Factoring can be a difficult skill to
acquire. However, it can be mastered with pattern recognition and
practice. (Sections marked advanced are intended for algebra 2
students)
First of all, integer squares and cubes
along with their square and cube roots need to be memorized, cold.
You cannot be sidetracked 'figuring out' the cube root of 27,
for example, nor do you need to be calculating the square of 3. This
will become apparent later.
Here are the squares and the cubes of
the numbers from 1 to 15,
memorize them. (cubes > 10 are
seldom needed)
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x
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1
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2
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3
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4
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5
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6
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7
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8
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9
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10
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11
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12
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13
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14
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15
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x2
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1
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4
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9
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16
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25
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36
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49
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64
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81
|
100
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121
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144
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169
|
196
|
225
|
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x3
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1
|
8
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27
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64
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125
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216
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343
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512
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729
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1000
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|
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Knowing the multiplication
and addition tables cold will also help considerably. (from 1 to 15,
both tables) If you don't know them, go back and memorize them
before you proceed. As a side note, apparently the trend in
grade school math is to NOT memorize these basic facts, but to
'figure them out' as needed. The goal may be to foster a creative,
intuitive approach to problem solving. I doubt there is but little
anecdotal evidence to support this no matter how good it sounds in
theory. Doing this will for sure impede the development of factoring
skills. In the end, as a parent it's up to you.
Factoring literally means to
find the factors of an arithmetic expression that when multiplied
together results in that expression. For a polynomial, the factors
are used to find the zeroes of the polynomial. These zeroes help
graphing the polynomial, and we graph the polynomial to better
understand how two variables from a real world problem vary
with respect to each other. Graphing polynomials is covered in
graphing polynomials.
First let's review the
greatest common
divisor.
Another name for this
divisor is greatest common factor.
Consider the numbers 6 and
15.
6
= 3
· 2
and 15 = 3
· 5
The
greatest common factor of 6 and 15 is 3.
Now here's the key.
If
we add 6 to 15 we get 21.
(
6 + 15 ) = 21 but 3 · 7 = 21 as well.
Now
go back four lines, do you see the 2 and
the 5? They add up to 7.
In
other words 3 · ( 2 + 5 ) = 21. We've factored the 3 out of
the sum
(6
+ 15). We now apply this numerical example to expressions with
variables.
All
factors imply products and/or quotients.
x
and y
are factors of the product xy.
x
and
are
factors of the
quotient
.
x
and y
are not
factors of the sum (x
+ y) nor
the quotient
.
x
is a factor of
since this is the product x
·
.
is
a factor of
since
this is the product
·
(x+y) .
In
both of these last two examples
and (x
+ y) are, respectively,
factors as well. Take another look at this. We have the sum
(x + y) as a factor.
Remember, the parenthesis makes this sum a
term.
We use the distributive
property of multiplication over addition/subtraction as the main
pattern for factoring:
a(b + c) = ab + ac
now in reverse:
ab + ac →
a( b + c )
we say
that ab + ac has been factored into a(b + c )
a
is one factor and the sum (b + c) is the other factor.
another
look: ab + ac
The
common pattern is a in both
terms ab and ac,
so we
may "pull out" the a
from both terms and get a(b + c).
Stop
here until you fully understand the pattern above.
Review the previous
numerical example now 6 + 15 = 3(2 + 5).
Continuing,
the
expression above is factored when a
is the greatest common factor.
Now some
examples:
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Example 1
factor 4x + 4y
4(x + y)
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4 is the common factor
of these two terms, to check, multiply
4(x + y) = 4x + 4y √
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Example 2
factor
4x + 8y
4x
+ 4 • 2y
4( x + 2y )
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Look for factors in
these terms that are related. 4 and 8 are related, 4 • 2 =
8.
The common pattern is
4 in
these terms.
check: 4( x + 2y ) =
4x + 4 • 2y
= 4x + 8y √
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Example 3
(advanced)
factor 3axy2
- 27a2y
3axy2
- 27a2y
3axy2
- 3 • 9 a2y
3(
axy2 - 9a2y )
3(
axy2 - 9a2y
)
3(
a ( xy2 - 9ay
) )
3a
(
xy2 - 9ay )
3a
(
xy2
- 9ay )
3a
(
y(
xy - 9a
) )
3a
y(
xy - 9a
)
3ay(
xy - 9a )
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As expressions become
more complex look for common factors one at a time.
