Transcript Sec. 6.1: Factoring polynomials: GCF, grouping

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Section 6.1
Introduction to Factoring Polynomials:
Greatest Common Factor (GCF),
Factoring by Grouping
Review of factoring
integer numbers:
•
Factors
• When an integer is written as a product of prime integers, each
of the integers in the product is a factor of the original number.
• Example 1: Factor 18 into a product of primes
Solution: 18 = 2*9 = 2*3*3 = 2*32
• Example 2: Factor 420 into a product of primes
Solution: 420 = 10*42 = 2*5*2*21 = 2*5*2*3*7
= 22*3*5*7
Factoring Polynomials:
• Factoring – writing a polynomial as a product of
polynomials. When a polynomial is written as a
product of polynomials, each of the polynomials in the
product is a factor of the original polynomial.
• The process of factoring is similar to the work you did in
section 5.3 (multiplying polynomials, ) but in reverse:
Multiplication: (2x + 1)(x – 5) = 2x2 – 9x - 5
Therefore the factoring of 2x2 – 9x – 5 is (2x + 1)(x – 5)
The first step in factoring a polynomial is to
always look for the Greatest Common
Factor – the largest quantity that is a factor
of all the terms of the polynomials involved.
Finding the GCF of a List of Integers:
1) Factor each integer into its prime factors.
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
12 = 2 • 2 • 3
8=2•2•2
So the GCF is 2 • 2 = 4.
2) 7 and 20
7 = prime (can’t be broken down further)
20 = 2 • 2 • 5
There are no common prime factors so the
GCF is 1.
Example
Find the GCF of 6, 8 and 46.
.
6=2•3
8=2•2•2
46 = 2 • 23
So the GCF is 2.
Finding the GCF of a list of variables:
1). Identify common variables (letters).
2). Look for the smallest power of that variable
in your list.
Examples:
1. Find the GCF of x3 and x7
x3 = x • x • x
x7 = x • x • x • x • x • x • x
So the GCF is x • x • x = x3
2. Find the GCF of a2b7c and b2c3d5
Answer: b2c1
Example from today’s homework:
y2z3
Finding the GCF of a list of terms:
1). Find the common factor of the coefficients
2). Find the common variables and powers
Example
Find the GCF of the following list of terms.
a3b2, a2b5 and a4b7
• The coefficient of all three terms is 1, so the
common coefficient is 1 and need not be written.
• All of the terms contain an “a” term and a “b”
term, so the look for the smallest power of each
variable:
The GCF is = a2b2
Notice that the GCF of terms containing variables will
use the smallest exponent found among the
individual terms for each variable.
Example
Find the GCF of each list of terms.
30x3y2 and 45x7y
30x3y2 = 2 • 3 • 5 • x • x • x • y • y
45x7y= 3 • 3• 5 • x • x • x • x • x • x • x • y
So the GCF is 3 • 5 • x • x • x •y = 15x3y
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)
The first step in factoring any kind of
polynomial ALWAYS is to see if you can find
a GCF (other than 1) of all its terms.
If we can find such a GCF, 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 this is written in parentheses after
the GCF.
Example: Factor out the GCF in 6x3 – 9x2 + 12x:
SOLUTION:
GCF = 3x
Now divide each term by 3x:
6x3 = 2x2
-9x2 = -3x
12x = 4
3x
3x
3x
ANSWER: 3x(2x2 – 3x + 4)
HOW WOULD YOU CHECK THIS ANSWER????
Multiply back out using the distributive property
and see if you get back to the original polynomial.
ALWAYS DO THIS!!!
• Note that a question that says “find the GCF” of a
list of terms only requires typing in the GCF for the
answer.
• Problems that say “factor out the GCF” from a
polynomial require you to write the GCF followed by
another polynomial in parentheses.
