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 factor 12 = (3)(4)3 & 4 are factors 12 = (6)(2)6 & 2 are factors 12 = (12)(1)12 and 1 are factors 12 = (2)(2)(3)2, 2, and 3 are factors In 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 = (2)(3)(5), 12= (2)(2)(3), 18 = (2)(3)(3) GCF = (2)(3) = 6

Find the GCF of the terms:

Polynomial A finite sum of terms –Term: a number, a variable, or any product of numbers and variables Examples of polynomials:

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) = 6x 2 + x – 15 If we were asked to factor 6x 2 + x – 15 we would write it as: (2x – 3)(3x + 5) So we would say that (2x – 3) and (3x + 5) are factors of 6x 2 + 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 + 3x 2 – 1 would be written: 3x 2 + 2x – 1 3xy 2 + y 3 + 4x 2 y – 3 could be written: y 3 + 3xy 2 + 4x 2 y – 3 (powers of y) 4x 2 y + 3xy 2 + y 3 – 3 (powers of x)

5 Steps in Completely Factoring a Polynomial (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: 3x 2 + 2x – 1 y 3 + 3xy 2 + 4x 2 y – 3

Factoring the Greatest Common Factor from Polynomials 9y 5 + y 2 y 2 ( ) y 2 (9y 3 + 1) 6x 2 t + 8xt + 12t 2t( ) 2t(3x 2 + 4x + 6)

Factoring the Greatest Common Factor from Polynomials 14m 4 (m + 1) – 28m 3 (m + 1) – 7m 2 (m + 1) 7m 2 (m + 1)( ) 7m 2 (m + 1)(2m 2 – 4m – 1)

5 Steps in Completely Factoring a Polynomial (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)

5 Steps in Completely Factoring a Polynomial (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 “ax 2 + bx + c”. If of this form, use “Trial and Error FOIL” or “abc Grouping” 2 terms: Always PRIME UNLESS they are: “difference of squares”:a 2 – b 2 “difference of cubes”:a 3 – b 3 “sum of cubes”: a 3 + b 3 In each of these cases factor by a formula

5 Steps in Completely Factoring a Polynomial (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 itout 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) (4) Once again count the terms in each of the new polynomial factors and return to step 4. 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)

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]

Factoring by Grouping Example Continued (3)Now determine if the underlined and factored groups contain a common factor, if they do, factor it out: (4) Once again count the terms in each of the new polynomial factors and return to step 4. 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)

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 Grouping Without First Rearranging Factor: (1) Group the terms by underlining (Try 2 and 2): (2)Factor each underlined group as if it were a factoring problem by itself:.

Now Try Same Problem by Rearranging Factor: Rearrange: (1) Group the terms by underlining: (2)Factor each underlined group as if it were a factoring problem by itself:.

Factoring by Grouping Example Continued (3)Now factor out the common factor: (4) Once again count the terms in each of the new polynomial factors and return to step 4. 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)

Additional Note on Factoring 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 ax 2 + 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

Continuing Steps in Trial and Error FOIL Given a trinomial of the form: Next use the idea that “last times last = last” to get possible answers for last term of each binomial

Continuing Steps in Trial and Error FOIL Given a trinomial of the form: 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

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:

Example of Factoring by Trial and Error FOIL Factor:12x x – 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) = 12x 2 – 59x – 5 (12x – 1)(x + 5) = 12x x – 5 (12x + 5)(x – 1) = 12x 2 – 7x – 5 (12x – 5)(x + 1) = 12x 2 + 7x – 5 (6x + 1)(2x – 5) = 12x 2 – 28x – 5 (6x – 1)(2x + 5) = 12x 2 +28x – 5 (6x + 5)(2x – 1) = 12x 2 + 4x – 5 (6x – 5)(2x + 1) = 12x 2 – 4x – 5 (4x + 1)(3x – 5) = 12x 2 – 17x – 5 (4x – 1)(3x + 1) = 12x 2 + x – 5 (4x + 5)(3x – 1) = 12x 2 +11x – 5 (4x – 5)(3x + 1) = 12x 2 -11x – 5

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

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

Homework Problems Section: 5.2 Page:340 Problems:Odd: 11 – 39, 43 – 59, 63 – 71 MyMathLab Homework Assignment 5.2 for practice MyMathLab Quiz 5.2 for grade