In this case notice
that 3 and 27 are related, 3 • 9 = 27.
Factor out this 3.
We can do better, note
a is common.
a2 = a ·
a
Pull out this a.
Now remove outer
parenthesis.
We can still do
better, y is common to both
terms.
y2 = y ·
y
Pull out the y.
Remove the outer
parenthesis.
There are no other
common patterns inside the parenthesis, so we're done. Since we
cannot factor any further, these two factors , 3ay and (xy - 9a)
are the prime factors of the original expression.
check:
3ay(
xy - 9a ) =
3ay xy - 3ay 9a =
3axy2
-
27a2y
√
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|
Example 3 revisited
factor
3axy2
- 27a2y
3
( )
3a
( )
3ay(
xy - 9a )
Write down the 3, then
the a , then the y leaving the factored out difference in the
parenthesis.
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The above solution can
be simplified by writing parenthesis as shown and pulling out
each greatest common factor, one at a time, and writing them in
front of the parenthesis and leaving what remains behind.
This is done in one step. The four steps shown are
only to explain the technique. It is best to do the previous one
step at a time approach until it becomes natural to use this
one-step technique.
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Special
binomial products:
Recall
the special binomial products:
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1) (a
+
b)2
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a2
+
2ab + b2
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2) (a + b)(a − b)
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a2
− b2
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3) (a −
b)2 = (a − b)(a − b)
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a2
− 2ab + b2
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4) (a − b) = −
( b − a)
(a − b
) = − (− a
+ b)
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This is a special
pattern involving the distribution of −1 across a binomial.
Learn it cold, it is used very often.
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RULE
4 the 'sign change' rule.
You can always 'pull' out a −1
from any expression changing the sign of each term within the
expression and reorder them as needed.
(a + b) = −1( −a −
b ) = −1( −b − a ) The −1 is usually
implied like so:
(a + b) = −( −a −
b ) = −( −b − a )
(a − b) = −( −a
+ b ) = −( b − a )
Multiply the following to
convince yourself:
−1(
−a − b ) =
−1(
−b − a ) =
−1(
−a + b ) =
−1(
b − a ) =
example why we do this
factor 2x − y + 2xy
− 4x2
2x − y + (2xy
− 2x · 2x)
2x − y + 2x(y
− 2x)
(2x − y) + 2x(−1)(2x
− y) ← sign change rule applied here
1(2x
− y) − 2x(2x
− y)
(1
− 2x ) (2x − y)
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The
patterns to recognize are the ones in the second column. (Compare the
patterns with the red + and - signs.)
Some
more examples:
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Example 4
factor x2 -
y2
immediately using
pattern 2
we have
(x + y)(x - y)
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"The difference of two perfect squares is
the product of the sum and the difference of their roots."
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Example 5
factor 36y2
- 9x2
(6y
+ 3x)(6y - 3x)
Notice all four terms
have something in common, namely 3,
so we can factor out
the 3 from both factors like so:
(
3(2y + x) ) (
3(2y - x) )
3·3 ·
(2y + x) ( 2y - x)
9(2y+x)(2y-x)
factor: 36y2
- 9x2
9(4y2 - x2)
apply pattern 3
directly:
9(2y + x)(2y - x)
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The square root of 36
is 6, the square root of 9 is 3, so immediately we use pattern 2.
Remember, since we
have a product the
exponent can be distributed, that
is
36y2 =
62 y2
= (6y)2
Now factor within the
grouping symbols (shown in red above.)
Remove the red
grouping symbols and rearrange all 4 factors, the two 3s and the
2 binomials.
Another solution:
(shorter)
9 is common to both,
so we factor it out first.
Pattern 2 requires the
difference of two squares. 4y2 is the square of 2y
and x2 is the square of x.
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Example 4
factor x2 +
2xy + y2
(x + y)(x + y)
or (x + y)2 is ok for an answer
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This expression matches pattern 1 exactly (a=x,
b=y)
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Example 6
factor 4x2
+ 4xy + y2
is 4x2 a
perfect square? yes 2x is its root
now, 4xy = 2(2x)y
see the pattern?
