Intermediate Algebra Exam 1 Material Factoring and Applications

Intermediate Algebra

Exam 1 Material

Factoring and Applications

Factoring Numbers

• To factor a number is to write it as a product of two or more other numbers, each of which is called a factor12 = (3)(4) 3 & 4 are factors12 = (6)(2) 6 & 2 are factors12 = (12)(1) 12 and 1 are factors12 = (2)(2)(3) 2, 2, and 3 are factorsIn the last case we say the 12 is “completely factored” because all the factors are prime numbers

Prime Numbers

• Numbers, not including 1, whose only factors are themselves and 1

2, 3, 5, 7, 11, 13, 17, 19, 23, etc.

Factoring Groups of Numbers to Find the Greatest Common Factor

• Factor each number completely

• Construct the Greatest Common Factor by including all factors that are common to all groups

• Example: Find the GCF of: 30, 12, and 18

30 = (2)(3)(5), 12= (2)(2)(3), 18 = (2)(3)(3)

GCF = (2)(3) = 6

Find the GCF of the terms:23242 12 ,27 ,18 yxyxyx

18 2 yx

2427 yx

2312 yx

yxx 332

yyxxxx 333

yyxxx 322

GCF yx23

Polynomial

• A finite sum of terms– Term: a number, a variable, or any product of

numbers and variables

• Examples of polynomials:156 2 xx

8155 342 yxxyyx

63 278 yx

Factoring Polynomials• To factor a polynomial is to write it as a product of two or more

other polynomials, each of which is called a factor• In a sense, factoring is the opposite of multiplying polynomials:

We have learned that:(2x – 3)(3x + 5) = 6x2 + x – 15

If we were asked to factor 6x2 + x – 15 we would write it as: (2x – 3)(3x + 5)

So we would say that (2x – 3) and (3x + 5) are factors of 6x2 + x – 15

Prime Polynomials

• A polynomial is called prime, if it is not 1, and if its only factors are itself and 1

• Just like we learn to identify certain numbers as being prime we will learn to identify certain polynomials as being prime

• We will also completely factor polynomials by writing them as a product of prime polynomials

Importance of Factoring

• If you don’t learn to factor polynomials you can’t pass college algebra or more advanced math classes

• It is essential that you memorize the following procedures and become proficient in using them

5 Steps in Completely Factoring a Polynomial

• (1) Write the polynomial in descending powers of one variable (if there is more than one variable, pick any one you wish)

2x + 3x2 – 1 would be written:3x2 + 2x – 1

3xy2 + y3 + 4x2y – 3 could be written:y3 + 3xy2 + 4x2y – 3 (powers of y)4x2y + 3xy2 + y3 – 3 (powers of x)


• (2) Look at each term of the polynomial to see if every term contains a common factor other than 1, if so, use the distributive property in reverse to place the greatest common factor outside a parentheses and other terms inside parentheses that give a product equal to the original polynomial

• In the previous two examples what was the greatest common factor found in all terms:3x2 + 2x – 1y3 + 3xy2 + 4x2y – 3

GCF out thefactor tonecessary not sit' so ,1

Factoring the Greatest Common Factor from Polynomials

• 9y5 + y2

y2( )

y2(9y3 + 1)

• 6x2t + 8xt + 12t

2t( )

2t(3x2 + 4x + 6)

GCF? theisWhat 2y

GCF? theisWhat 2t

Factoring the Greatest Common Factor from Polynomials

• 14m4(m + 1) – 28m3(m + 1) – 7m2(m + 1)

7m2(m + 1)( )

7m2(m + 1)(2m2 – 4m – 1)

7m2(m + 1)(2m2 – 4m – 1)

GCF? theisWhat 17m2 m


• (3) After factoring out the greatest common factor, look at the new polynomial factors to determine how many terms each one contains

(The fourth step will depend on the number of terms in each of the factors)


• (4) Use the method appropriate to the number of terms in the polynomial:

4 or more terms: “Factor by Grouping”

3 terms: PRIME UNLESS they are of the form “ax2 + bx + c”. If of this form, use “Trial and Error FOIL” or “abc Grouping”

2 terms: Always PRIME UNLESS they are:“difference of squares”: a2 – b2

“difference of cubes”: a3 – b3

“sum of cubes”: a3 + b3

In each of these cases factor by a formula


• (5) Cycle through step 4 as many times as necessary until all factors are “prime”

Factor by Grouping(Used for 4 or more terms)

(1) Group the terms by underlining:

If there are exactly 4 terms try:

2 & 2 grouping, 3 & 1 grouping, or 1 & 3 grouping

If there are exactly 5 terms try:

3 & 2 grouping, or 2 & 3 grouping

Factoring by Grouping(2) Factor each underlined group as if it were a factoring problem by itself

(3) Now determine if the underlined and factored groups contain a common factor,

if they contain a common factor, factor it out

if they don’t contain a common factor, try other groupings, if none work, the polynomial is prime

(4) Once again count the terms in each of the new polynomial factors and return to step 4.

