Eureka Math Homework Helper 2015–2016 Algebra II Module 1
Eureka Math Homework Helper2015–2016Algebra IIModule 1Lessons 1–40Eureka Math, A Story of Functions Published by the non-profit Great Minds.Copyright 2015 Great Minds. No part of this work may be reproduced, distributed, modified, sold, orcommercialized, in whole or in part, without consent of the copyright holder. Please see our User Agreement formore information. “Great Minds” and “Eureka Math” are registered trademarks of Great Minds.
20M1-16A Story of Functions15Homework HelperALGEBRA IILesson 1: Successive Differences in PolynomialsInvestigate Successive Differences in SequencesCreate a table to find the third differences for the polynomial 45𝑥𝑥 10𝑥𝑥 2 3𝑥𝑥 3 for integer values of 𝑥𝑥 from0 to 4.I calculate first differences by subtracting theoutputs for successive inputs. The seconddifferences are the differences of successive firstdifferences, and the third differences are differencesbetween successive second differences.𝑥𝑥First 𝟐𝟐𝟐𝟑𝟑𝟑𝟑 𝟎𝟎 𝟑𝟑𝟑𝟑𝟕𝟕𝟕𝟕 𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑𝟏𝟏𝟏𝟏𝟏𝟏 𝟕𝟕𝟕𝟕 𝟓𝟓𝟓𝟓𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏 𝟖𝟖𝟖𝟖SecondDifferences𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑 𝟐𝟐𝟓𝟓𝟓𝟓 𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏𝟖𝟖𝟖𝟖 𝟓𝟓𝟓𝟓 𝟑𝟑𝟑𝟑Third Differences𝟏𝟏𝟏𝟏 ( 𝟐𝟐) 𝟏𝟏𝟏𝟏𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏I notice the third differences areconstant, which should be true for apolynomial of degree 3.Lesson 1: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015Successive Differences in Polynomials1
20M1-16A Story of Functions15Homework HelperALGEBRA IIWrite Explicit Polynomial Expressions for Sequences by Investigating Their Successive DifferencesShow that the set of ordered pairs (𝑥𝑥, 𝑦𝑦) in the table below satisfies a quadratic relationship. Find theequation of the form 𝑦𝑦 𝑎𝑎𝑥𝑥 2 𝑏𝑏𝑏𝑏 𝑐𝑐 that all of the ordered pairs 2345478𝒚𝒚 𝒂𝒂𝒙𝒙𝟐𝟐 𝒃𝒃𝒃𝒃 𝒄𝒄𝒂𝒂 𝟓𝟓 and 𝒄𝒄 𝟔𝟔𝟗𝟗 𝒂𝒂(𝟏𝟏)𝟐𝟐 𝒃𝒃(𝟏𝟏) 𝒄𝒄𝟗𝟗 𝟓𝟓(𝟏𝟏)𝟐𝟐 𝒃𝒃 𝟔𝟔𝟗𝟗 𝟏𝟏𝟏𝟏 𝒃𝒃𝒃𝒃 𝟐𝟐𝒚𝒚 𝟓𝟓𝒙𝒙𝟐𝟐 𝟐𝟐𝒙𝒙 𝟔𝟔Lesson 1: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015222345I know that if the seconddifferences are constant, and thefirst differences are not, then therelationship is quadratic.478First �𝟑𝟑𝟑𝟑Second DifferencesThird �𝟏𝟏To model the ordered pairs with a quadratic equation, I need tofind values of 𝑎𝑎, 𝑏𝑏, and 𝑐𝑐 in the equation 𝑦𝑦 𝑎𝑎𝑥𝑥 2 𝑏𝑏𝑏𝑏 𝑐𝑐.I found the second differences to be 10, and I know that thesecond differences are equal to 2𝑎𝑎, so it must be that 𝑎𝑎 5.I know the 𝑦𝑦-intercept is equal to 𝑐𝑐 and 5(0)2 𝑏𝑏(0) 𝑐𝑐 6,so 𝑐𝑐 6.I solved 𝑦𝑦 5𝑥𝑥 2 𝑏𝑏𝑏𝑏 6 when 𝑥𝑥 1and 𝑦𝑦 9 to find the value of 𝑏𝑏.Successive Differences in Polynomials2