(2x)2
+ 2(2x)y
+ (y)2
the pattern:
a2
+ 2 a
b + b2
a is 2x and b is y so
we substitute into pattern 1 and get:
(2x + y)2
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This example is
typical of factoring problems using these patterns.
strategy:
1) check if you have
perfect squares to replace a and b in the pattern.
if not, you cannot
uses these patterns.
2) now see if the
square roots multiplied by 2 is the middle term.
If so, directly
factor.
Stop here until you fully understand this example.
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Example 7
factor
x2 ― 8xy +16y2
notice
the middle term is negative so
using
pattern 3:
a2
― 2
a b
+ b2
x2
― 2 x(4y)
+ (4y)2
(x ―
4y) 2
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As in the previous
example
check for perfect
squares:
x2 = (x)2
and 16y2 = (4y)2
now check the middle
term,
8xy = 2 (x · 4y)
so a = x and b = 4y
and using
pattern 3 we get the
answer
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Example 8
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Study this example, it
is heavily used in advanced factoring problems.
Use pattern 4 in the
denominator.
Cancel the like term
(x ― y)
and get -1.
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Factoring
by regrouping terms: (advanced)
Sometimes
you need to group terms to factor, consider:
(a +
b)(c + d) = ac + ad + bc + bd
Now, in
these cases, it doesn't matter how you group the terms, you'll end up
with the same result.
For
example: (ac + ad) + (bc + bd)
factor
an a out of the first sum then a b out of the second sum and get
a(c
+ d) + b (c + d )
Notice
now that each product has a common factor (c + d),
factoring
that out we get
(c
+ d) (a + b)
Think of the (c + d) as
one letter say k and what we have is
a · k
+ b · k = k(
a + b ) = (c + d)(a
+ b)
Let's look at this a different way
Factor: ac + ad + bc + bd
Let's group the terms with a common c and a
common d
(ac + bc)
+ (ad +
bd)
In the first sum we can factor out a c and in
the second sum we can factor out a d like so
c ( a + b )
+ d ( a + b )
Now (a + b) is a factor of both terms, we can
factor it out and get
( a + b ) ( c + d )
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Example 9
factor:
xw - vw + xz - vz
w(
x - v ) + z ( x - v )
( x - v ) ( w + z )
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W
is common to the 1st and 2nd term and z is
common to the 3rd and 4th terms.
(x-v) is common to both
terms so pull it out.
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Example 9
factor: sq – 2sp
+ 5q – 10p
s(
q – 2p ) + 5 ( q – 2p )
( q – 2p
) (s + 5)
(solution 2)
sq
+ 5q - 2sp
– 10p
q(s
+ 5) – 2p(s + 5)
q(s
+ 5) – 2p(s
+ 5)
(s + 5) ( q – 2p )
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Look for the patterns:
The first two terms
have a common s, and the second terms have a common 5, let's give
this a shot.
(q - 2p) is common now
so pull it out.
We can also do this,
group the terms with q together and the terms with s together. Be
careful here, factoring out a -2 from the last two terms changes
the last sign to a +. Now factor out the common (s + 5).
A closer look at: ― 2sp ―10p
Multiply the following
-1
( 2sp + 10p) = -1(2sp)
+ -1(10p)
=
-2sp
- 10p
look
again:
-1(
2sp + 10p)
-1(2)(sp
+ 5p) pull out the 2
-2
(p)(s + 5) pull out the p
-2p(s+5)
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Example 10
factor:
x2 +
6x +9 – 9y2
(
x2 – 9y2 ) + ( 6x + 9 )
Applying
rule 2 on the first group:
(x
+ 3y)(x - 3y) + 3(2x + 3)
hmmm....
not getting anywhere, looks worse than what we had, so let's
regroup
x2
+6x + 9 – 9y2
(x
+ 3)2 – 9y2
now
apply rule 2
(
(x + 3) + 3y
) ( (x + 3) – 3y
)
( x + 3y + 3) (
x - 3y + 3 )
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This example takes
factoring one step further.
We have 2 perfect
squares, x2 and 9y2
so try grouping these
two fist. It looks like we've finished but we we still have two
terms, we need a product of terms to be done.
Here's pattern 1.
