```Factoring Polynomials
Finding GCF
If you remember multiplication with real
numbers, then you should remember the
following facts:
If 6 • 2 = 12, then we know
• 12 is the product of 6 and 2
• 6 and 2 are factors, or divisors, of 12.
• the quotient of 12 divided by 6 is 2.
• the quotient of 12 divided 2 is 6.
So, to factor a number is to write it as the product
of two or more numbers, usually natural numbers.
Factoring and division are closely related.
Review: a factor - is a number
that is multiplied by another
number to produce a product. A
prime number - is any natural
number greater than 1 whose
only factors are 1 and itself. A
composite number - is a number
greater than 1 that has more
than two factors.
The prime factorization - is the
factorization of a natural
number that contains only prime
numbers or powers of prime
numbers.
The figure below shows 24 square tiles
arranged to form a rectangle. (4 tiles wide, 6
tiles long)
Sketch other ways that the 24 tiles can be
arranged to form a rectangle.
If you notice each rectangle has an area of 24,
which includes (4 x 6, 3 x 8, 2 x 12, 1 x 24).
Each of the numbers involved in these
multiplications is a factor of 24. There are no
other natural number pairs that have a
product of 24, so 1, 2, 3, 4, 6, 8, 12, and 24 are
the only factors of 24.
Here are two examples
Write 3 different factorizations of 16. Use
natural numbers.
16 divided by
1
2
3
4
5
6
7
8
Natural number
16
8
no
4
no
no
no
2
1 x 16
2x8
4x4
8x2
Finding Greatest
Common Factor
GCF’s
Let’s make a list of factors of 36, written in
order from least to greatest.
36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Make a list of factors for the number 54.
54: 1, 2, 3, 6, 9, 18, 27, 54
Examine the two lists for factors that appear
in both list.
36: 1, 2, 3, 6, 9, 12, 18, 36
54: 1, 2, 3, 6, 9, 18, 27, 54
common factors: 1, 2, 3, 6, 9, 18
GCF: 18
What if the numbers in our previous example
were expressions 36c3 and 54c2?
36c3
2∙2∙3∙3∙c∙c∙c
54c2
2∙3∙3∙3∙c∙c
21 ∙ 32 ∙ c2
18 ∙ c2
21 ∙32 ∙c2
18 ∙ c2
GCF 18c2
Try again
Find the GCF:
a) 18d5 and 108d
b) 18d and 5
c) 3m3n3 and 9m2n2
d) 4mn3, 4m2n3, and 16m2n2
Factoring a monomial
from a polynomial
Using GCF
Factoring a Monomial
from a Polynomial
Find the GCF of the terms of:
4x3 + 12x2 – 8x
List the prime factors of each term.
4x3 = 2 · 2 · x · x x
12x2 = 2 · 2 · 3 · x · x
8x = 2 · 2 · 2 · x
The GCF is 2 · 2 · x or 4x.
Factoring a polynomial
reverses the
multiplication process.
To factor a monomial
from a polynomial, first
find the greatest
common factor (GCF) of
its terms.
Find the GCF of the terms of each polynomial.
a) 5v5 + 10v3
b) 3t2 – 18
c) 4b3 – 2b2 – 6b
d) 2x4 + 10x2 – 6x
Use the GCF to factor each polynomial.
a) 8x2 – 12x
b) 5d3 + 10d
c) 6m3 – 12m2 – 24m
d) 4x3 – 8x2 + 12x
Try to factor mentally by
scanning the coefficients of
each term to find the GCF.
Next, scan for the least power
of the variable.
Factoring Out a
Monomial
Factor 3x3 – 12x2 + 15x
Step 1
Find the GCF
3x3 = 3 · x · x · x
12x2 = 2 · 2 · 3 · x · x
15x = 3 · 5 · x
The GCF is 3 · x or 3x
To factor a polynomial
completely, you must factor
until there are no common
factors other than 1.
Step 2
Factor out the GCF
3x3 – 12x2 + 15x
= 3x(x2) + 3x(-4x) + 3x(5)
= 3x(x2 – 4x + 5)
Factoring x2 + bx + c
Factoring x2 + bx + c when c is
positive
Observe the two columns below, the
multiplication of binomials on the left and the
products on the right.