20M1-16A Story of Functions15Homework HelperALGEBRA IILesson 2: The Multiplication of PolynomialsUse a Table to Multiply Two Polynomials1. Use the tabular method to multiply (5𝑥𝑥 2 𝑥𝑥 3)(2𝑥𝑥 3 5𝑥𝑥 2 4𝑥𝑥 1), and combine like terms.The right-hand column containsthe terms of the secondpolynomial.I multiply theterms in thecorrespondingrow andcolumn to fillin each cell inthe table.The top row contains the terms of thefirst polynomial.I add the liketerms found alongthe upper right tolower leftdiagonals in thetable. Forexample, 6𝑥𝑥 3 35𝑥𝑥 20𝑥𝑥 21𝑥𝑥 3 .𝟏𝟏𝟏𝟏𝒙𝒙𝟓𝟓 𝒙𝟓𝟓 ��𝒙𝟑𝟑 ��𝟑𝟐𝟐𝒙𝒙𝟑𝟑 𝟓𝟓𝒙𝒙𝟑𝟑 𝟏𝟏𝟏𝟏𝒙𝒙𝟐𝟐 �𝒙𝒙For example,𝑥𝑥 2𝑥𝑥 3 2𝑥𝑥 4 .Also,3 ( 5𝑥𝑥 2 ) 15𝑥𝑥 2 . 𝟓𝟓𝒙𝒙𝟐𝟐 𝒙𝒙 𝟑𝟑 𝟐𝟐𝒙𝒙𝟑𝟑 𝟓𝟓𝒙𝒙𝟐𝟐 𝟒𝟒𝟒𝟒 𝟏𝟏 𝟏𝟏𝟏𝟏𝒙𝒙𝟓𝟓 𝟐𝟐𝟐𝟐𝒙𝒙𝟒𝟒 𝟐𝟐𝟐𝟐𝒙𝒙𝟑𝟑 𝟔𝟔𝒙𝒙𝟐𝟐 𝟏𝟏𝟏𝟏𝒙𝒙 𝟑𝟑Now that I have combined like terms, I write theproduct as the sum of the terms from highest tolowest degree.Lesson 2: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015The Multiplication of Polynomials3
20M1-16A Story of Functions15Homework HelperALGEBRA IIUse the Distributive Property to Multiply Two Polynomials2. Use the distributive property to multiply (6𝑥𝑥 2 𝑥𝑥 2)(7𝑥𝑥 5), and then combine like terms.I distribute each term in the first polynomial to thesecond polynomial. Then I multiply each term of thefirst polynomial by each term in the second polynomial. 𝟔𝟔𝒙𝒙𝟐𝟐 𝒙𝒙 𝟐𝟐 (𝟕𝟕𝟕𝟕 𝟓𝟓) 𝟔𝟔𝒙𝒙𝟐𝟐 (𝟕𝟕𝟕𝟕 𝟓𝟓) 𝒙𝒙(𝟕𝟕𝟕𝟕 𝟓𝟓) 𝟐𝟐(𝟕𝟕𝟕𝟕 𝟓𝟓) 𝟔𝟔𝒙𝒙𝟐𝟐 (𝟕𝟕𝟕𝟕) 𝟔𝟔𝒙𝒙𝟐𝟐 (𝟓𝟓) 𝒙𝒙(𝟕𝟕𝟕𝟕) 𝒙𝒙(𝟓𝟓) 𝟐𝟐(𝟕𝟕𝟕𝟕) 𝟐𝟐(𝟓𝟓) 𝟒𝟒𝟒𝟒𝒙𝒙𝟑𝟑 𝟑𝟑𝟑𝟑𝒙𝒙𝟐𝟐 𝟕𝟕𝒙𝒙 𝟐𝟐 𝟓𝟓𝟓𝟓 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟒𝟒𝟒𝟒𝒙𝒙𝟑𝟑 𝟑𝟑𝟑𝟑𝒙𝒙𝟐𝟐 𝟗𝟗𝟗𝟗 𝟏𝟏𝟏𝟏Apply Identities to Multiply PolynomialsMultiply the polynomials in each row of the table.I recognize this product as a difference of squares:(𝑎𝑎 𝑏𝑏)(𝑎𝑎 𝑏𝑏) 𝑎𝑎2 𝑏𝑏 2 . In this case, 𝑎𝑎 is 𝑥𝑥 and 𝑏𝑏 is 2𝑦𝑦 2 .If I recognize anidentity, I canuse the formulaassociated withit to multiplypolynomials.If I do notrecognize anidentity, I canmultiplypolynomialsusing the tabularmethod ordistributiveproperty.3. (𝑥𝑥 2𝑦𝑦 2 )(𝑥𝑥 2𝑦𝑦 2 )(𝒙𝒙)𝟐𝟐 (𝟐𝟐𝒚𝒚𝟐𝟐 )𝟐𝟐 𝒙𝒙𝟐𝟐 𝟒𝟒𝒚𝒚𝟒𝟒24. (3𝑥𝑥 7𝑦𝑦)(𝟑𝟑𝒙𝒙)𝟐𝟐 𝟐𝟐(𝟑𝟑𝒙𝒙)( 𝟕𝟕𝒚𝒚) ( 𝟕𝟕𝒚𝒚)𝟐𝟐5. 𝟗𝟗𝒙𝒙𝟐𝟐 𝟒𝟒𝟒𝟒𝒙𝒙𝒙𝒙 𝟒𝟒𝟒𝟒𝒚𝒚𝟐𝟐I know that (𝑎𝑎 𝑏𝑏)2 𝑎𝑎2 2𝑎𝑎𝑎𝑎 𝑏𝑏 2 .In this case, 𝑎𝑎 is 3𝑥𝑥 and 𝑏𝑏 is 7𝑦𝑦.(𝑥𝑥 1)(𝑥𝑥 6 𝑥𝑥 5 𝑥𝑥 4 𝑥𝑥 3 𝑥𝑥 2 𝑥𝑥 1)𝒙𝒙𝟔𝟔 𝟏𝟏 𝟏𝟏 𝒙𝒙𝟕𝟕 𝟏𝟏I recognize this pattern:(𝑥𝑥 1)(𝑥𝑥 𝑛𝑛 𝑥𝑥 𝑛𝑛 1 𝑥𝑥 𝑛𝑛 2 𝑥𝑥 1) 𝑥𝑥 𝑛𝑛 1 1Lesson 2: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015The Multiplication of Polynomials4