Factor using this pattern.
Now we have the
difference of 2 perfect squares, see them? In red, (x + 3)2
and in blue, 9y2 = (3y)2
Now we can apply rule
2 with a = (x+3) and b = 3y.
Simplify, listing
variables first.
(At first glance,
maybe a 3 can be factored from each factor, but the x must have a
multiple of 3 as a factor and it does not.)
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Example 11
factor:
4x2 – 2xy – 7yz + 14xz
(4x2
– 2xy ) – (7yz –
14xz)
sign
changed here ^
2(2x2
– xy ) – 7(yz – 2xz)
2x(
2x – y) – 7z( y – 2x)
2x(
2x – y)
– 7z(–1)( 2x – y)
(2x
– y) (2x
+ 7z)
the
answer: (2x – y) (2x + 7z)
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Look
for known patterns. We only have one perfect square, 4x2,
so we cannot use any of our rules. Now group terms with common
factors. The 7 and 14 looks promising and the 4 and 2 look
promising, let's pair those groups.
Watch
the sign changes! (rule 4)!
(–
7yz + 14xy) = –
(7yz – 14xy)
Continue
factoring out common factors.
(rule
4 again! We need to get the same order for the two factors (2x −
y) and (y − 2x) to factor
it out.)
(
− 7z(− 1)) = +7z
We're
almost there. Use rule 4 now to get the common factor (2x –
y)
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Polynomials like Integers have unique prime
factors. However with a polynomial we talk about monomials,
binomials, etc, all of which can be factors of a larger polynomial;
they are prime if they, themselves, cannot be factored further with
all integer coefficients.
A
polynomial is factored completely if its factors are prime.
x + 2 is prime
2x + 4ax is not prime
2( x + b ) has 2 factors, 2 and ( x + b ), both
of which
are prime. So 2x + 2b in this form is factored
completely.
The
sum and difference of cubes: (advanced)
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x3 +
y3
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( x + y)(x2 –
xy + y2)
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x3 – y3
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( x – y)(x2
+ xy + y2)
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Exercise:
Multiply the products on the right as an exercise to convince
yourself that they they equal the sum and difference of cubes on the
left.
Examples:
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Example 12
factor a3 +
b3
(a + b) ( a2 - ab + b2)
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direct substitution into the sum of cubes pattern
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Example 13
factor z3 –
y3
(z – y)
( z2 +
zy + y2)
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direct substitution into the difference of cubes
pattern
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Example 14
factor 125x3
– 729y3
53x3
– 93y3
(5x)3
– (9y)3
(5x
– 9y)
( (5x)2
+
5x9y + (9y)2)
(5x
– 9y)
( 52x2
+45xy
+ 92y2)
(5x –
9y)(25x2
+ 45 xy + 81y2)
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Knowing the cubes from
1 to 15 speeds this process up, considerably, since immediately
we know 53 is 125 and 93 is 729 and we can
proceed with the difference of cubes pattern!
Recall how exponents
distribute.
53x3
=
(5x)3
and
(5x)2
=
52x2
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Example 15
factor
x6 – 1
2
• 3
= 6 the exponent of x, so we can treat this as the difference of
cubes or the difference of squares.
as
difference of cubes:
(x2)3
– 1
(x2
– 1
) ((x2)2 +
x2 +
1)
(x
+ 1) (x – 1)(x4 +
x2 +
1)
as
difference of squares:
(x3)2
– 1
(x3 +
1)(x3
– 1)
(x
+1)(x2
– x
+ 1)(x
– 1)(x2+x
+1)
(x
+1)(x – 1)(x2
– x + 1)(x2
+ x +1)
Technically,
this last answer is correct since it is a product of prime
factors, see to the right. ( x4 + x2 + 1)
can be factored.
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With higher power
expressions try writing them in terms of squares or cubes. In
this example we will do both.
However,
(x4
+
x2
+
1) is not prime.