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 ax 2 + bx + c) When a polynomial is of this form: ax 2 + 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: ax 2 + mx + nx + c (5)Factor these four terms by 2 and 2 grouping

Example of Factoring by abc Grouping Factor:12x x – 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:12x x – 4x – 5 (5)Factor by grouping:12x x – 4x – 5 3x(4x + 5) – 1(4x + 5) (4x + 5)(3x – 1)

Example of Factoring by abc Grouping (with two variables) Factor:35x 2 – 12y 2 – 13xy 35x 2 – 13xy – 12y 2 (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 = (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 = and m + n = - 13 (4)Rewrite as four terms:35x xy – 28xy – 12y 2 (5)Factor by grouping:35x xy – 28xy – 12y 2 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”

Homework Problems Section: 5.3 Page:347 Problems:Odd: 21 – 83 MyMathLab Homework Assignment 5.3 for practice MyMathLab Quiz 5.3 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 have also learned that we should try to factor trinomials of the form ax 2 +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)

Factoring Binomials by Formula Factor by using formula appropriate for the binomial: “difference of squares”: a 2 – b 2 = (a – b)(a + b) “difference of cubes”: a 3 – b 3 = (a – b)(a 2 + ab + b 2 ) Trinomial is prime “sum of cubes”: a 3 + b 3 = (a + b)(a 2 – ab + b 2 ) Trinomial is prime If none of the formulas apply, the binomial is prime BINOMIALS ARE PRIME UNLESS THEY ARE ONE OF THESE

Example of Factoring Binomials Factor:25x 2 – 9y 2 Note that this binomial is a difference of squares: (5x) 2 – (3y) 2 Using formula gives: (5x – 3y)(5x + 3y)

Example of Factoring Binomials Factor:8x 3 – 27 Note that this is a difference of cubes: (2x) 3 – (3) 3 Using formula gives: (2x – 3)(4x 2 + 6x + 9)

Example of Factoring Binomials Factor:4x Note that this is not a difference of squares, difference of cubes, or sum of cubes, therefore it is prime (4x 2 + 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 MyMathLab Homework Assignment 5.4 for practice MyMathLab Quiz 5.4 for grade

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:2x 3 – 8x + 2x 6 – 8x 4 (1) Write the polynomial in descending powers of one variable (if there is more than one variable, pick any one you wish) 2x 6 – 8x 4 + 2x 3 – 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 parentheses 2x(x 5 – 4x 3 + x 2 – 4)

Example of Factoring Polynomials Using Five Steps 2x(x 5 – 4x 3 + x 2 – 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 polynomial Since there are 4 terms we will try “factor by grouping”: x 5 – 4x 3 + x 2 – 4 x 3 (x 2 – 4) + 1(x 2 – 4) (x 2 – 4)(x 3 + 1) So far we have factored the original polynomial as: 2x(x 2 – 4)(x 3 + 1)

Example of Factoring Polynomials Using Five Steps 2x(x 2 – 4)(x 3 + 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)(x 2 – x + 1) COMPLETELY FACTORED!

Homework Problems Section: Summary Exercises on Factoring Page:358 Problems:Odd: 1 – 79 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:

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:

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:

Identifying Quadratic Equations Technical Definition: any equation in one variable that can be written in the form: ax 2 + 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: 5x = – 4x 9x 2 = 4 2x(x – 3) = x – 1

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 by Zero Factor Method Put in standard form: Factor non-zero side: Apply zero factor principle: Solve the equations:

More Notes on Solving 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?

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:

Homework Problems Section: 5.5 Page:366 Problems:Odd: 11 – 23, 27 – 81 MyMathLab Homework Assignment 5.5 for practice MyMathLab Quiz 5.5 for grade

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

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 + 5 How would the length of the third be named? 2(x + 5) Word List of Unknowns:Algebra Names: Length of firstx Length of secondx + 5 Length of third2(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 unknowns What other information is given in the problem that has not been used? Total length of pipe is 31 inches How 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 question This is a linear equation so solve using the appropriate steps: x + (x + 5) + 2(x + 5) = 31 x + x x + 10 = 31 4x + 15 = 31 4x = 16 x = 4 Is 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? Unknowns L Rec W Rec L Box W Box.

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. Unknowns First Odd Int Next Odd Int.

Homework Problems Section: 5.6 Page:375 Problems:Odd: 7 – 29 MyMathLab Homework Assignment 5.6 for practice MyMathLab Quiz 5.6 for grade