20IIM1-16A Story of Functions15Homework HelperALGEBRA IILesson 3: The Division of PolynomialsUse a Table to Divide Two PolynomialsUse the reverse tabular method to find the quotient: (2𝑥𝑥 5 7𝑥𝑥 4 22𝑥𝑥 2 3𝑥𝑥 12) (𝑥𝑥 3 4𝑥𝑥 2 3).I need to write the terms of the dividend along the outside of thetable aligned with the upper-right to lower-left diagonals, includingplaceholders. These terms represent the diagonal sums.These are the termsof the divisor.Step 1𝒙𝒙𝟑𝟑𝟐𝟐𝒙𝒙𝟓𝟓I need to use aplaceholder herebecause the divisordoes not contain alinear �𝟐𝟐𝟐𝟐𝒙𝒙𝟐𝟐I know theleadingterm in thedividend isequal to thediagonalsum and,therefore,belongs inthis cell. 𝟑𝟑𝒙𝒙𝟏𝟏𝟏𝟏Step �𝟐𝟐Lesson 3: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015 𝟑𝟑𝒙𝒙The Division of Polynomials𝟏𝟏𝟏𝟏𝟑𝟑I need to findan expressionthat is equal to2𝑥𝑥 5 when it ismultiplied by𝑥𝑥 3 . I can thenmultiply thisexpression bythe terms inthe divisor tocomplete thefirst column ofthe table.5
20IIM1-1641𝑥𝑥A Story of Functions15Homework HelperALGEBRA III need to find an expression that equals 1𝑥𝑥 4I know this value, addedto 8𝑥𝑥 4 , must equal thediagonal sum, 7𝑥𝑥 4 �𝟕𝒙𝒙𝟒𝟒expression by the remaining terms in thedivisor to complete the second column of thetable.Step 3 𝒙𝒙𝒙𝒙𝟑𝟑 𝟐𝟐𝟐𝒙𝒙𝟐𝟐𝟎𝟎𝒙𝒙 ��𝒙𝟑𝟑when multiplied by 𝑥𝑥 3 . I can multiply thisI need to find an expression that equals 4𝑥𝑥 3when multiplied by 𝑥𝑥 3 . I can multiply thisexpression by the remaining terms in thedivisor to complete the third column of thetable and confirm the constant term.I know this value, added to 4𝑥𝑥 3 0𝑥𝑥 3 , must equal thediagonal sum, 0𝑥𝑥 3 .Step �𝟓 𝟒𝟑𝟑 ��𝒙𝟐𝟐 𝟒𝟒𝒙𝒙𝟑𝟑 𝟑𝟑𝒙𝒙 𝟑𝟑𝒙𝒙 𝟏𝟑𝟑The expression across the top row of the table is the quotient. 𝟐𝟐𝒙𝒙𝟓𝟓 𝟕𝟕𝒙𝒙𝟒𝟒 𝟐𝟐𝟐𝟐𝒙𝒙𝟐𝟐 𝟑𝟑𝒙𝒙 𝟏𝟏𝟏𝟏 𝒙𝒙𝟑𝟑 𝟒𝟒𝒙𝒙𝟐𝟐 𝟑𝟑 𝟐𝟐𝒙𝒙𝟐𝟐 𝒙𝒙 𝟒𝟒Lesson 3: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015The Division of Polynomials6