Both answers are
equivalent.
notice that
(x2
– x
+ 1)(x2+x
+1) =
(x4 +
x2 +
1)
multiplying
the blue and red trinomials:
(x2)(x2)
= x4
notice
(– x
)(
+1)
+ (+x)(
+
1 )
= 0x
and
(
x2 )(
x ) +
(– x
)(
x2
)
= 0x3
which
leaves
(x2
)(+1 ) +
(+
1 )(
x2 )
+
(– x
)
(+x
)
= 2x2
–
x2
= x2
and
(+
1 )(
+1)
= 1
Adding
the results we get
x4 +
0x + 0x3 +
x2 + 1
x4 +
x2 + 1
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Completing
the Square (advanced)
Certain quadratics can be factored by a
technique called completing the square.
Recall that (a + b)2 = a2
+ 2ab + b2
and (a –
b)2 = a2 –
2ab + b2
Example:
x4 + 2x2 + 9
If instead of 2x2, we had 6x2
then we could directly factor. Since adding 0 to any expression does
not change that expression we can add 4x2 and subtract 4x2
to get
x4 + 2x2 + 4x2
+ 9 – 4x2
x4 + 6x2 + 9 – 4x2
(x2
+ 3)2
– 4x2
we have the difference
of 2 square now
((x2
+ 3) + 2x) ((x2
+ 3) – 2x)
simplify to
get
( x2
+ 2x + 3)( x2
- 2x + 3)
Another
example.
factor x4
– 19x2
+ 25
from the
pattern: a=x2,
b = 5, -2ab = -10x2
here we have -19x2
so to get
-10x2 we
need to add 9x2 and
subtract 9x2
x4
– 19x2
+ 9x2
+ 25 – 9x2
x4
– 10x2
+ 25
– 9x2
(x2
– 5)2
– 9x2
((x2
– 5) + 3x)((x2
– 5) – 3x)
(x2
+ 3x – 5)(x2
– 3x – 5)
Higher
order polynomials and the sum and difference of two terms with like odd
exponents. (advanced)
Example
factor:
x3 –
x2
– 5x + 2
This is not
one of our patterns. Grouping terms and completing the square will
not work here.
Recall that
the factors of a polynomial are the zeroes of that polynomial.
For example,
x2
+ 4x – 12
= (x – 2)( x + 6)
If
we substitute +2 into x2
– 4x – 12
we
get (2)2 +
4(2) – 12 = 4 + 8 + 12 = 0
If
we substitute –6 into x2
– 8x + 12
we
get (– 6)2
+ 4(– 6) + 12 = 36 – 24
– 12 = 0
The
point here is the the values of x that make these factors 0 when
substituted back into the polynomial make the variable terms add up
to the negative
of the the polynomial constant.
In
this case when x = 2 or x = – 6 x2
+ 4x add to 12 which is – (–
12)
We
can use this fact to look for factors of x3
– x2
– 5x + 2
We
need a value of x that substituted into x3
– x2
– 5x equals -2
trial and
error:
x
= 1: 1 – 1 – 5 = – 5 ≠ -2
x
= -1: -1 – 1 + 5 = 2 ≠ -2
x
= 2: 8 – 4 –10 = –6 ≠ -2
x
= -2: -8 – 4 + 10 = –2 this
is it!
so,
(x + 2)
is a factor (if x=-2 then
x+2 = -2+2 = 0, ie, (x+2) = 0)
x3
– x2 –
5x + 2 = (x +
2)( ? )
Synthetic Division
is
the quickest method to get the other factor:
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1
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-1
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-5
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2
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-(+2)
=
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-2
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-2
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6
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-2
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1
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-3
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1
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0
|
If (x+2) is a
factor, this remainder MUST
be 0!
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x3
– x2
– 5x + 2 = (x +
2)( x2
– 3x + 1)
The second
factor is a quadratic and a quick check on the discriminate gives
(-3)2
- 4(1)(1) = 5 and 5 is not a square
so that factor has no integer roots and it is therefore prime. The
answer is:
(x
+ 2)( x2 – 3x + 1)
This approach
can be time consuming but you only need to consider all factors of
the constant term only. In the previous case the factors of 2 are ±1
and ±2.
Example
factor 3x3
+ 5x2 – 5x + 1
In this
example the leading coefficient of x3 is not 1 but 3. We
can divide the terms by 3 to get the coefficient to be 1. First we
consider the factors of the constant term 1 (± 1), then divide
and consider the factors of 1/3 (±1/3).
