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Calculator Function –
How to take the cube root of a number
To take the cube root of a
number, press MATH, then
select option 4.
Example: What is
3
13824
24
?
The Greatest Common
Factor
Factors (either numbers or polynomials)
When an integer is written as a product of
integers, each of the integers in the
product is a factor of the original number.
When a polynomial is written as a product
of polynomials, each of the polynomials in
the product is a factor of the original
polynomial.
Factoring – writing a polynomial as a product
of polynomials.
Greatest common factor – largest quantity
that is a factor of all the integers or
polynomials involved.
Finding the GCF of a List of Integers or Terms
1) Prime factor the numbers.
2) Identify common prime factors.
3) Take the product of all common prime factors.
• If there are no common prime factors, GCF
is 1.
Example Find the GCF of each list of numbers.
1) 12 and 8
2) 7 and 20
Example Find the GCF of each list of numbers.
1) 12 and 8
12 = 2 · 2 · 3
8=2·2·2
So the GCF is 2 · 2 = 4.
2) 7 and 20
7=1·7
20 = 2 · 2 · 5
There are no common prime factors so the
GCF is 1.
Example Find the GCF of each list of numbers.
1) 6, 8 and 46
2) 144, 256 and 300
.
Example Find the GCF of each list of numbers.
1) 6, 8 and 46
6=2·3
8=2·2·2
46 = 2 · 23
So the GCF is 2.
2) 144, 256 and 300
144 = 2 · 2 · 2 · 3 · 3
256 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2
300 = 2 · 2 · 3 · 5 · 5
So the GCF is 2 · 2 = 4.
Example Find the GCF of each list of terms.
1)
x3 and x7
2)
6x5 and 4x3
Example Find the GCF of each list of terms.
1)
x3 and x7
x3 = x · x · x
x7 = x · x · x · x · x · x · x
So the GCF is x · x · x = x3
2)
6x5 and 4x3
6x5 = 2 · 3 · x · x · x
4x3 = 2 · 2 · x · x · x
So the GCF is 2 · x · x · x = 2x3
Example Find the GCF of the following list of terms.
a3b2, a2b5 and a4b7
Example Find the GCF of the following list of terms.
a3b2, a2b5 and a4b7
a3b2 = a · a · a · b · b
a2b5 = a · a · b · b · b · b · b
a4b7 = a · a · a · a · b · b · b · b · b · b · b
So the GCF is a · a · b · b = a2b2
Notice that the GCF of terms containing variables will use the
smallest exponent found amongst the individual terms for each
variable.
The first step in factoring a polynomial is to
find the GCF of all its terms.
Then we write the polynomial as a product by
factoring out the GCF from all the terms.
The remaining factors in each term will form a
polynomial.
Example
Factor out the GCF in each of the
following polynomials.
1) 6x3 – 9x2 + 12x =
2) 14x3y + 7x2y – 7xy =
Example
Factor out the GCF in each of the
following polynomials.
1) 6x3 – 9x2 + 12x =
3 · x · 2 · x2 – 3 · x · 3 · x + 3 · x · 4 =
3x(2x2 – 3x + 4)
2) 14x3y + 7x2y – 7xy =
7 · x · y · 2 · x2 + 7 · x · y · x – 7 · x · y · 1 =
7xy(2x2 + x – 1)
Example
Factor out the GCF in each of the
following polynomials.
3) 6(x + 2) – y(x + 2) =
4) xy(y + 1) – (y + 1) =
Example
Factor out the GCF in each of the
following polynomials.
3) 6(x + 2) – y(x + 2)
= 6 · (x + 2) – y · (x + 2)
= (x + 2)(6 – y)
4) xy(y + 1) – (y + 1)
= xy · (y + 1) – 1 · (y + 1)
= (y + 1)(xy – 1)
Remember that factoring out the
GCF from the terms of a
polynomial should always be the
first step in factoring a polynomial.
This will usually be followed by
additional steps in the process.
Example
Factor 90 + 15y2 – 18x – 3xy2.