Example of Factoring by Grouping

Factor: ax + ay + 6x + 6y

(1) Group the terms by underlining (start with 2 and 2 grouping): ax + ay + 6x + 6y

(2) Factor each underlined group as if it were a factoring problem by itself:a(x + y) + 6(x + y)

[notice sign between groups gets carried down]

Factoring by Grouping Example Continued

(3) Now determine if the underlined and factored groups contain a common factor, if they do, factor it out:a(x + y) + 6(x + y)

(x + y)(a + 6)

ax + ay + 6x + 6y = (x + y)(a + 6)


Each of these polynomial factors contains two terms, return to step 4 to see if these will factor (SINCE WE HAVE NOT YET DISCUSSED FACTORING POLYNOMIALS WITH TWO TERMS WE WILL NOT CONTINUE AT THIS TIME)

factored Now

Example of Factoring by Grouping

Factor:

(1) Group the terms by underlining (Try 2 and 2 grouping):

(2) Factor each underlined group as if it were a factoring problem by itself:

[notice sign between groups gets carried down and you have to be careful with this sign]

yxyxx 3232 2

yxyxx 3232 2

3232 xyxx


(3) Now determine if the underlined and factored groups contain a common factor, if they do, factor it out:


Each of these polynomial factors contains two terms, return to step 4 to see if these will factor (AGAIN WE HAVE LEARNED TO FACTOR BINOMIALS YET, SO WE WON’T CONTINUE ON THIS EXAMPLE)

factored Now

3232 xyxx

yxx 32

Note on Factoring by Grouping

• It was noted in step 3 of the factor by grouping steps that sometimes the first grouping, or the first arrangement of terms might not result in giving a common factor in each term – in that case other groupings, or other arrangements of terms, must be tried

• Only after we have tried all groupings and all arrangement of terms can we determine whether the polynomial is factorable or prime

Try Factoring by GroupingWithout First Rearranging

Factor:

(1) Group the terms by underlining (Try 2 and 2):


.

49312 xyyx

49312 xyyx

49143 xyyx

13

continue? tong with tryiproblem thesWhat'

groups! underlined twoin thefactor common No

Now Try Same Problemby Rearranging

Factor: Rearrange:

(1) Group the terms by underlining:


.

49312 xyyx

431433 yyx

13 x

now? factoring continue Can we

groups! underlined twoin thefactor common a is thereYes,

43129 yxxy

43129 yxxy


(3) Now factor out the common factor:


Each of these polynomial factors contains two terms, return to step 4 to see if these will factor (AGAIN WE TO WAIT UNTIL WE LEARN TO FACTOR BINOMIALS BEFORE WE CAN CONTINUE)

possible! factoring made gRearrangin

43 y

431433 yyx

1343 xyHELP! ALWAYS TDOESN'

Additional Note onFactoring by Grouping

• In all of our examples we tried this method on polynomials with four terms and used only “2 and 2 grouping”

• Sometimes with four terms we must use “3 and 1 grouping” or “1 and 3 grouping”

• On polynomials with more than four terms other combinations of groupings must be tried

• In any case, all combinations of grouping must be tried before we can determine of the polynomial is factorable or is prime

• We will not deal with any of these situations at this time

Homework Problems

• Section: 5.1

• Page: 334

• Problems: Odd: 1 – 5, 9 – 19, 23 – 57, 59 – 63, 67 – 85

• MyMathLab Homework Assignment 5.1 for practice

• MyMathLab Quiz 5.1 for grade

More on Factoring

• In the overall scheme of “factoring polynomials completely” we need to know how to factor polynomials containing various numbers of terms

• Thus far we have learned that regardless of the number of terms, we should always attempt, as a first step, to factor out the GCF

• We have also learned that for a polynomial with four or more terms we can try the “factor by grouping” method

