Today’s daily quiz will be given after today’s lecture, during the last 10 minutes of class. 1.

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Today’s daily quiz will be given after today’s lecture, during the last 10 minutes of class. 1

Transcript of Today’s daily quiz will be given after today’s lecture, during the last 10 minutes of class. 1.

Page 1: Today’s daily quiz will be given after today’s lecture, during the last 10 minutes of class. 1.

Today’s daily quiz will be given after today’s lecture,

during the last 10 minutes of class.

1

Page 2: Today’s daily quiz will be given after today’s lecture, during the last 10 minutes of class. 1.

Please

CLOSE

YOUR LAPTOPS,and turn off and put away your

cell phones,

and get out your note-taking materials.

Page 3: Today’s daily quiz will be given after today’s lecture, during the last 10 minutes of class. 1.

Section 6.1Introduction to Factoring Polynomials:

Greatest Common Factor (GCF), Factoring by Grouping

Page 4: Today’s daily quiz will be given after today’s lecture, during the last 10 minutes of class. 1.

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

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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)

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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.

Page 7: Today’s daily quiz will be given after today’s lecture, during the last 10 minutes of class. 1.

Example

Find the GCF of each list of numbers.

1) 12 and 812 = 2 • 2 • 3 8 = 2 • 2 • 2So the GCF is 2 • 2 = 4.

2) 7 and 20 7 = prime (can’t be broken down further)20 = 2 • 2 • 5There are no common prime factors so the

GCF is 1.

Page 8: Today’s daily quiz will be given after today’s lecture, during the last 10 minutes of class. 1.

Example

Find the GCF of 6, 8 and 46.

. 6 = 2 • 3 8 = 2 • 2 • 246 = 2 • 23

So the GCF is 2.

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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

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Example from today’s homework:

y2z3

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Finding the GCF of a list of terms:

1). Find the common factor of the coefficients

2). Find the common variables and powers

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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.

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1) 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

Example

Find the GCF of each list of terms.

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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.

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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!!!

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• 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)

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Example:

GCF = 7xy

Now divide each term by 7xy:

14x3y = 2x2 7x2y = x -7xy = -1 7xy 7xy 7xy

ANSWER: 7xy(2x2 + x – 1)

Factor out the GCF in 14x3y + 7x2y – 7xy

NOW CHECK: Multiply back out using the distributive property and see if you get back to the original polynomial.

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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)

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)

How would you check this?

Now check this!

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• 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 separately:

xy + y = y(x+1)

2x + 2 = 2(x+1)

So then y(x + 1) + 2(x + 1) = (x + 1)(y + 2)

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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!)

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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)

Example Factor x3 + 4x + x2 + 4 by grouping.

How would you check this?

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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. 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) Now check this answer!

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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)

Example Factor 2x – 9y + 18 – xy by grouping.

Now check this answer!

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Warning: Some polynomials are PRIME.

This means that they can’t be factored, just like a PRIME NUMBER.

Example: 2ac2 + 2a2c – 3c + cStep 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.

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The assignment on this material

(HW 6.1) is due at the start of the next class session.

You can now open your laptops and work on this assignment until it’s time to take the quiz.

(You’ll have 9 minutes to take the quiz, so that you will have time to check your answers.)

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Please open your laptops, log in to the MyMathLab course web site, and open Quiz

5.6B.

You may use the pink formula sheet on this quiz – please don’t write on this sheet, and remember to hand it back in with your quiz answer sheet.

You have 9 minutes to take the quiz, so take time to

check your answers.