20M1-16A Story of Functions15Homework HelperALGEBRA IILesson 4: Comparing Methods—Long Division, Again?1. Is 𝑥𝑥 5 a factor of 𝑥𝑥 3 125? Justify your answer using the long division algorithm.𝒙𝒙 𝟓𝟓𝒙𝒙 𝟓𝟓𝒙𝒙 𝟓𝟓𝒙𝒙 𝟓𝟓𝒙𝒙𝟑𝟑𝒙𝒙𝟑𝟑 (𝒙𝒙𝟑𝟑𝒙𝒙𝟑𝟑 (𝒙𝒙𝟑𝟑𝒙𝒙𝟑𝟑 (𝒙𝒙𝟑𝟑 𝟎𝟎𝒙𝒙𝟐𝟐 𝟎𝟎𝒙𝒙𝒙𝒙𝟐𝟐 𝟎𝟎𝒙𝒙𝟐𝟐 𝟎𝟎𝒙𝒙 𝟓𝟓𝒙𝒙𝟐𝟐 𝟎𝟎𝒙𝒙 𝟓𝟓𝒙𝒙𝟐𝟐 )𝒙𝒙𝟐𝟐 𝟎𝟎𝒙𝒙𝟐𝟐 𝟓𝟓𝒙𝒙𝟐𝟐 ) 𝟓𝟓𝒙𝒙𝟐𝟐I know that, just like with longdivision of integers, the divisoris a factor of the dividend if theremainder is zero. 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏 𝟓𝟓𝟓𝟓 𝟎𝟎𝒙𝒙 𝟏𝟏𝟏𝟏𝟏𝟏 𝟎𝟎𝒙𝒙 ( 𝟓𝟓𝒙𝒙𝟐𝟐 𝟐𝟐𝟐𝟐𝟐𝟐)𝟐𝟐𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏𝒙𝒙𝟐𝟐 𝟓𝟓𝟓𝟓 𝟐𝟐𝟐𝟐 𝟎𝟎𝒙𝒙𝟐𝟐 𝟓𝟓𝒙𝒙𝟐𝟐 ) 𝟓𝟓𝒙𝒙𝟐𝟐 ( 𝟓𝟓𝒙𝒙𝟐𝟐 𝟎𝟎𝒙𝒙I need to find an expression thatmultiplied by 𝑥𝑥 will result in 𝑥𝑥 3 , the firstterm in the dividend. Then I multiply thisexpression by the terms of the divisor,subtract them from the dividend, andbring down the next term from thedividend. 𝟏𝟏𝟏𝟏𝟏𝟏 𝟎𝟎𝒙𝒙 𝟐𝟐𝟐𝟐𝟐𝟐)𝟐𝟐𝟐𝟐𝟐𝟐 (𝟐𝟐𝟐𝟐𝟐𝟐I repeat the steps in the divisionalgorithm until the remainder hasa degree less than that of thedivisor. 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏) 𝟐𝟐𝟐𝟐𝟐𝟐Because long division does not result in a zero remainder, I know that 𝒙𝒙 𝟓𝟓 is not a factor of 𝒙𝒙𝟑𝟑 𝟏𝟏𝟏𝟏𝟏𝟏.Lesson 4: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015Comparing Methods—Long Division, Again?7
20M1-16A Story of Functions15Homework HelperALGEBRA IILesson 5: Putting It All TogetherOperations with PolynomialsFor Problems 1–2, quickly determine the first and last terms of each polynomial if it was rewritten in standardform. Then rewrite each expression as a polynomial in standard form.1.For each quotient, I can find the term with the highestdegree by dividing the first term in the numerator bythe first term in the denominator. Since the highestdegree terms are like terms, I combine them to findthe first term of the polynomial.𝑥𝑥 3 8𝑥𝑥 3 𝑥𝑥 2 10𝑥𝑥 8 𝑥𝑥 2𝑥𝑥 4The first term is𝒙𝒙𝟑𝟑𝒙𝒙 𝒙𝒙𝟑𝟑𝒙𝒙 𝒙𝒙𝟐𝟐 𝒙𝒙𝟐𝟐 𝟐𝟐𝒙𝒙𝟐𝟐 ,and the last term is 𝟖𝟖 𝟖𝟖 𝟐𝟐 𝟒𝟒 𝟒𝟒 𝟐𝟐 𝟔𝟔.I can use the same process with the last terms of thequotients to find the last term of the polynomial.To rewrite the polynomial, I can use the reverse tabular method or division algorithm (fromLessons 3–4) to find each quotient and then add the resulting polynomials.𝒙𝒙 𝟐𝟐𝒙𝒙𝟑𝟑 (𝒙𝒙𝟑𝟑𝒙𝒙𝟐𝟐 𝟎𝟎𝒙𝒙𝟐𝟐 𝟐𝟐𝒙𝒙𝟐𝟐 )𝟐𝟐𝒙𝒙𝟐𝟐 (𝟐𝟐𝒙𝒙𝟐𝟐 𝟐𝟐𝟐𝟐 𝟎𝟎𝟎𝟎 𝟒𝟒𝟒𝟒)𝟒𝟒𝟒𝟒 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015𝒙𝒙𝟑𝟑 (𝒙𝒙𝟑𝟑𝒙𝒙𝟐𝟐 𝟑𝟑𝟑𝟑 𝒙𝒙𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏𝟑𝟑𝒙𝒙𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏 𝟒𝟒𝒙𝒙𝟐𝟐 ) (𝟑𝟑𝒙𝒙𝟐𝟐 𝟖𝟖 𝟏𝟏𝟏𝟏𝟏𝟏)𝟐𝟐𝟐𝟐 (𝟐𝟐𝟐𝟐 𝟖𝟖)𝟎𝟎𝒙𝒙𝟑𝟑 𝟖𝟖𝒙𝒙𝟑𝟑 𝒙𝒙𝟐𝟐 𝟏𝟏𝟏𝟏𝒙𝒙 𝟖𝟖 𝒙𝒙 𝟐𝟐𝒙𝒙 𝟒𝟒Lesson 5:𝒙𝒙 𝟒𝟒 𝟖𝟖 𝟎𝟎𝟎𝟎 (𝟒𝟒𝟒𝟒Standard form: 𝟒𝟒 𝟐𝟐 𝟖𝟖 𝟖𝟖 𝟖𝟖)𝟎𝟎 𝒙𝒙𝟐𝟐 𝟐𝟐𝒙𝒙 𝟒𝟒 𝒙𝒙𝟐𝟐 𝟑𝟑𝒙𝒙 𝟐𝟐 𝟐𝟐𝒙𝒙𝟐𝟐 𝟓𝟓𝒙𝒙 𝟔𝟔Putting It All Together8