90 + 15y2 – 18x – 3xy2 = 3(30 + 5y2 – 6x – xy2) =
3(5 · 6 + 5 · y2 – 6 · x – x · y2) =
3(5(6 + y2) – x (6 + y2)) =
3(6 + y2)(5 – x)
Example
90 + 15y2 – 18x – 3xy2 = 3(30 + 5y2 – 6x – xy2)
= 3(5 · 6 + 5 · y2 – 6 · x – x · y2)
= 3(5(6 + y2) – x (6 + y2))
= 3(6 + y2)(5 – x)
Factoring Trinomials of the
2
Form x + bx + c
Recall by using the FOIL method that
F
O
I
L
(x + 2)(x + 4) = x2 + 4x + 2x + 8
= x2 + 6x + 8
To factor x2 + bx + c into (x + one #)(x + another #),
note that b is the sum of the two numbers and c is
the product of the two numbers.
So we’ll be looking for 2 numbers whose product is c
and whose sum is b.
Note: there are fewer choices for the product, so
that’s why we start there first.
Example Factor the polynomial x2 + 13x + 30.
Since our two numbers must have a product of
30 and a sum of 13, the two numbers must both
be positive.
Positive factors of 30 Sum of Factors
1, 30
31
2, 15
17
3, 10
13
Note, there are other factors, but once we find a pair
that works, we do not have to continue searching.
So x2 + 13x + 30 = (x + 3)(x + 10).
Example Factor the polynomial x2 – 11x + 24.
Since our two numbers must have a product of
24 and a sum of -11, the two numbers must
both be negative.
Negative factors of 24 Sum of Factors
– 1, – 24
– 25
– 2, – 12
– 14
– 3, – 8
So x2 – 11x + 24 = (x – 3)(x – 8).
– 11
Prime Polynomials
Example Factor the polynomial x2 – 6x + 10.
Since our two numbers must have a product of 10
and a sum of – 6, the two numbers will have to
both be negative.
Negative factors of 10 Sum of Factors
– 1, – 10
– 11
– 2, – 5
–7
Since there is not a factor pair whose sum is – 6,
x2 – 6x +10 is not factorable and we call it a prime
polynomial.
When polynomials contain four terms, it is
sometimes easier to group like terms in order to
factor.
Your goal is to create a common factor.
You can also move terms around in the polynomial to
create a common factor.
Practice makes you better in recognizing common
factors.
» FACTOR: 3xy - 21y + 5x – 35
Factor the first two terms:
3xy - 21y = 3y (x – 7)
Factor the last two terms:
+ 5x - 35 = 5 (x – 7)
The green parentheses are the same so it’s the common
factor
Now you have a common factor
(x - 7) (3y + 5)
» FACTOR: 6mx
– 4m + 3rx – 2r
Factor the first two terms:
6mx – 4m = 2m (3x - 2)
Factor the last two terms:
+ 3rx – 2r = r (3x - 2)
The green parentheses are the same so it’s the common
factor
Now you have a common factor
(3x - 2) (2m + r)
» FACTOR:
15x – 3xy + 4y –20
Factor the first two terms:
15x – 3xy = 3x (5 – y)
Factor the last two terms:
+ 4y –20 = 4 (y – 5)
The green parentheses are opposites so change the
sign on the 4 ; - 4 (-y + 5) or – 4 (5 - y)
Now you have a common factor
(5 – y) (3x – 4)
When trinomials have a degree of “2”, they are
known as quadratics.
We learned earlier to use the “diamond” to factor
trinomials that had a “1” in front of the squared
term.
x2 + 12x + 35
(x + 7)(x + 5)
When there is a coefficient larger than “1” in front
of the squared term, we can use a modified
diamond or square to find the factors.
Always remember to look for a GCF before you do
ANY other factoring.
Let’s try this example
3x2 + 13x + 4
Make a box
Write the factors of the first term.
Write the factors of the last term.
Multiply on the diagonal and add to see if you get
the middle term of the trinomial. If so, you’re
done!
When factoring using a difference of
squares, look for the following three things:
˃ only 2 terms
˃ minus sign between them
˃ both terms must be perfect squares
If all 3 of the above are true, write two
( ), one with a + sign and one with a – sign :
( + ) ( - ).
» 1.
» 2.
» 3.
» 4.
» 5.
» 6.
a2 – 16
x2 – 25
4y2 – 16
9y2 – 25
3r2 – 81
2a2 + 16
When factoring using perfect square trinomials,
look for the following three things:
˃ 3 terms
˃ last term must be positive
˃ first and last terms must be perfect squares
If all three of the above are true, write one ( )2
using the sign of the middle term.
» 1.
» 2.
» 3.
» 4.
» 5.
