Transcript Math 50 - North Carolina Central University

MATH 1000 /11
Chapter 6
6. 1. Greatest Common Factor and Factoring
by Grouping
6. 2. Factoring Trinomials of the form a
x2 + bx + c
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• 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.
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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.
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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.
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Example
Find the GCF of each list of numbers.
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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.
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Example
Find the GCF of each list of terms.
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 • x • x
4x3 = 2 • 2 • x • x • x
So the GCF is 2 • x • x • x = 2x3
1)
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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.
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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.
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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)
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Example
Factor out the GCF in each of the following
polynomials.
1) 6(x + 2) – y(x + 2) =
6 • (x + 2) – y • (x + 2) =
(x + 2)(6 – y)
2) xy(y + 1) – (y + 1) =
xy • (y + 1) – 1 • (y + 1) =
(y + 1)(xy – 1)
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Factoring polynomials often involves additional
techniques after initially factoring out the GCF.
One technique is factoring by grouping.
Example
Factor xy + y + 2x + 2 by grouping.
Notice that, although 1 is the GCF for all four
terms of the polynomial, the first 2 terms have a
GCF of y and the last 2 terms have a GCF of 2.
xy + y + 2x + 2 = x • y + 1 • y + 2 • x + 2 • 1 =
y(x + 1) + 2(x + 1) =
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Factoring a four-term Polynomial by Grouping
1) Arrange the terms so that the first two terms have
a common factor and the last two terms have a
common factor.
2) For each pair of terms, use the distributive
property to factor out the pair’s greatest common
factor.
3) If there is now a common binomial factor, factor
it out.
4) If there is no common binomial factor in step 3,
begin again, rearranging the terms differently.
•
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If no rearrangement leads to a common binomial
factor, the polynomial cannot be factored.
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Example
Factor each of the following polynomials by grouping.
1) x3 + 4x + x2 + 4 = x • x2 + x • 4 + 1 • x2 + 1 • 4 =
x(x2 + 4) + 1(x2 + 4) =
(x2 + 4)(x + 1)
2) 2x3 – x2 – 10x + 5 = x2 • 2x – x2 • 1 – 5 • 2x – 5 • (-1) =
x2(2x – 1) – 5(2x – 1) =
(2x – 1)(x2 – 5)
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Example
Factor 2x – 9y + 18 – xy by grouping.
Neither pair has a common factor (other than 1).
So, rearrange the order of the factors.
2x + 18 – 9y – xy = 2 • x + 2 • 9 – 9 • y – x • y =
2(x + 9) – y(9 + x) =
2(x + 9) – y(x + 9) =
(make sure the factors are identical)
(x + 9)(2 – y)
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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)) =
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Section 6.2
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.
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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).
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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
-11
So x2 – 11x + 24 = (x – 3)(x – 8).
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Example
Factor the polynomial x2 – 2x – 35.
Since our two numbers must have a product of -35 and a
sum of -2, the two numbers will have to have
different signs.
Factors of -35
Sum of Factors
-1, 35
34
1, -35
-34
-5, 7
2
5, -7
-2
So x2 – 2x – 35 = (x + 5)(x – 7).
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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
Now we have a problem, because we have exhausted
all possible choices for the factors, but have not found
a pair whose sum is -6.
2 – 6x +10 is not factorable and we call it a prime
So
x
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polynomial.
Example
Factor the polynomial x2 – 11xy + 30y2.
We look for two terms whose product is 30y2 and whose
sum is –11y. The two terms will have to both be
negative.
Note: each term will contain the variable y, for the sum
to be –11y.
Negative factors of 30y2 Sum of Factors
-y, -30y
-31y
-2y, -15y
-17y
-3y, -10y
-13y
-5y, -6y
-11y
So x2 – 11xy + 30y2 = (x – 5y)(x – 6y).
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Example
Factor the polynomial 3x6 + 30x5 + 72x4.
First we factor out the GCF.
3x6 + 30x5 + 72x4 = 3x4(x2 + 10x + 24)
Then we factor the trinomial.
Positive factors of 24
Sum of Factors
1, 24
25
2, 12
14
3, 8
11
4, 6
10
So 3x6 + 30x5 + 72x4 = 3x4(x2 + 10x + 24)
= 3x4(x + 4)(x + 6).
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• You should always check your factoring
results by multiplying the factored
polynomial to verify that it is equal to the
original polynomial.
• Many times you can detect computational
errors or errors in the signs of your numbers
by checking your results.
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