20M1-16A Story of Functions15Homework HelperALGEBRA II2. (𝑥𝑥 5)2 (2𝑥𝑥 1)(2𝑥𝑥 1)I can multiply the first terms in each product and addthem to find the first term of the polynomial. I canmultiply the constant terms in each product and addthem to find the last term.The first term is 𝒙𝒙𝟐𝟐 𝟒𝟒𝟒𝟒𝟐𝟐 𝟓𝟓𝒙𝒙𝟐𝟐 ; the last term is 𝟐𝟐𝟐𝟐 𝟏𝟏 𝟐𝟐𝟐𝟐. 𝒙𝒙𝟐𝟐 𝟐𝟐(𝒙𝒙)( 𝟓𝟓) ( 𝟓𝟓)𝟐𝟐 (𝟐𝟐𝟐𝟐)𝟐𝟐 (𝟏𝟏)𝟐𝟐 𝒙𝒙𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 𝟒𝟒𝒙𝒙𝟐𝟐 𝟏𝟏 𝟓𝟓𝒙𝒙𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐I can use the identities (𝑎𝑎 𝑏𝑏)2 𝑎𝑎2 2𝑎𝑎𝑎𝑎 𝑏𝑏 2and (𝑎𝑎 𝑏𝑏)(𝑎𝑎 𝑏𝑏) 𝑎𝑎2 𝑏𝑏 2 to multiply thepolynomials.Lesson 5: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015Putting It All Together9
20ALGEBRA IILesson 6: Dividing by 𝒙𝒙 𝒂𝒂 and by 𝒙𝒙 𝒂𝒂Compute each quotient.1.I can factor the numerator using thedifference of squares identity:9𝑥𝑥 2 253𝑥𝑥 5(𝟑𝟑𝟑𝟑)𝟐𝟐 𝟓𝟓𝟐𝟐 (𝟑𝟑𝒙𝒙 𝟓𝟓)(𝟑𝟑𝒙𝒙 𝟓𝟓) 𝟑𝟑𝒙𝒙 𝟓𝟓𝟑𝟑𝟑𝟑 𝟓𝟓𝟑𝟑𝒙𝒙 𝟓𝟓2.64𝑥𝑥 3 274𝑥𝑥 3(𝟒𝟒𝟒𝟒)𝟑𝟑 𝟑𝟑𝟑𝟑 (𝟒𝟒𝒙𝒙 𝟑𝟑) (𝟒𝟒𝒙𝒙)𝟐𝟐 𝟑𝟑(𝟒𝟒𝒙𝒙) 𝟑𝟑𝟐𝟐 𝟒𝟒𝟒𝟒 𝟑𝟑𝟒𝟒𝒙𝒙 𝟑𝟑 𝟏𝟏𝟏𝟏𝒙𝒙𝟐𝟐 𝟏𝟏𝟏𝟏𝒙𝒙 𝟗𝟗3.8𝑥𝑥 3 11 2x𝑥𝑥 2 𝑎𝑎2 (𝑥𝑥 𝑎𝑎)(𝑥𝑥 𝑎𝑎)I can rewrite the numerator as(4𝑥𝑥)3 33 . Then I can factor thisexpression using the difference of cubesidentity:𝑥𝑥 3 𝑎𝑎3 (𝑥𝑥 𝑎𝑎)(𝑥𝑥 2 𝑎𝑎𝑎𝑎 𝑎𝑎2 ).(𝟐𝟐𝟐𝟐)𝟑𝟑 𝟏𝟏𝟑𝟑 (𝟐𝟐𝒙𝒙 𝟏𝟏) (𝟐𝟐𝒙𝒙)𝟐𝟐 𝟏𝟏(𝟐𝟐𝒙𝒙) 𝟏𝟏𝟐𝟐 𝟒𝟒𝒙𝒙𝟐𝟐 𝟐𝟐𝒙𝒙 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐𝟐𝟐𝒙𝒙 𝟏𝟏I recognize that I can use an identity to factor the numerator for Problems 4 and 5:4.𝑥𝑥 9 1𝑥𝑥 1𝑥𝑥 𝑛𝑛 𝑎𝑎𝑛𝑛 (𝑥𝑥 𝑎𝑎)(𝑥𝑥 𝑛𝑛 1 𝑎𝑎𝑥𝑥 𝑛𝑛 2 𝑎𝑎2 𝑥𝑥 𝑛𝑛 3 𝑎𝑎𝑛𝑛 2 𝑥𝑥1 𝑎𝑎𝑛𝑛 1 )(𝒙𝒙 𝟏𝟏) 𝒙𝒙𝟖𝟖 𝒙𝒙𝟕𝟕 𝒙𝒙𝟔𝟔 𝒙𝒙𝟓𝟓 𝒙𝒙𝟒𝟒 𝒙𝒙𝟑𝟑 𝒙𝒙𝟐𝟐 𝒙𝒙 𝟏𝟏 𝒙𝒙 𝟏𝟏 𝒙𝒙𝟖𝟖 𝒙𝒙𝟕𝟕 𝒙𝒙𝟔𝟔 𝒙𝒙𝟓𝟓 𝒙𝒙𝟒𝟒 𝒙𝒙𝟑𝟑 𝒙𝒙𝟐𝟐 𝒙𝒙 𝟏𝟏5.𝑥𝑥 5 32𝑥𝑥 2𝒙𝒙𝟓𝟓 𝟐𝟐𝟓𝟓 (𝒙𝒙 𝟐𝟐) 𝒙𝒙𝟒𝟒 𝟐𝟐𝒙𝒙𝟑𝟑 𝟐𝟐𝟐𝟐 𝒙𝒙𝟐𝟐 𝟐𝟐𝟑𝟑 𝒙𝒙 𝟐𝟐𝟒𝟒 𝒙𝒙 𝟐𝟐𝒙𝒙 𝟐𝟐 𝒙𝒙𝟒𝟒 𝟐𝟐𝒙𝒙𝟑𝟑 𝟒𝟒𝒙𝒙𝟐𝟐 𝟖𝟖𝒙𝒙 𝟏𝟏𝟏𝟏Lesson 6: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015Dividing by 𝑥𝑥 𝑎𝑎 and by 𝑥𝑥 𝑎𝑎106M1-1A Story of Functions15Homework Helper
20ALGEBRA IILesson 7: Mental MathUse Polynomial Identities to Perform Arithmetic1. Using an appropriate polynomial identity, quickly compute the product 52 28. Show each step. Be sureto state your values for 𝑥𝑥 and 𝑎𝑎.𝒙𝒙 𝟓𝟓𝟓𝟓 𝟐𝟐𝟐𝟐 𝟒𝟒𝟒𝟒𝟐𝟐𝒂𝒂 𝟓𝟓𝟓𝟓 𝟒𝟒𝟒𝟒 𝟏𝟏𝟏𝟏𝟓𝟓𝟓𝟓 𝟐𝟐𝟐𝟐 (𝟒𝟒𝟒𝟒 𝟏𝟏𝟏𝟏)(𝟒𝟒𝟒𝟒 𝟏𝟏𝟏𝟏) 𝟒𝟒𝟒𝟒𝟐𝟐 𝟏𝟏𝟏𝟏𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏2. Rewrite 121 36 as a product of two integers.I can rewrite the product 52 28 as thedifference of two squares, 𝑥𝑥 2 𝑎𝑎2 ,where 𝑥𝑥 is the arithmetic mean of thefactors (numbers being multiplied) and 𝑎𝑎is the positive difference between eitherfactor and 𝑥𝑥. This means that 𝑥𝑥 40and 𝑎𝑎 12.