+ 1: 9 –
5 – 5 = 9
–
1 :
-9 + 2 + 5 = -2
+ 1/3: 1/9 +
5/9 – 5/3 = – 1 this is a root
–
1/3:
-1/9 + 5/9 + 5/3 = 19/9
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Since
x = 1/3 is a root this means
(3x
- 1) is a factor
|
x = 1/3
3x = 3(1/3)
3x = 1
-1 +3x = 1 - 1
3x - 1 = 0
(3x
– 1) = 0
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Let's
factor this by inspection. (I've provided detailed
explanation, in practice you would write (3x – 1)( x2
x ) as the second step and fill in the blanks.)
3x3
+ 5x2 – 5x + 1
(3x
– 1)( ? x2 ? x ? )
The result
must be a quadratic: why? We started with a degree 3 polynomial and
we've found one factor that involves x; This x must multiply every
term in the factor we are looking for, so the highest power that
factor can have is 2 otherwise we'll end up with terms with a degree
higher than the 3 we started with.
There is only
one way to get 3x3
from 3x and that is to
multiply by x2.
so far we have
(3x
– 1)( x2 ? x ? )
Now let's
consider the constant term (red question mark next.)
(3x –
1)(
x2 ? x ? )
here's the
original polynomial:
3x3
+ 5x2 – 5x + 1
The constant
term +1 is the result of the -1
in our factor multiplied by the constant in the unknown
factor which means this constant must be -1 as well.
(Only those
two terms result in a constant when multiplied. -1 • -1 = 1)
Now we have:
(3x –
1)( x2 ? x –
1)
Our original
polynomial has a +5x2 and a –5x.
The question
to ask is, what do I need as a coefficient for ?
x that will provide these terms?
The blue terms
multiplied plus the red terms multiplied must equal -5x.
(3x
–
1)(
x2 ? x
– 1)
so we have -3x
+ (-1)?x
= -5x
The question
mark must be 2. - 3 +
(-1)2
= -3 -2 = -5
Now we have:
(3x –
1)( x2 + 2x – 1)
Verify that we
get 5x2 using this 2
|
(3x
–
1)( x2
+ 2x –
1)
3x2x
– x2 = 6x2 – x2 = 5x2 √
|
Recall
the ONLY way we can get an x2 from this product is by
multiplying the red terms and adding
them to the blue terms multiplied.
|
One last
check: the discriminate = 22 - 4(1)(-1) = 4 + 4 = 8 not a
square, quadratic is prime.
answer: (3x
– 1)(
x2 + 2x – 1)
Sum
and difference of like odd powers. (advanced)
Example:
x5
+ 32
First
recognize that 25
= 32 (worth remembering; 210
= 1024 as well.)
so we factor X5
+ 25
Recall that a
factor of a polynomial is also a zero of that polynomial.
0
= x5 +
25
if x5 =
-32 then the sum is 0 so x = -2 is a root of this polynomial and (x+2) is a factor.
(x
+2)( P(x) ) = x5
+ 25
where P(x) is
the other factor. Uisng the following formula we get:
P(x)
= x4 –
2x3
+ (2)2x2
–(2)3x
+(–2)4
=
x4 –
2x3
+ 4x2
– 8x + 16
putting it
together
x5
+ 32 = (x+2)(x4
– 2x3
+ 4x2
– 8x + 16)
Sum
of like odd powers: (advanced)
Once
you know that (x + c) is factor of xn
+ cn
with n odd,
then the polynomial can be factored into
(x
+ c)(xn-1
– xn-2c
+ xn-3c2
– xn-4c3
+ ... + cn-1)
Notice how the exponents in each term add to n.
Example
(sum):
factor:
x7 + 1
x
= -1 is a zero for this polynomial so (x + 1) is a factor.
Here's
a method to use the general formula above.
1)
The x terms decrease in their exponents starting from 7 - 1 = 6
and
the signs alternate, so write these down (leaving room as shown):
(x
+ 1)( x6 – x5 +x4 – x3
+x2 – x1 + )
2) Now
write the k term next to each x term skipping the first (x6)
and increase its power until it reaches 6 like so
(x
+ 1)( x6 – x5
11
+x4 12
– x3 13
+x2 14
– x1 15
+ 16 )
simplify
(x
+ 1)( x6 – x5 + x4 – x3
+ x2 – x + 1 )
and
we're done.