What do you notice about the two list?
1. (x + 5)(x + 6)
[x2 + 11x + 30]
2. (x + 3)(x + 10)
[x2 + 13x + 30]
3. (x + 2)(x + 15)
[x2 + 17x + 30]
4. (x + 1)(x + 30)
[x2 + 31x + 30]
The product of the constants = 30
The sum of the constants = the coefficient of
the x-terms
1. (x + 5)(x + 6)
[x2 + 11x + 30]
2. (x + 3)(x + 10)
[x2 + 13x + 30]
3. (x + 2)(x + 15)
[x2 + 17x + 30]
4. (x + 1)(x + 30)
[x2 + 31x + 30]
TRY THIS
Remember the patterns you say earlier.
x2 + 17x + 30
Write the binomial multiplication that gives
this product.
(
)(
The sum of what
two of those
constants give you
17?
)
What two constants
multiplied together
gives you 30?
In other words, to factor x2 + bx + c, look for
the factor pairs (the two numbers) whose product
is c. Then choose the pair whose sum is b.
Factor x2 + 5x + 6
1. c = 6: write all (+ and -) the factor pairs of 6.
1•6
2•3
-1 • -6
-2 • -3
2. b = 5: choose the pair whose sum is 5.
1 + 6 = 7 2 + 3 = 5 -1 - 6 = -7 -2 – 3 = -5
3. Write the product using 2 and 3. Thus
(x + 2)(x + 3) = x2 + 5x + 6
TRY THIS
a2 + 9a + 20
NEXT
Factor y2 – 10y + 24
1. c = 24 Write all (+ and -) factor pairs of 24.
1 • 24
-1 • -24
4•6
-4 • -6
2 • 12
-2 • -12
3•8
-3 • -8
2. b = -10: choose the pair whose sum is -10.
1 + 24 = 25 -1 +(-24) = -25 4 + 6 = 10 -4 + (-6) = -10
3. Write the product using -4 and -6.
(y – 4)(y – 6) = y2 -10y + 24
TRY THIS
n2 -13n + 36
Practice and Problem
Solving.
1. t2 + 7t + 10 =
(t + 2)(t + □)
2. x2 – 8x + 7 =
(x – 1)(x - □)
3. x2 + 9x + 18 =
(x + 3)(x + □)
Factor each expression
1. r2 +4r + 3
2. n2 – 3n + 2
3. k2 + 5k + 6
4. x2 – 2x + 1
5. y2 + 6y + 8
Factoring x2 + bx + c
Factoring x2 + bx + c, when c is
negative
Find each product.
What do you notice about the c term in each?
(x – 5)(x + 6)
(x – 3)(x + 10)
(x – 2)(x + 15)
(x – 1)( x + 30)
x2
x2
x2
x2
+
+
+
+
x – 30
7x – 30
13x – 30
29x – 30
(x + 5)(x - 6)
(x + 3)(x - 10)
(x + 2)(x – 15)
(x + 1)(x – 30)
x2
x2
x2
x2
–
–
–
–
x – 30
7x – 30
13x – 30
29x - 30
What do you
in each
expression?
What do you
and the
coefficient of
the x-term?
You can also factor x2 + bx – c
Factor x2 + x – 20
1. c = -20: write all (+ and -) factors of -20.
(-1) • 20 1 • (-20) (-2) • 10 2 • (-10) (-4) • 5 4 • (-5)
2. b = 1: find the pair whose sum is 1.
(-1) • 20 1 • (-20) (-2) • 10 2 • (-10) (-4) • 5 4 • (-5)
3. Write the product using -4 and 5 = 1
(x – 4)(x + 5) = x2 + x - 20
TRY THIS
n2 + 3n - 40
Factor z2 – 4z – 12
1. c = -12: write all (+ and -)factors of -12.
(-1) • 12 1 • (-12) (-2) • 6 2 • (-6) (-3) • 4 3 • (-4)
2. b = -4: find the factor pair of -4.
(-1) • 12 1 • (-12) (-2) • 6 2 • (-6) (-3) • 4
3. write the product using 2 and -6.
(x + 2)(x – 6) = x2 – 4z - 12
TRY THIS
n2 – 3n - 40
3 • (-4)
Not every polynomial of the form x2 + bx + c
is factorable.
