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# Grade Scale - PowerPoint PPT Presentation

Quiz 3 Results: Average class score after partial credit: __________ Commonly missed questions : #_________________. Grade Scale.

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Presentation Transcript

• Quiz 3 Results:

• Average class score after partial credit: __________

• Commonly missed questions: #_________________

• If you got less than 70% on Quiz 3, make sure to go over your quiz with me or a TA as soon as possible. This material will be covered again on Test 2, as well as on the final exam.

• Four sections on factoring polynomials

• VERY important for Math 120!! (and beyond…)

• Next quiz (Quiz 4) covers sections 5.5-5.8 on factoring and section 8.2 on the quadratic formula.

• After Quiz 4, there will be a review day and then you’ll take the second 100-point test on all material covered since Test 1 (i.e. Test 2 covers Chapter 5 and sections 4.1 and 8.2)

CLOSE

and turn off and put away your cell phones.

Sample

Problems

(Dr. Bruce Johnston)

### Section 5.5

• Note: Today’s lecture will be longer than usual because we will be going over a large variety of examples to help you prepare for doing the homework.

• You are welcome to stay after class to work on your homework in the open lab next door, since you may not get time to get started on it in class today.

• The lab is open every day we have class, from 8:00 in the morning to 6:30 at night.

The Greatest Common Factor

Factoring by Grouping

• Factors

• When an integer is written as a product of 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 – 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 like the work you did in section 5.4 (multiplying polynomials, ) but in reverse:

So if (2x + 1)(x – 5) = 2x2 – 9x - 5 , then 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) 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.

Find the GCF of each list of numbers.

• 12 and 8

12 = 2• 2 • 3

8 = 2 • 2 • 2

So the GCF is 2 • 2 = 4.

• 7 and 20

7 = 1 • 7

20 = 2 • 2 • 5

There are no common prime factors so the GCF is 1.

Find the GCF of each list of numbers.

• 6, 8 and 46

6 = 2• 3

8 = 2 • 2 • 2

46 = 2 • 23

So the GCF is 2.

• 144, 256 and 300

144 = 2• 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.

Find the GCF of 630 and 385

(You are welcome to work with others at your table on this problem.)

• Solution:

• Factors of 630: 2, 3, 3, 5, 7

• Factors of 385: 5, 7, 11

• Common factors: 5, 7

• GCF = 5∙7 = 35

• Solution:

• Factors of 630: 2, 3, 3, 5, 7

• Factors of 385: 5, 7, 11

• Common factors: 5, 7

• GCF = 5∙7 = 35

• 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

1). Find the common factor of the coefficients

2). Find the common variables and powers

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.

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

• 6x5 and 4x3

6x5 = 2• 3 • x• x • x

4x3 = 2• 2 • x• x • x

So the GCF is 2 • x• x • x = 2x3

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.

• Note that a question that says “ of all its terms. 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.

• 2. Factor out the GCF of 2x3 + 10x2 + 4x.

• Answer: 2x (x2 + 5x +2)

Example: of all its terms.

Factor out the GCF in 6x3 – 9x2 + 12x:

SOLUTION:

GCF = 3x

Now divide each term by 3x:

6x3 = 2x2-9x2 = -3x12x = 4

3x 3x 3x

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!!!

Example: of all its terms.

Factor out the GCF in

14x3y + 7x2y – 7xy

GCF = 7xy

Now divide each term by 7xy:

14x3y = 2x27x2y = 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 of all its terms.

Factor out the GCF in each of the following polynomials.

• 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)

• 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!

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.

First, we check all four terms to see if there are any common numbers or variables. There aren’t, but we can still factor it by grouping the four terms into two groups of two:

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)

Factoring a after initially factoring out the GCF.four-term Polynomial by Grouping

• Arrange the terms so that the first two terms have a common factor and the last two terms have a common factor.

• For each pair of terms, use the distributive property to factor out the pair’s greatest common factor.

• If there is now a common binomial factor, factor it out.

• If there is no common binomial factor in step 3, begin again, rearranging the terms differently.

• If no rearrangement leads to a common binomial factor, the polynomial cannot be factored.

Example after initially factoring out the GCF.

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?

Example from today’s homework: after initially factoring out the GCF.

Example after initially factoring out the GCF.

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

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)

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)

Reminder: polynomial should

This homework assignment

on Section 5.5 is due

at the start of

next class period.

(This assignment has 30 problems, and may take you a little longer than usual to complete.)

You may now polynomial should

OPEN