Example
(difference):
factor:
x5
– 243
243
must be the 5th power of some number, otherwise this polynomial
cannot be factored, it turns out to be 3.
(A
side note, prior to calculators, slide rules and log tables were used
to find these values using mental arithmetic and pencil and paper...
the common logarithms from 2 to 9 were memorized. Knowing the log of
2 and the log of 3 you'd estimate the log of 2.43 (a little less than
0.4) from 2, say 2.38 then add 2 (the log of 100) to get 2.38, divide
that by 5 (5th root) to get 0.37. Knowing the log of 3 is
2.3776, the 5th root must be 3. A quick check shows that to be the
case. Yes, those days did require a bit more mental effort and
concentration for calculation. Good or bad, you make the call.)
So
x=3 is a root, that is (x – 3) is a factor.
write
the result: (red factors and signs first, then blue factors)
(x + (- 3))(
x4
–
x3(-3)
+ x2(-3)2
–
x(-3)3
+
(-3)4)
simplify
(x
– 3)(x4
+
3 x3
+
9x2
+
27 x + 81)
Notice
in this example all the signs are positive. This will always be the
case when you are factoring the difference of like odd powers.
Rational Roots of
Polynomials (advanced)
The Rational Zeros
Theorem states that if a rational number p/q is a zero of a
the polynomial (where p and q are relatively prime) then p
is a factor of a0 and q is a factor of
an.
Example: find the
possible rational zeros of 15x6 + 47x2
+ 2
possible factors of +2
for p: +1, -1, +2, -2
possible factors of 15
for q: 1, -1, 3, -3, 5, -5, 15, -15
possible ratios p/q
are:
1, -1, 1/3, -1/3, 1/5,
-1/5, 2/3, -2/3, 2/5, -2/5, 1/15, -1/15, 2/15,
-2/15
Can any of these ratios
be a zero of this polynomial?
Notice the the powers
of this polynomial are even which means all possible roots
raised to these powers become positive values and the
constant term is positive as well, so this polynomial can
never become zero. The answer is therefore
no.
Let's change the
constant term +2 to -62: 15x6 + 47x2 -
62
possible factors of +2
for p: +1, -1, +2, -2, +31, -31, 62,
-62
possible factors of 15
for q: 1, -1, 3, -3, 5, -5, 15, -15
possible ratios p/q
are:
1, -1, 1/3, -1/3, 1/5,
-1/5, 2/3, -2/3, 2/5, -2/5, 1/15, -1/15, 2/15, -2/15, +31,
-31, +31/3, -31/3, +31/5, -31/5, +31/15, -31/15, 62,
-62
use synthetic division
to check each possible zero
|
(x+1)
|
15
|
0
|
0
|
0
|
47
|
0
|
-62
|
|
|
-1
|
|
-15
|
15
|
-15
|
15
|
-62
|
62
|
|
|
|
15
|
-15
|
15
|
-15
|
62
|
-62
|
0
|
YES
|
Inspecting the
polynomial we can see x = 1 is a root, as is x =
-1
|
(x+1/3)
|
15
|
0
|
0
|
0
|
47
|
0
|
-62
|
|
|
-1/3
|
|
-5
|
5/3
|
-5/9
|
5/27
|
-47/3
|
-15
2/3
|
|
|
|
15
|
-5
|
5/3
|
-5/9
|
47
|
-15
2/3
|
-77
2/3
|
no
|
This becomes
cumbersome, access to a spreadsheet makes this check in
short order.
As you can see only 1
and -1 are rational roots of this polynomial. So, from the
synthetic division table we obtain the other
factors:
(x + 1) (
15x5 -
5x4+5/3 x3
-5/9x2 + 47x - 15
2/3)
|
(x -
1)
|
15
|
0
|
0
|
0
|
47
|
0
|
-62
|
|
|
1
|
|
15
|
15
|
15
|
15
|
62
|
62
|
|
|
|
15
|
15
|
15
|
15
|
62
|
62
|
0
|
check
|
(x - 1) (
15x5 -
15x4+ 15 x3
+ 15x2 + 62x +
62)
This product may not be
the prime factorization; we would need to carry out the
procedure with the second factor repeatedly until no further
rational zeroes can be found. (This second step is
illustrated in the next example.)