Factor x2 + 3x – 1
c = -1: the only factors of -1, are 1 and -1.
b = 3: because -1 + 1 ≠ 3,
x2 + 3x – 1 cannot be factored
TRY THIS
t2 + 5t - 8
Factor:
1. y2 + 10y – 11
2. x2 – x – 42
3. b2 – 17b – 38
4. s2 + 4s – 5
5. y2 + 2y – 63
Factoring ax2 + bx + c
Factoring ax2 + bx + c, when c is
positive
Before we tackle this factoring let’s go back and
review the F
First Terms
Outer Terms
O I L method of how the product of
two binomials works.
(2x + 3)(5x + 4) =
Last Terms
Inner Terms
10x2 + (8x +15x) + 12 =
10x2 + 23x + 12
Notice what happens
when the multiply the
Outer Terms and Inner
Terms.
NEXT
10x2 + 23x + 12
To factor it, think of 23x as 8x + 15x.
8x + 15x = 23x
10x2 + 23x + 12 = 10x2 + 8x + 15x + 12
Where did we get 8x and 15x?
Notice that multiplying (a) 10 and (c) 12 gives you 120,
which is the product of the x2-coefficient (10) and the
constant term (12).
In the form of
ax2 + bx + c
8 and 15 are factors of 120
1•120 2•60 3•40 4•30 5•24 6•20 8•15 10•12
also 8x + 15x = 23
This example suggest that, to factor a trinomial, you
should look for factors of the product ac that have a sum
of b.
Let’s see if it works.
Consider the trinomial 6x2 + 23x + 7. To factor it,
think of 23x as 2x + 21x.
Where did we get 2 and 21?
If we multiply 6 and 7 we get 42, which is the product
of the x2-coefficient (6) and the constant (7).
2 and 21 are factor of 42
1•42 2•21 3•14 6•7
and
2x + 21x = 23x
Yes it does work!
So we must find the product of ac that have the sum b.
NEXT
Now we must rewrite the trinomial using the factors
you found for b. (2 and 21)
6x2 + 2x + 21x + 7
Now we are going to find the GCF by grouping terms.
(Remember the Associative Property)
FACTOR
(6x2 + 2x) + (21x + 7)
2x(3x + 1) + 7(3x + 1)
FACTOR
What do you notice
the parenthesis?
Now lets use the Distributive Property to write the
two binomials.
(2x + 7)(3x + 1)
Now we have factored
6x2 + 23x + 7 to
(2x + 7)(3x + 1)
Step 1:
Find factors of ac that
have a sum b.
Factor 5x2 + 11x + 2
Factors of 10
Sums of factors
1 x 10
2 x 5
11
7
Since ac = 10 and b = 11,
find the positive factors of 10
that have a sum 11.
5x2 + 11x + 2 = 5x2 + 1x + 10x + 2 Rewrite bx: 11x = 1x + 10x.
= (5x2 + 1x)(10x + 2) Group terms, Associative Property.
= x(5x + 1) + 2(5x + 1) Factor GCF of each pair of terms.
= (x + 2)(5x + 1) Use Distributive Property to write factored terms.
Step2:
To factor the trinomial,
use the factors you found
(1 + 10) to rewrite bx.
TRY ANOTHER
6x2 + 13x + 5
What is the factored form of 6x2 + 13x + 5?
6x2 + 13x + 5 =
6x2 + 3x + 10x + 5 Rewrite bx: 13x = 3x + 10x.
(6x2 + 3x)(10x + 5) Group terms together to factor.
3x(2x + 1) + 5(2x + 1) Factor GCF of each pair of terms.
(3x + 5)(2x + 1) Use Distributive Property to write binomials.
Factor:
1) 2n2 + 11n + 5
2) 5x2 + 34x + 24
3) 2y2 – 23y + 60
4) 4y2 + 62y + 30
5) 8t2 + 26t + 15
Factoring ax2 + bx +c
Factoring when ac is negative
Can we apply the same steps we
have learned to factor
trinomials that contain negative
numbers?
Yes. Your goal is still to find
factors of ac that have sum b.
Because ac < 0 (less than), the
factors must have different signs.