Examples:
1. Find the GCF of 2x3, 10x2, and 4x.
Answer: 2x
2. Factor out the GCF of 2x3 + 10x2 + 4x.
Answer: 2x (x2 + 5x +2)
Example:
Factor out the GCF in 14x3y + 7x2y – 7xy
GCF = 7xy
Now divide each term by 7xy:
14x3y = 2x2 7x2y = x -7xy = -1
7xy
7xy
7xy
ANSWER: 7xy(2x2 + x – 1)
NOW CHECK: Multiply back out using
the distributive property and see if you get
back to the original polynomial.
Example
Factor out the GCF in each of the following
polynomials.
1) 6(x + 2) – y(x + 2):
They both have an (x+2), so that’s the common
factor. Pull that part out and just see what’s left:
(x + 2)(6 – y) How would you check this?
2) xy(y + 1) – (y + 1):
First write the – as a -1:
xy(y + 1) – 1(y + 1)
Now pull out the (y + 1) from both parts:
(y + 1)(xy – 1) Now check this!
•
•
Factoring polynomials often involves additional techniques
after initially factoring out any GCF.
One technique is factoring by grouping., which is especially
useful for 4-term polynomials.
Example: Factor xy + y + 2x + 2 by grouping.
First, we check all four terms to see if there are any common
numbers or variables other than 1.
There aren’t, but we can still factor it by grouping the four terms
into two groups of two:
xy + y + 2x + 2 = xy + y + 2x + 2
Now look for the common factor in each pair:
xy + y = y(x+1)
2x + 2 = 2(x+1)
So then y(x + 1) + 2(x + 1) = (x + 1)(y + 2)
Recap:
Problem: Factor xy + y + 2x + 2 by grouping.
Answer: (x + 1)(y + 2)
How would you check this answer?
You should always check answers of factoring problems
by multiplying the factors back out to see if you get back
to the original polynomial given in the problem.
(Yes, you do have the check answer button on homework
problems, but remember you won’t have that on quizzes
and tests!)
Example
Factor x3 + 4x + x2 + 4 by grouping.
SOLUTION:
• First, look for a GCF. (Always do this first!)
• There isn’t one, so now separate the four terms into two
groups of two:
x3 + 4x + x2 + 4
• Now factor each pair:
x3 + 4x = x(x2 + 4)
x2 + 4 = 1(x2 + 4)
• Now rewrite the groups and pull out the common factor:
x(x2 + 4) + 1(x2 + 4) = (x2 + 4)(x + 1)
How would you check this?
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.
All of the coefficients are divisible by 3, so first factor out the 3:
90 + 15y2 – 18x – 3xy2 = 3(30 + 5y2 – 6x – xy2)
Now factor the part inside by grouping:
5(6 + y2) – x (6 + y2)
(6 + y2)(5 – x)
Don’t forget to include the GCF in your final answer:
ANSWER: 3(6 + y2)(5 – x)
Example
Factor 2x – 9y + 18 – xy by grouping.
Neither pair has a common factor (other than 1).
Don’t give up yet: try rearranging the order of the
factors.
2x + 18 – 9y – xy
Now factor each pair:
2x + 18 = 2(x + 9)
-9y – xy = -y(9 + x)
This gives 2(x + 9) – y(9 + x), but the factors don’t
look the same.
Note that (x + 9) and (9 + x) are really the same thing,
so we can write it as 2(x + 9) – y(x + 9)
Now factor out the (x + 9) to get (x + 9)(2 – y)
Now check this answer!
Warning:
Some polynomials are PRIME.
This means that they can’t be factored, just like a PRIME NUMBER.
Example: 2ac2 + 2a2c – 3c + c
Step 1: Does this polynomial have a GCF that can be
pulled out of all four factors?
Answer: NO (ALWAYS check this first!!)
Step 2: Next, try factoring by grouping, including
rearranging the terms if it doesn’t work in the order given.
Result: You’ll find that there’s no way to arrange
these terms so that you can get the same binomial factor
from each group. Therefore this polynomial is PRIME.
Expect this assignment (HW 32) to take at
least an hour and a half, so don’t leave it
to the last minute!
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