• We next learn methods for factoring polynomials with three terms (trinomials)

Factoring Trinomials by

Trial and Error FOIL(Used for 3 terms of form ax2 + bx + c)

• Given a trinomial if this form, experiment to try to find two binomials that could multiply to give that trinomial

• Remember that when two binomials are multiplied:

First times First = First Term of Trinomial

Outside times Outside + Inside times Inside = Middle Term of Trinomial

Last times Last = Last Term of Trinomial

Steps in Using Trial and Error FOIL

• Given a trinomial of the form:

• Write two blank parentheses that will each eventually contain a binomial

• Use the idea that “first times first = first” to get possible answers for first term of each binomial

cbxax 2

Continuing Steps in Trial and Error FOIL


• Next use the idea that “last times last = last” to get possible answers for last term of each binomial

cbxax 2

Continuing Steps in Trial and Error FOIL


• Finally use the idea that “Outside times Outside + Inside times Inside = Middle Term of Trinomial” to get the final answer for two binomials that multiply to give the trinomial

cbxax 2

Prime Trinomials

• A trinomial is automatically prime if it is not of the form:

• However, a trinomial of this form is also prime if all possible combinations of “trial and error FOIL” have been tried, and none have yielded the correct middle term

• Example: Why is this prime?• The only possible combinations that give the

correct first and last terms are:

• Neither gives the correct middle term:

cbxax 2

13 13 xxandxx

352 xx

32 32 22 xxandxx

Example of Factoring byTrial and Error FOIL

• Factor: 12x2 + 11x – 5• Using steps on previous slides, we see all the possibilities that give the

correct first and last terms on the left and the result of multiplying them on the right (we are looking for the one that gives the correct middle term):(12x + 1)(x – 5) = 12x2 – 59x – 5 (12x – 1)(x + 5) = 12x2 + 59x – 5 (12x + 5)(x – 1) = 12x2 – 7x – 5(12x – 5)(x + 1) = 12x2 + 7x – 5 (6x + 1)(2x – 5) = 12x2 – 28x – 5 (6x – 1)(2x + 5) = 12x2 +28x – 5 (6x + 5)(2x – 1) = 12x2 + 4x – 5 (6x – 5)(2x + 1) = 12x2 – 4x – 5 (4x + 1)(3x – 5) = 12x2 – 17x – 5 (4x – 1)(3x + 1) = 12x2 + x – 5 (4x + 5)(3x – 1) = 12x2 +11x – 5(4x – 5)(3x + 1) = 12x2 -11x – 5

FactoringCorrect Only

Factoring“Perfect Square Trinomials”

• A trinomial is a “perfect square trinomial” if it has resulted from squaring a binomial:

• Perfect square trinomials have the characteristic that the first and last terms are perfect squares, and the middle term is twice the product of the square roots of the first and last terms

• When this characteristic is seen in a trinomial, we automatically factor it as a binomial squared being careful to place the correct sign in the middle

222222 2 2 bababaandbababa

Example of Factoring“Perfect Square Trinomials”

• Which of these is a perfect square trinomial?

• Only the second has first and last terms that are perfect squares and a middle term that is twice the product of the square roots of the first and last terms

• The second can be factored by writing a binomial, with a square on it, whose first and last terms are the square roots of the first and last terms, that has a middle sign that matches the sign of the middle term of the trinomial

4129 354 22 xxorxx

22 4129 xx 23 x

Homework Problems

• Section: 5.2

• Page: 340

• Problems: Odd: 11 – 39, 43 – 59, 63 – 71



A Second Method of Factoring Trinomials

• While the “Trial and Error FOIL” method can always be used in attempting to factor trinomials, and is usually best when first and last terms have “small coefficients,” there is a second method that is usually best to use when first and last coefficients are “larger”

• We call the second method: “abc grouping”

Factoring Trinomials by

abc Grouping(Used for 3 terms of form ax2 + bx + c)

• When a polynomial is of this form:ax2 + bx + c

(1) Identify “a”, “b”, and “c”(2) Multiply “a” and “c”(3) Find two numbers “m” and “n”, that multiply to

give “ac” and add to give “b” (If this can not be done, the polynomial is already prime)

(4) Rewrite polynomial as: ax2 + mx + nx + c(5) Factor these four terms by 2 and 2 grouping