» 6.
a2 – 8a + 16
x2 + 10x + 25
4y2 + 16y + 16
9y2 + 30y + 25
3r2 – 18r + 27
2a2 + 8a - 8
Factoring Trinomials of the
2
Form ax + bx + c
Returning to the FOIL method,
F
O
I L
(3x + 2)(x + 4) = 3x2 + 12x + 2x + 8
= 3x2 + 14x + 8
To factor ax2 + bx + c into (#1·x + #2)(#3·x + #4),
note that a is the product of the two first coefficients, c
is the product of the two last coefficients and b is the
sum of the products of the outside coefficients and
inside coefficients.
Note that b is the sum of 2 products, not just 2
numbers, as in the last section.
Example
Factor the polynomial 25x2 + 20x + 4.
Possible factors of 25x2 are {x, 25x} or {5x, 5x}.
Possible factors of 4 are {1, 4} or {2, 2}.
We need to methodically try each pair of factors until we find
a combination that works, or exhaust all of our possible pairs
of factors.
Keep in mind that, because some of our pairs are not identical
factors, we may have to exchange some pairs of factors and
make 2 attempts before we can definitely decide a particular
pair of factors will not work.
Example
Factor the polynomial 25x2 + 20x + 4.
We will be looking for a combination that gives the sum of the
products of the outside terms and the inside terms equal to 20x.
Factors of Factors of
25x2
4
Resulting
Binomials
Product of Outside
Terms
Product of Inside
Terms
Sum of
Products
{x, 25x} {1, 4} (x + 1)(25x + 4)
4x
25x
29x
(x + 4)(25x + 1)
x
100x
101x
{x, 25x} {2, 2} (x + 2)(25x + 2)
2x
50x
52x
{5x, 5x} {2, 2} (5x + 2)(5x + 2)
10x
10x
20x
Example Continued
Check the resulting factorization using the FOIL method.
F
O
I
L
(5x + 2)(5x + 2) = 5x(5x) + 5x(2) + 2(5x) + 2(2)
= 25x2 + 10x + 10x + 4
= 25x2 + 20x + 4
So our final answer when asked to factor 25x2 + 20x + 4
will be (5x + 2)(5x + 2) or (5x + 2)2.
Example Factor the polynomial 21x2 – 41x + 10.
Since the middle term is negative, possible factors of 10
must both be negative: {-1, -10} or {-2, -5}.
We need to methodically try each pair of factors until
we find a combination that works, or exhaust all of our
possible pairs of factors.
Example Factor the polynomial 21x2 – 41x + 10.
Factors Factors Resulting
Product of
Product of
Sum of
of 21x2 of 10 Binomials Outside Terms Inside Terms Products
{3x, 7x}{1, 10}(3x – 1)(7x – 10)
30x
7x
37x
(3x – 10)(7x – 1)
3x
70x
73x
{3x, 7x} {2, 5} (3x – 2)(7x – 5)
15x
14x
29x
(3x – 5)(7x – 2)
6x
35x
41x
Example Continued
Check the resulting factorization using the FOIL method.
F
(3x – 5)(7x – 2) =
O
I
L
3x(7x) + 3x(-2) - 5(7x) - 5(-2)
= 21x2 – 6x – 35x + 10
= 21x2 – 41x + 10
So our final answer when asked to factor 21x2 – 41x + 10
will be (3x – 5)(7x – 2).
Now that we’ve learned all the types of
factoring, we need to remember to use
them all.
Whenever it says to factor, you must
break down the expression into the
smallest possible
factors.
Let’s review all the ways to factor.
1. Look for GCF first.
2. Count the number of terms:
a) 4 terms – factor by grouping
b) 3 terms – look for perfect square trinomial
– if not, try diamond or box
c) 2 terms – look for difference of squares
If any ( ) still has an exponent of 2 or more, see if you can
factor again.
We know that an equation must be solved for the
unknown.
Up to now, most of what we have only solved were
equations with a degree of 1.
2x + 8 = 4x +6
-2x + 8 = 6
-2x = -2
x=1
If an equation has a degree of 2 or higher, we
cannot solve it until it has been factored.
You must first get “0” on one side of the = sign
before you try any factoring.
Once you have “0” on one side, use all your
rules for factoring to make 2 ( ) or factors.