𝟏𝟏𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏𝟐𝟐 𝟔𝟔𝟐𝟐 (𝟏𝟏𝟏𝟏 𝟔𝟔)(𝟏𝟏𝟏𝟏 𝟔𝟔) 𝟏𝟏𝟏𝟏 𝟓𝟓I know that 𝑥𝑥 2 𝑎𝑎2 (𝑥𝑥 𝑎𝑎)(𝑥𝑥 𝑎𝑎).In this case, 𝑥𝑥 11 and 𝑎𝑎 6.3. Quickly compute the difference of squares 842 162 .𝟖𝟖𝟖𝟖𝟐𝟐 𝟏𝟏𝟏𝟏𝟐𝟐 (𝟖𝟖𝟖𝟖 𝟏𝟏𝟏𝟏)(𝟖𝟖𝟖𝟖 𝟏𝟏𝟏𝟏) 𝟔𝟔𝟔𝟔 𝟏𝟏𝟏𝟏𝟏𝟏 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔Use Polynomial Identities to Determine if a Number is Prime4. Is 1729 prime? Use the fact that 123 1728 and an identity to support your answer.𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏 𝟏𝟏𝟐𝟐𝟑𝟑 𝟏𝟏𝟑𝟑 (𝟏𝟏𝟏𝟏 𝟏𝟏) 𝟏𝟏𝟐𝟐 (𝟏𝟏)(𝟏𝟏𝟏𝟏) 𝟏𝟏𝟐𝟐 (𝟏𝟏𝟏𝟏)(𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟏𝟏) 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏The number 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 is not prime because itcan be factored as 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟏𝟏.Lesson 7: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015Mental MathI know that𝑥𝑥 3 𝑎𝑎3 (𝑥𝑥 𝑎𝑎)(𝑥𝑥 2 𝑎𝑎𝑎𝑎 𝑎𝑎).In this case, 𝑥𝑥 12 and 𝑎𝑎 1.116M1-1A Story of Functions15Homework Helper
20ALGEBRA II5. Show that 99,999,951 is not prime without using a calculator or computer.𝟗𝟗𝟗𝟗, 𝟗𝟗𝟗𝟗𝟗𝟗, 𝟗𝟗𝟗𝟗𝟗𝟗 𝟏𝟏𝟏𝟏𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎, 𝟎𝟎𝟎𝟎𝟎𝟎 𝟒𝟒𝟒𝟒𝟒𝟒 𝟐𝟐𝟖𝟖𝟐𝟐 𝟏𝟏𝟎𝟎 𝟒𝟒𝟒𝟒 𝟏𝟏𝟏𝟏 𝟕𝟕 𝟏𝟏𝟏𝟏𝟒𝟒 𝟕𝟕 𝟏𝟏𝟏𝟏𝟒𝟒 𝟕𝟕 𝟏𝟏𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗Use Polynomial Identities to Determine DivisibilityI need to rewrite99,999,951 either as adifference of perfectsquares or as a sum ordifference of perfectcubes.6. Find a value of 𝑏𝑏 so that the expression 𝑏𝑏 𝑛𝑛 1 is always divisible by 11 for any positive integer 𝑛𝑛. Explainwhy your value of 𝑏𝑏 works for any positive integer 𝑛𝑛.We can factor 𝒃𝒃𝒏𝒏 𝟏𝟏 as follows:𝒃𝒃𝒏𝒏 𝟏𝟏 (𝒃𝒃 𝟏𝟏) 𝒃𝒃𝒏𝒏 𝟏𝟏 𝒃𝒃𝒏𝒏 𝟐𝟐 𝒃𝒃𝒏𝒏 𝟑𝟑 𝒃𝒃 𝟏𝟏 .Choose 𝒃𝒃 𝟏𝟏𝟏𝟏 so that 𝒃𝒃 𝟏𝟏 𝟏𝟏𝟏𝟏.Since 𝒃𝒃 𝟏𝟏 𝟏𝟏𝟏𝟏, then 𝒃𝒃𝒏𝒏 𝟏𝟏 will have 𝟏𝟏𝟏𝟏 as a factorand therefore will always be divisible by 𝟏𝟏𝟏𝟏.I need to find a value 𝑏𝑏 so 11 is afactor of 𝑏𝑏 𝑛𝑛 1. I can do this byfinding a value of 𝑏𝑏 so that (𝑏𝑏 1) isdivisible by 11. One way to do this isto set 𝑏𝑏 1 11, so that 𝑏𝑏 12.Note: Any value of 𝑏𝑏 where (𝑏𝑏 1) is a multiple of 11 will produce a valid solution. Possible values of𝑏𝑏 include 12, 23, 34, and 45.7. Find a value of 𝑏𝑏 so that the expression 𝑏𝑏 𝑛𝑛 1 is divisible by both 3 and 11 