Example of Factoring byabc Grouping

• Factor: 12x2 + 11x – 5(1) Identify “a”, “b”, and “c”

a = 12, b = 11, c = - 5

(2) Multiply “a” and “c”ac = - 60

(3) Find two numbers “m” and “n”, that multiply to give “ac” and add to give “b” (If this can not be done, the polynomial is already prime)m = 15 and n = - 4, because mn = -60 and m + n = 11

(4) Rewrite as four terms: 12x2 + 15x – 4x – 5

(5) Factor by grouping: 12x2 + 15x – 4x – 5 3x(4x + 5) – 1(4x + 5)(4x + 5)(3x – 1)

60ac

11 nm

4 51

Example of Factoring byabc Grouping (with two variables)

• Factor: 35x2 – 12y2 – 13xy35x2 – 13xy – 12y2 (descending powers of x)

(1) Identify “a”, “b”, and “c” (Ignore y variable)a = 35, b = - 13, c = - 12

(2) Multiply “a” and “c”ac = - 420

(3) Find two numbers “m” and “n”, that multiply to give “ac” and add to give “b” (If this can not be done, the polynomial is already prime)m = 15 and n = - 28, because mn = - 420 and m + n = - 13

(4) Rewrite as four terms: 35x2 + 15xy – 28xy – 12y2

(5) Factor by grouping: 35x2 + 15xy – 28xy – 12y2 5x(7x + 3y) – 4y(7x + 3y)(7x + 3y)(5x – 4y)

Another Comment on Prime Trinomials

• A trinomial is prime if it is not of the form:

• A trinomial of this form is also prime, if it can not be factored by the “abc grouping method”

cbxax 2

Homework Problems

• Section: 5.3

• Page: 347

• Problems: Odd: 21 – 83



More on Factoring

• In the overall scheme of “factoring polynomials completely” we need to know how to factor polynomials containing various numbers of terms

• Thus far we have learned that regardless of the number of terms, we should always attempt, as a first step, to factor out the GCF

• We have also learned that for a polynomial with four or more terms we can try the “factor by grouping” method

• We have also learned that we should try to factor trinomials of the form ax2+bx+c by either “trial and error FOIL” or “abc grouping”

• We next learn methods for factoring binomials

Note on Factoring Binomials

• Binomials are factorable only if they are a:– Difference of Squares:– Difference of Cubes:– Sum of Cubes:

• In each of these cases, factoring is done by means of a formula that needs to be memorized

• All other binomials are prime (In saying this, we assume that any GCF has already been factored out)

22 ba 33 ba 33 ba

Factoring Binomials by Formula

• Factor by using formula appropriate for the binomial:

“difference of squares”:a2 – b2 = (a – b)(a + b)

“difference of cubes”:a3 – b3 = (a – b)(a2 + ab + b2) Trinomial is prime

“sum of cubes”:a3 + b3 = (a + b)(a2 – ab + b2) Trinomial is prime

• If none of the formulas apply, the binomial is primeBINOMIALS ARE PRIME UNLESS THEY ARE ONE OF THESE

Example of Factoring Binomials

• Factor: 25x2 – 9y2

• Note that this binomial is a difference of squares:

(5x)2 – (3y)2

• Using formula gives:

(5x – 3y)(5x + 3y)


• Factor: 8x3 – 27

• Note that this is a difference of cubes:

(2x)3 – (3)3

• Using formula gives:

(2x – 3)(4x2 + 6x + 9)


• Factor: 4x2 + 9

• Note that this is not a difference of squares, difference of cubes, or sum of cubes, therefore it is prime

• (4x2 + 9)

• To show factoring of a polynomial that is prime, put it inside parentheses

Homework Problems

• Section: 5.4

• Page: 356

• Problems: Odd: 7 – 29, 59 – 81



Completely Factoring Polynomials

• We now have all the skills necessary to either factor polynomials completely, or to determine if they are prime

• Summary of steps:– Arrange polynomial in descending powers of one variable– Factor out the GCF (also factor out a negative if highest degree

term has a negative coefficient)– For each polynomial factor in the expression, try to factor it by

using the method appropriate for the number of terms it has– Continue factoring each new polynomial factor until all

polynomial factors are prime

• We now apply this procedure in completely factoring polynomials

Example of Factoring Polynomials Using Five Steps

• Factor: 2x3 – 8x + 2x6 – 8x4

• (1) Write the polynomial in descending powers of one variable (if there is more than one variable, pick any one you wish)2x6 – 8x4 + 2x3 – 8x