Next, set each factor = 0 and solve for the
unknown.
x2 + 12x = 0 Factor GCF
x(x + 12) = 0
(set each factor = 0, & solve)
x=0
x + 12 = 0
x = - 12
You now have 2 answers, x = 0 and x = -12.
Q. Factor the polynomial x 3 – 400. If the polynomial
cannot be factored, write prime.
Answer: The first term is a perfect cube, but the second
term is not. It is a prime polynomial.
As with some quadratic equations, factoring a
polynomial equation is one way to find its real
roots.
Recall the Zero Product Property. You can
find the roots, or solutions, of the polynomial
equation P(x) = 0 by setting each factor
equal to 0 and solving for x.
Every polynomial function with degree n
greater than or equal to 1 has exactly n
complex zeros, including multiplicities
The polynomial 3x5 + 18x4 + 27x3 = 0 has two
multiple roots, 0 and –3. The root 0 is a factor three
times because 3x3 = 0.
The multiplicity of root r is the number of times
that x – r is a factor of P(x).
When a real root has even multiplicity, the graph of y = P(x)
touches the x-axis but does not cross it.
When a real root has odd multiplicity greater than 1,
the graph “bends” as it crosses the x-axis.
You cannot always determine the multiplicity of a
root from a graph. It is easiest to determine
multiplicity when the polynomial is in factored
form.
Solve the polynomial equation by factoring.
4x6 + 4x5 – 24x4 = 0
4x6 + 4x5 – 24x4 = 0
Factor out the GCF, 4x4.
4x4(x2 + x – 6) = 0
Factor the quadratic.
4x4(x + 3)(x – 2) = 0
Set each factor equal to 0.
4x4 = 0 or (x + 3) = 0 or (x – 2) = 0
Solve for x.
x = 0, x = –3, x = 2
The roots are 0, –3, and 2.
Check Use a graph.
The roots appear to be located at x = 0, x = –3, and
x = 2.
Solve the polynomial equation by factoring.
x4 + 25 = 26x2
x4 + 25 = 26x2
Set the equation equal to 0.
x4 – 26 x2 + 25 = 0
Factor the trinomial in quadratic
form.
(x2 – 25)(x2 – 1) = 0
Factor the difference of two
squares.
(x – 5)(x + 5)(x – 1)(x + 1)
Solve for x.
x – 5 = 0, x + 5 = 0, x – 1 = 0, or x + 1 =0
x = 5, x = –5, x = 1 or x = –1
The roots are 5, –5, 1, and –1.
Identify the roots of each equation. State the
multiplicity of each root.
x3 + 6x2 + 12x + 8 = 0
Since this is difficult to
factor, use a graph. A
calculator graph shows
a bend near (–2, 0).
The root –2 has a multiplicity of 3. Therefore
x + 2 is a factor three times.
Check to see if that is true!
(x + 2)(x + 2)(x + 2)= x3 + 6x2 + 12x + 8
Identify the roots of each equation. State the
multiplicity of each root.
x4 + 8x3 + 18x2 – 27 = 0
x4 + 8x3 + 18x2 – 27 = 0
A calculator graph shows a
bend near (–3, 0) and
crosses at (1, 0).
The root 1 has a multiplicity of 1. The root –3 has
a multiplicity of 3. Therefore (x – 1) is a factor
once, and (x + 3) is a factor three times.
The root 1 has a multiplicity of 1. The root –3 has
a multiplicity of 3. Therefore (x – 1) is a factor
once, and (x + 3) is a factor three times.
(x – 1)(x + 3)(x + 3)(x + 3)=x4 + 8x3 + 18x2 – 27
Location Principle
If P is a polynomial function and P(x1) and P(x2) have
opposite signs, then there is a real number r between
x1 and x2 that is a zero of P, that is, P(r) = 0
P( x)
x
4
3x
3
5x
2
13x 6
Lesson Quiz
Identify the roots of each equation. State the
multiplicity of each root.
1. 5x4 – 20x3 + 20x2 = 0
2. x3 – 12x2 + 48x – 64 = 0
3. x3 + 9 = x2 + 9x
Lesson Quiz
Identify the roots of each equation. State the
multiplicity of each root.
1. 5x4 – 20x3 + 20x2 = 0
0 and 2 each with
multiplicity 2
2. x3 – 12x2 + 48x – 64 = 0
4 with multiplicity 3
3. x3 + 9 = x2 + 9x
–3, 3, 1