for any positive integer 𝑛𝑛.Explain why your value of 𝑏𝑏 works for any positive integer 𝑛𝑛.We can factor 𝒃𝒃𝒏𝒏 𝟏𝟏 as follows:𝒃𝒃𝒏𝒏 𝟏𝟏 (𝒃𝒃 𝟏𝟏)(𝒃𝒃𝒏𝒏 𝟏𝟏 𝒃𝒃𝒏𝒏 𝟐𝟐 𝒃𝒃𝒏𝒏 𝟑𝟑 𝒃𝒃 𝟏𝟏).Choose 𝒃𝒃 𝟑𝟑𝟑𝟑 so that 𝒃𝒃 𝟏𝟏 𝟑𝟑𝟑𝟑.Since 𝒃𝒃 𝟏𝟏 𝟑𝟑𝟑𝟑, then 𝒃𝒃𝒏𝒏 𝟏𝟏 will have 𝟑𝟑𝟑𝟑 as a factor, which is a multiple of both 𝟑𝟑 and 𝟏𝟏𝟏𝟏.Therefore, 𝒃𝒃𝒏𝒏 𝟏𝟏 will always be divisible by 𝟑𝟑 and 𝟏𝟏𝟏𝟏.Note: Any value of 𝑏𝑏 where (𝑏𝑏 1) is a multiple of 33 will produce a valid solution. Possible values of 𝑏𝑏include 34, 67, and 100Lesson 7: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015Mental Math126M1-1A Story of Functions15Homework Helper
20ALGEBRA IILesson 8: The Power of Algebra—Finding PrimesApply Polynomial Identities to Factor Composite Numbers1. Factor 512 1 in three different ways: using the identity (𝑥𝑥 𝑛𝑛 𝑎𝑎𝑛𝑛 )(𝑥𝑥 𝑛𝑛 1 𝑎𝑎𝑥𝑥 𝑛𝑛 2 𝑎𝑎𝑛𝑛 1 ), thedifference of squares identity, and the difference of cubes identity.i.ii.iii.𝟓𝟓𝟏𝟏𝟏𝟏 𝟏𝟏 (𝟓𝟓 𝟏𝟏) 𝟓𝟓𝟏𝟏𝟏𝟏 𝟓𝟓𝟏𝟏𝟏𝟏 𝟓𝟓𝟗𝟗 𝟓𝟓𝟖𝟖 𝟓𝟓𝟏𝟏 𝟏𝟏 I know that 512 1 can be written as adifference of squares because it can bewritten in the form 𝑥𝑥 2𝑛𝑛 𝑎𝑎2𝑛𝑛 , where𝑥𝑥 5, 𝑎𝑎 1 and 𝑛𝑛 6.𝟐𝟐𝟓𝟓𝟏𝟏𝟏𝟏 𝟏𝟏 𝟓𝟓𝟔𝟔 𝟏𝟏𝟐𝟐 (𝟓𝟓𝟔𝟔 𝟏𝟏)(𝟓𝟓𝟔𝟔 𝟏𝟏)𝟑𝟑𝟓𝟓𝟏𝟏𝟏𝟏 𝟏𝟏 𝟓𝟓𝟒𝟒 𝟏𝟏𝟑𝟑𝟒𝟒𝟒𝟒 𝟐𝟐I can then apply the difference of squaresidentity:𝑥𝑥 2𝑛𝑛 𝑎𝑎2𝑛𝑛 (𝑥𝑥 𝑛𝑛 𝑎𝑎𝑛𝑛 )(𝑥𝑥 𝑛𝑛 𝑎𝑎𝑛𝑛 ).𝟒𝟒I know that 512 1 can be written as adifference of cubes because it can bewritten in the form 𝑥𝑥 3𝑛𝑛 𝑎𝑎3𝑛𝑛 , where 𝑥𝑥 5,𝑎𝑎 1 and 𝑛𝑛 4.𝟐𝟐 𝟓𝟓 𝟏𝟏 𝟓𝟓 𝟏𝟏 𝟓𝟓 𝟏𝟏 𝟓𝟓𝟒𝟒 𝟏𝟏 (𝟓𝟓𝟖𝟖 𝟓𝟓𝟒𝟒 𝟏𝟏)I can then apply the difference of cubesidentity:𝑥𝑥 3𝑛𝑛 𝑎𝑎3𝑛𝑛 (𝑥𝑥 𝑛𝑛 𝑎𝑎𝑛𝑛 )(𝑥𝑥 2𝑛𝑛 𝑎𝑎𝑛𝑛 𝑥𝑥 𝑛𝑛 𝑎𝑎2𝑛𝑛)Apply Polynomial Identities to Analyze Divisibility2. Explain why if 𝑛𝑛 is odd, the number 3𝑛𝑛 1 will always be divisible by 4.For any odd value of 𝒏𝒏,For odd numbers 𝑛𝑛, I can use the identity𝑥𝑥 𝑛𝑛 𝑎𝑎𝑛𝑛 (𝑥𝑥 𝑎𝑎)(𝑥𝑥 𝑛𝑛 1 𝑎𝑎𝑥𝑥 𝑛𝑛 2 𝑎𝑎2 𝑥𝑥 𝑛𝑛 3 𝑎𝑎𝑛𝑛 1 ).𝟑𝟑𝒏𝒏 𝟏𝟏 𝟑𝟑𝒏𝒏 𝒂𝒂𝒏𝒏 (𝟑𝟑 𝟏𝟏) 𝟑𝟑𝒏𝒏 𝟏𝟏 (𝟏𝟏)𝟑𝟑𝒏𝒏 𝟐𝟐 𝟏𝟏𝟐𝟐 𝟑𝟑𝒏𝒏 𝟑𝟑 𝟏𝟏𝒏𝒏 𝟑𝟑 𝟑𝟑𝟐𝟐 𝟏𝟏𝒏𝒏 𝟐𝟐 𝟑𝟑 𝟏𝟏𝒏𝒏 𝟏𝟏 𝟒𝟒 𝟑𝟑𝒏𝒏 𝟏𝟏 (𝟏𝟏)𝟑𝟑𝒏𝒏 𝟐𝟐 𝟏𝟏𝟐𝟐 𝟑𝟑𝒏𝒏 𝟑𝟑 𝟏𝟏𝒏𝒏 𝟑𝟑 𝟑𝟑𝟐𝟐 𝟏𝟏𝒏𝒏 𝟐𝟐 𝟑𝟑 𝟏𝟏𝒏𝒏 𝟏𝟏 .Since 𝟒𝟒 is a factor of 𝟑𝟑𝒏𝒏 𝟏𝟏 for any odd value of 𝒏𝒏, the number 𝟑𝟑𝒏𝒏 𝟏𝟏 is divisible by 𝟒𝟒.Lesson 8: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015The Power of Algebra—Finding Primes136M1-1A Story of Functions15Homework Helper