• (2) Look at each term of the polynomial to see if every term contains a common factor, if so, use the distributive property in reverse to place the greatest common factor outside a parentheses2x(x5 – 4x3 + x2 – 4)

Example of Factoring Polynomials Using Five Steps

2x(x5 – 4x3 + x2 – 4)• (3) After factoring out the greatest common factor, look at the

new polynomial factors to determine how many terms each one contains

• The polynomial in parentheses has 4 terms• (4) Use the method appropriate to the number of terms in the

polynomialSince there are 4 terms we will try “factor by grouping”: x5 – 4x3 + x2 – 4x3(x2 – 4) + 1(x2 – 4)(x2 – 4)(x3 + 1)

So far we have factored the original polynomial as:2x(x2 – 4)(x3 + 1)

Example of Factoring Polynomials Using Five Steps 2x(x2 – 4)(x3 + 1)

• (5) Cycle through step 4 as many times as necessary until all factors are “prime” (count terms and use appropriate method)The first binomial is a difference of squares, and the second is a sum of cubes so they must be factored by formulas to get the final complete factoring of:

• 2x(x – 2)(x + 2)(x + 1)(x2 – x + 1)COMPLETELY FACTORED!

Homework Problems

• Section: Summary Exercises on Factoring

• Page: 358


• MyMathLab Homework Factoring Summary Assignment for practice

• MyMathLab Quiz Factoring Summary for grade

Solving Equations

• You have previously learned to identify and solve linear equations

• The next objective is to learn to identify and solve “quadratic equations”

• Before beginning the new goal, we will quickly review identifying and solving linear equations

Linear Equations in One Variable

• Linear Equation: any polynomial equation in one variable where, after parentheses are gone, the highest degree term is “1”

• Examples:

1253 xxx

xxx4

37.14

Solving Linear Equations

• Get rid of parentheses• Get rid of fractions and decimals by multiplying

both sides by LCD• Collect like terms• Decide which side will keep variable terms and

get rid of variable terms on other side• Get rid of non-variable terms on variable side• Divide both sides by the coefficient of variable

Solve the Equation

• Identify the type of equation:

• Get rid of parentheses:

• Get rid of fractions and decimals by multiplying both sides by LCD:

2

17.

3

22

xxx

linear! isIt

2

17.

3

42 xxx

:is 2 and 10, 3, of LCD 30

2

1307.

3

4230 xxx

1530214060 xxx

Example Continued

• Collect like terms:

• Decide which side will keep variable terms and get rid of variable terms on other side:

• Get rid of non-variable terms on variable side:

• Divide both sides by coefficient of variable:

1530214060 xxx

15304039 xx

right on the thoseof ridget you willleft on variableskeep tochoseyou If15409 x

559 x

9

55x

Identifying Quadratic Equations

• Technical Definition: any equation in one variable that can be written in the form: ax2 + bx + c = 0 where “a”, “b”, and “c” are real and a ≠ 0 (This form is called the “standard form”)

• Practical Definition: any polynomial equation in one variable where, after parentheses are gone, the highest degree term is “2”

• Examples:5x2 + 7 = – 4x 9x2 = 42x(x – 3) = x – 1

form? standardin theseofany Areform. standardin put be could allbut No,

Solving Quadratic Equations by Zero Factor Method

• Put equation in standard form (one side zero other side in descending powers)

• Factor non-zero side(If it won’t factor this method won’t work!)

• Use zero factor property that says, “if two numbers multiply to get zero, one of them is zero:”ab = 0 if and only if a = 0 or b = 0

• Set each factor equal to zero• Solve resulting equations

Solving byZero Factor Method

Put in standard form:

Factor non-zero side:

Apply zero factor principle:

Solve the equations:

352 xx

352 2 xx0352 2 xx

0312 xx

12 x

03 x012 x

2

1x

3x

OR

More Notes onSolving Quadratic Equations

• All quadratic equations can be solved, but the “zero factor method” works only when the non-zero side of the quadratic equation can be factored

• In college algebra we will learn methods of solving quadratic equations that can not be solved by the “zero factor method”

• Example: Why can’t this be solved by the zero factor method? 0252 xx

factored bet can' and prime is 252 xx

Solving Polynomial Equations with Degree Higher than 2

• We have said that a quadratic equation is a second degree polynomial equation and that such an equation can sometimes be solved by the “Zero Factor Method”