Example: find
the rational zeros for 5x4 - 4x3 +
19x2 -16x - 4
possible p: 1, -1, 2,
-2, 4, -4
possible q: 1, -1, 5,
-5
possible p/q: 1, -1,
1/5, -1/5, 2, -2, 2/5, -2/5,4, -4, 4/5,
-4/5
we can eliminate
choices by inspection
x = 1, 5 – 4 +19
-16 - 4 = 0 yes, a root
x = -1 5 +4 + 19 + 16
– 4 = 40 not a root
|
(x -
1/5)
|
5
|
-4
|
19
|
-16
|
-4
|
|
|
1/5
|
|
1
|
-3/5
|
92/25
|
-308/25
|
|
|
|
5
|
-3
|
18
2/5
|
-308/25
|
-328/25
|
no
|
/pthis is crap
|
(x +
1/5)
|
5
|
-4
|
19
|
-16
|
-4
|
|
|
-1/5
|
|
-1
|
1
|
-4
|
4
|
|
|
|
5
|
-5
|
20
|
-20
|
0
|
yes
|
The rational roots are x = 1, and x =
-1/5
We'll use synthetic division as before to
find the other factor for (x -1)
|
(x - 1
)
|
5
|
-4
|
19
|
-16
|
-4
|
|
|
1
|
|
5
|
1
|
20
|
4
|
|
|
|
5
|
1
|
20
|
4
|
0
|
|
Finally we have (x – 1) (
5x3 + x2+ 20
x + 4)
and
( x + 1/5)( 5x3
- 5x2+ 20x - 20)
This is all well and good, but are we
done factoring?
Can we factor (5x3 +
x2 + 20 x + 4) or (5x3 - 5x2 +
20x - 20) ?
Well repeat the process
5x3 + x2 + 20 x +
4
p: 1, -1, 2, -2, 4, -4
q: 1, -1, 5, -5
p/q: 1, -1, 1/5, -1/5, 2, -2, 2/5, -2/5,
4, -4, 4/5, -4/5
inspecting:
all positive possibilities will not work
(coefficients are all positive)
so this leaves: -1, -1/5, -2, -2/5, -4
and -4/5 to check
-1: -5 + 1 -20 + 4 = -20 no
-2: -40 + 4 – 40 + 4 = -48
no
-4: -320 + 16 -80 + 4 = -380
no
now let's use synthetic
division
|
(x + 1/5)
|
5
|
1
|
20
|
4
|
|
|
-1/5
|
|
-1
|
0
|
-4
|
|
|
|
5
|
0
|
20
|
0
|
yes
|
|
(x + 2/5)
|
5
|
1
|
20
|
4
|
|
|
-2/5
|
|
-2
|
2/5
|
-84/25
|
|
|
|
5
|
-1
|
42/5
|
-26/25
|
no
|
|
(x + 4/5)
|
5
|
1
|
20
|
4
|
|
|
-4/5
|
|
-4
|
12/5
|
-208/25
|
|
|
|
5
|
-3
|
52/5
|
-108/25
|
no
|
so we have (x + 1/5)( 5x2 +
20)
and this can further be factored to 5(x +
1/5)(x2 + 4)
So the original polynomial can be
factored into
5(x + 1/5)(x2+ 4)(x
– 1)
These are prime factors so this is the
polynomial is now completely factored.
This begs yet another question, how
about
( x + 1/5)( 5x3 -
5x2 + 20x - 20) ?
Since we found the prime factors, this product must be
equivalent.
Notice in the second factor the common 5.
Let's factor that out.
(x + 1/15) 5 (x3 –
x2 + 4x – 4)
Now if you multiply the (x2+ 4)(x
-1) from the factored equation just
above,
you get (x3 –
x2 + 4x – 4)! So this second product factors
into the same prime factors.
We expect this since the prime
factorization of a polynomial is unique.
In summary, the Rational Zeros Theorem
allows us to find rational factors of polynomials, if those
factors exist. Synthetic division is very useful in this
process as is a spreadsheet. These rational zeroes may not
exist, but this does not mean real zeros do not exist. (There
are numerical techniques that will find these real roots if
they exist. college level mathematics)