer
above line in
above line in
next column
next column
List all 
coefficients
(numbers in1
front of x's) and the constant along
2
x

3
x

1


Is the quotient
the top. If a term is missing, x
putin3a0.
Put variables back in (one x was divided out in the process, so
the first term is x²).
1x +3 x - 1
1
CAVEAT……
Set divisor = 0 and
solve. Put answer here.
Find the quotient
0
0x
4 3
2
x
1 x  4x  6
x4
4
1
x - 4 = 0 so x = 4
0
-4
0
6
Add these
upbelow
Bring4
first number
theseAdd
line
up these Add
up these up
16downAdd
48 192
Multiply
Multiply
these
and
Multiply
these and
put
answer
these
and
This is the
3
2
put
answer
above
line in
put answer
remainder
above line in
next column
above
line in
next column
198 
 3
2
next column
x

4
x

12
x

48

List all coefficients
(numbers
in
front
of
x's) and the constant along

x terms.
4
 forget the 0's for missing
the top. Don't
Now put variables back in (remember one x was divided out in
the process
so first2term198
is x3). Is the quotient
2
1x +4 x +12 x + 48 198
x

 12 ( x  4) 
x4