• It is also true that some higher degree polynomial equations may be solved by this method

• To attempt this method:– Make one side zero– Factor other side– Apply Zero Factor Property (abcd=0 means a = 0,

b = 0, c = 0, or d = 0)

Example

• Solve: xxx 253 23 0253 23 xxx

0253 2 xxx

0213 xxx

02 013 0 xorxorx

13 x 2x

3

1x

Homework Problems

• Section: 5.5

• Page: 366

• Problems: Odd: 11 – 23, 27 – 81



Application Problems Involving Quadratic Equations

• In previous algebra courses you have learned to solve basic application (word) problems

• In this course we learn that some application problems translate to quadratic equations

• Before introducing those type of problems, we will review the basic approach to solving application problems

Application Problems

• General methods for solving an applied (word) problem:

1. Read problem carefully taking notes, drawing pictures, thinking about formulas that apply, making charts, etc.

2. Read problem again to make a “word list” of everything that is unknown

3. Give a variable name, such as “x” to the “most basic unknown” in the list (the thing that you know the least about)

4. Give all other unknowns in your word list an algebraic expression name that includes the variable, “x”

5. Read the problem one last time to determine what information has been given, or implied by the problem, that has not been used in giving an algebra name to the unknowns and use this information to write an equation about the unknowns

6. Solve the equation and answer the original question

Solve the Application Problem

A 31 inch pipe needs to be cut into three pieces in such a way that the second piece is 5 inches longer than the first piece and the third piece is twice as long as the second piece. How long should the third piece be?

1. Read the problem carefully taking notes, drawing pictures, thinking about formulas that apply, making charts, etc.Perhaps draw a picture of a pipe that is labeled as 31 inches with two cut marks dividing it into 3 pieces labeled first, second and third

1st 2nd 3rd

31

Example Continued

2. Read problem again to make a “word list” of everything that is unknown

What things are unknown in this problem?

The length of all three pieces (even though the problem only asked for the length of the third).

Word List of Unknowns:

Length of first

Length of second

Length of third

Example Continued

3. Give a variable name, such as “x” to the “most basic unknown” in the list (the thing that, if you knew its value, the other values could be found)

What is the most basic unknown in this list?

Length of first piece is most basic, because problem describes second in terms of the first, and third in terms of second

Give the name “x” to the length of first

Example Continued

4. Give all other unknowns in the word list an algebraic expression name that includes the variable, “x”How would the length of the second be named?x + 5How would the length of the third be named?2(x + 5)Word List of Unknowns: Algebra Names:Length of first xLength of second x + 5Length of third 2(x + 5)

Example Continued

5. Read the problem one last time to determine what information has been given, or implied by the problem, that has not been used in giving an algebra name to the unknowns and use this information to write an equation about the unknownsWhat other information is given in the problem that has not been used?Total length of pipe is 31 inchesHow do we say, by using the algebra names, that the total length of the three pieces is 31?x + (x + 5) + 2(x + 5) = 31

Example Continued

6. Solve the equation and answer the original questionThis is a linear equation so solve using the appropriate steps:x + (x + 5) + 2(x + 5) = 31 x + x + 5 + 2x + 10 = 31 4x + 15 = 31 4x = 16 x = 4Is this the answer to the original question?No, this is the length of the first piece.How do we find the length of the third piece?The length of the third piece is 2(x + 5):2(4 + 5) = (2)(9) = 18 inches = length of third piece

Example

A rectangular piece of metal is 2 inches longer than it is wide. Four inch squares are cut from each corner to make a box with a volume of 32 cubic inches. What were the original dimensions of the metal?

UnknownsL RecW RecL BoxW Box .

x2x

x

2x

4

4

4

4

82 x8x

4 Box Height

LWHV 48632 xx 4481432 2 xx

19256432 2 xx1605640 2 xx

40140 2 xx 4100 xx

04 OR 010 xx

4x

10x

Impossible

.12

.10

inL

inW

Example

The product of two consecutive odd integers is equal to negative one minus the sum of the two integers. Find all possible answers for the two integers.

UnknownsFirst Odd IntNext Odd Int

.

x2x

212 xxxx

031 xx

03 OR 01 xx

1x12 x 22122 xxx

22122 xxx3222 xxx

0342 xx

3x

12 x

Homework Problems

• Section: 5.6

• Page: 375




Documents

Intermediate Algebra Exam 1 Material Factoring and Applications