We need to use all combinations
The sums of positive and
of factors.
negative numbers gives
us our b.
Ex: factors of -15.
1•(-15) (-1)•15 3•(-5) (-3)•5
1+(-15)=-14 (-1)+15=14 3+(-5)=-2 (-3)+5 =2
Factor 3x2 + 4x – 15
Factors of -45
1, -45
-1, 45
3, -15
-3, 15
5, -9
-5, 9
Sums of factors
-44
44
-12
12
-4
4
Find factors of ac with the sum b.
Since ac = -45 and b = 4, find factors of -45.
3x2 -5x +9x – 15
Rewrite bx: 4x = -5x + 9x.
(3x2 -5x) + (9x – 15)
Group terms together to factor.
x(3x – 5) + 3(3x– 5)
(x + 3)(3x – 5)
Factor GCF of each pair of terms.
Use Distributive to rewrite binomials.
Factor:
1) 3k2 + 4k – 4
2) 5x2 + 4x – 1
3) 10y2 – 11y – 6
4) 6q2 – 7q – 49
5) 2y2 + 11y – 90
Not all expressions of the form
ax2 + bx – c can be factored.
This is especially common when
the polynomial contains
subtraction. Try this.
-10x2 + 21x - 5
TRY THIS
Geometry
The area of a rectangle is 2y2 – 13y – 7.
What are the possible dimensions of the
rectangle? Use factoring.
2y2 – 13y – 7 =
Simplifying before
factoring
Some polynomials can be factored
repeatedly.
Some polynomials can be factored repeatedly. This
means you can continue the process of factoring until
there are no common factors other than 1. If a
trinomial has a common monomial factor, factor it out
before trying to find binomial factors.
Take for example:
(6h + 2)(h + 5) and (3h + 1)(2h + 10)
Find each product.
6h2 + 32h + 10 and 6h+ 32h + 10
these two polynomials?
Can you factor a monomial
before factoring a
binomial?
NEXT
Factor 20x2 + 80x + 35 completely.
Factors of 28
1 x 28
2 x 14
4 x 7
Sum of factors
29
16
11
20x2 + 80x + 35 = Factor out GCF monomial.
5(4x2 + 16x + 7) =
Step 1:
Find factors of ac with
sum b.
Rewrite bx: 16x = 2x + 14x.
5[4x2 + 2x + 14x + 7)] =
Factor GCF of each pair of terms.
5[2x(2x + 1) + 7(2x + 1)] = 5(2x + 7)(2x + 1) Rewrite using the
Distributive Property.
Step 2:
Rewrite bx using
factors.
Solving Polynomial
Equations by Factoring
Finding x
If you remember graphing linear equations,
many times the line crossed the x-axis. You
could find the x- and y-intercepts of these
lines.
Polynomials have the same characteristics, but
quadratics can have no x-intercept, one xintercept or two x-intercepts.
x-intercept
y-intercept
x-intercept
x-intercept
y-intercept
Make a table of the polynomial shown below.
y = x2 + 2x – 3
Identify the numbers that appear to be the
x-intercepts.
Rewrite the equation by factoring the right side.
[y = (x + 3)(x – 1)]
If you notice the x-intercept are solutions to the
equation (x + 3)(x – 1) = 0
x-intercept = -3
x-intercept = 1
Remember the Standard Form of a Quadratic
Equation ax2 + bx + c, where a ≠ 0. The value of
the variable in a standard form equation is called
the solution, or the root, of the equation.
Let’s discuss the Multiplication Property of Zero
which states that if a = 0 or b = 0, then ab = 0.
We can use the Zero-Product Property to solve
has been factored into a product of two linear
factors.
Let’s see how we can use this property to solve
Solve (4x + 5)(3x – 2) = 0
Apply the ZeroProduct Property.
(4x + 5)(3x – 2) = 0 The quadratic expression has already been factored.
4x + 5 = 0 or 3x – 2 = 0 If (4x + 5)(x – 2) = 0, then (4x + 5) = 0 or 3x – 2) = 0.
4x = -5
3x = 2
x = -5/4
x = 2/3 Solve each equation for x.
x-intercept
x-intercept
TRY THIS
Solve (x + 5)(2x – 6) = 0
(x + 5)(2x – 6) = 0
x + 5 = 0 or 2x – 6 = 0 Use the Zero-Product Property.
2x = 6 Solve for x.
x = -5 or
x=3
Substitute – 5 for x.
(x + 5)(2x – 6) = 0
(-5 + 5)[2(-5) – 6] = 0
(0)(-16) = 0
Substitute 3 for x.
(x + 5)(2x – 6) = 0
(3 + 5)[2(3) – 6) = 0
(8)(0) = 0
Solve
1) (x + 7)(x – 4) = 0
2) (x – 3)(x – 7) = 0
3) (x + 4)(2x – 9) = 0
4) (2x + 3)(x – 4) = 0
5) (x + 3)(x + 5) = 0
You can also use the Zero-Product Property to solve
equations of the form ax2 + bx + c = 0, if the
quadratic expression ax2 + bx + c can be factored.
Solve x2 – 8x – 48 = 0
(x – 12)(x + 4) = 0
x – 12 = 0 or x + 4 = 0
x = 12 or x = -4
Factor x2 – 8x – 48.
Use the Zero-Product Property.
Solve for x.
Try This x2 + x – 12 = 0
Before solving a quadratic equation, you may
need to add or subtract terms in order to
write the equation in standard form. Then
Solve 2x2 – 5x = 88
2x2 – 5x – 88 = 0
(2x + 11)(x – 8) = 0
2x + 11 = 0 or x – 8 = 0
2x = -11 or x = -8
x = -5.5
Subtract 88 from each side.
Factor 2x2 – 5x – 88.
Use the Zero=Product Property.
Solve for x.
Try This x2 – 12x = -36
Solve
1) b2 + 3b – 4 = 0
2) y2 – 3y – 10 = 0
3) 2z2 – 10z = -12
4) n2 + n – 12 = 0
5) x2 + 8x = -15
6) 4y2 = 25
MORE
Write each equation in standard form. Then solve.
1) 2q2 + 22q = -60
2) 3a2 + 4a = 2a2 – 2a – 9
3) 4x2 + 20 = 10x + 3x2 – 4
4) 3t2 + 8t = t2 – 3t – 12
Solving Cubic Equations
by Factoring
Cubic Equations
ax3 +bx2 + cx + d = 0
The standard form of a cubic equation in x is any
equation that can be written in the form
ax3 + bx2 + cx + d = 0, where a ≠ 0.
Using an extension of the Zero-Product Property,
you can solve many cubic equations.
If a, b, and c represent real numbers and abc = 0,
[(5)(4)(0) = 0]
then a = 0, b = 0 or c = 0.
For example,
if x(x – 2)(3x + 4) = 0, you can write the following.
x = 0 or x – 2 = 0 or 3x + 4 = 0
Solving each for x, then x = 0, 2, and -4/3
Solve 2n3 + 8n2 – 42n = 0
2n3 + 8n2 – 42n = 0 Factor the GCF, 2n, from each term.
2n(n2 + 4n – 21) = 0 Factor n + 4n – 21.
2n(n + 7)(n – 3) = 0 Apply the Zero-Product Property.
2n = 0 or (n + 7) = 0 or (n – 3) = 0 Solve for n.
n = 0 or
n = - 7 or
n=3
2
The solutions are 0, -7, and 3.
Try 10k3 – 13k2 + 4k =0
Solve m3 + 22m2 + 121m = 0
m3 + 22m2 + 121m = 0 Factor for GCF, m, from each term.
m(m2 + 22m + 121) = 0 Factor m + 22m + 121.
m(m + 11)(m + 11) = 0 Apply the Zero-Product Property.
m = 0 or m + 11 = 0 or m + 11 = 0 Solve for n.
m=0
m = -11
m = - 11
2
The solution are 0 and -11.
Try z3 – 14z2 + 49z = 0
The steps for solving a polynomial by factoring
are:
Step 1: Write the equation in standard form.
Step 2: Factor the GCF, if one exists, from each term in
the equation.
Step 3: Factor the polynomial
Step 4: Apply the Zero-Product Property and set each
factor equal to zero.
Step 5: Solve for the variable.
Step 6: Check you solution(s) in the original equation.
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