20ALGEBRA II3. If 𝑛𝑛 is a composite number, explain why 4𝑛𝑛 1 is never prime.Since 𝒏𝒏 is composite, there are integers 𝒂𝒂 and 𝒃𝒃larger than 𝟏𝟏 so that 𝒏𝒏 𝒂𝒂𝒂𝒂. ThenI can use the identity𝑥𝑥 𝑛𝑛 1 (𝑥𝑥 1)(𝑥𝑥 𝑛𝑛 1 𝑥𝑥 𝑛𝑛 2 𝑥𝑥 𝑛𝑛 3 𝑥𝑥 1).𝟒𝟒𝒏𝒏 𝟏𝟏 𝟒𝟒𝒂𝒂𝒂𝒂 𝟏𝟏 (𝟒𝟒𝒂𝒂 𝟏𝟏) (𝟒𝟒𝒂𝒂 )𝒃𝒃 𝟏𝟏 (𝟒𝟒𝒂𝒂 )𝒃𝒃 𝟐𝟐 (𝟒𝟒𝒂𝒂 )𝒃𝒃 𝟑𝟑 (𝟒𝟒𝒂𝒂 )𝟏𝟏 𝟏𝟏 .Since 𝒂𝒂 𝟏𝟏, the expression 𝟒𝟒𝒂𝒂 𝟏𝟏 must be an integer greater than or equal to 𝟏𝟏𝟏𝟏, so 𝟒𝟒𝒂𝒂𝒂𝒂 𝟏𝟏 has afactor other than 𝟏𝟏. Therefore, the number 𝟒𝟒𝒏𝒏 𝟏𝟏 is composite.Apply the Difference of Squares Identity to Rewrite Numbers4. Express the numbers from 31 to 40 as the difference of two squares, if possible. The first four have beendone for you.Number FactorizationDifference of two squares162 – 152 256 225 313131 31 1 (16 15)(16 15)3333 11 3 (7 4)(7 4)72 42 49 16 33𝟑𝟑𝟑𝟑 𝟕𝟕 𝟓𝟓 (𝟔𝟔 𝟏𝟏)(𝟔𝟔 𝟏𝟏)𝟔𝟔𝟐𝟐 𝟏𝟏𝟐𝟐 𝟑𝟑𝟑𝟑 𝟏𝟏 𝟑𝟑𝟑𝟑323435363738394032 8 4 (6 2)(6 2)Can’t be done.𝟑𝟑𝟑𝟑 𝟔𝟔 𝟔𝟔 (𝟔𝟔 𝟎𝟎)(𝟔𝟔 𝟎𝟎)𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑 𝟏𝟏 (𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏)(𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏)62 22 36 4 32𝟔𝟔𝟐𝟐 𝟎𝟎𝟐𝟐 𝟑𝟑𝟑𝟑 𝟎𝟎 𝟑𝟑𝟑𝟑𝟐𝟐𝟐𝟐𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑Can’t be done𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏 𝟑𝟑 (𝟖𝟖 𝟓𝟓)(𝟖𝟖 𝟓𝟓)𝟒𝟒𝟒𝟒 𝟏𝟏𝟏𝟏 𝟒𝟒 (𝟕𝟕 𝟑𝟑)(𝟕𝟕 𝟑𝟑)Lesson 8: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015𝟖𝟖𝟐𝟐 𝟓𝟓𝟐𝟐 𝟔𝟔𝟔𝟔 𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑𝟕𝟕𝟐𝟐 𝟑𝟑𝟐𝟐 𝟒𝟒𝟒𝟒 𝟗𝟗 𝟒𝟒𝟒𝟒The Power of Algebra—Finding PrimesI need to writeeach number as aproduct of twowhole numbers. Ican rewrite theproduct as adifference of twosquares using themethod welearned in Lesson7.If I cannot factorthe number intotwo positiveintegers with aneven sum, thenumber cannotbe written as adifference ofsquares becausethe mean wouldbe a fraction.146M1-1A Story of Functions15Homework Helper
20ALGEBRA IILesson 9: Radicals and ConjugatesConvert Expressions to Simplest Radical FormExpress each of the following as a rational expression or in simplest radical form. Assume that the symbol 𝒙𝒙represents a positive number. Remember that a simplified radical expression has a denominator that is aninteger.1. 5𝑥𝑥 10 2𝑥𝑥 𝟓𝟓𝒙𝒙 𝟏𝟏𝟏𝟏 𝟓𝟓𝒙𝒙 𝟐𝟐𝟐𝟐 𝟓𝟓𝟓𝟓𝒙𝒙 𝟏𝟏𝟏𝟏𝒙𝒙𝟐𝟐 𝟐𝟐𝟐𝟐 𝟐𝟐𝒙𝒙 𝟏𝟏𝟏𝟏 𝒙𝒙𝟐𝟐 𝟐𝟐𝟐𝟐 𝟐𝟐𝒙𝒙 𝟏𝟏𝟏𝟏 𝒙𝒙𝟐𝟐 𝟓𝟓 𝟐𝟐𝒙𝒙 𝒙𝒙 𝟏𝟏𝟏𝟏323I can distribute the 5𝑥𝑥 to eachterm in the parentheses bymultiplying the radicands.Once I distribute the 5𝑥𝑥 term toeach term in the parentheses, Ican factor each radicand so thatone factor is a perfect square. Theperfect square factors can betaken outside the radical.3 12. 27 24 4𝟑𝟑I know that I can𝟑𝟑𝟑𝟑 𝟏𝟏𝟐𝟐 𝟐𝟐 𝟏𝟏𝟑𝟑 𝟑𝟑 𝟐𝟐𝟐𝟐 𝟑𝟑 𝟖𝟖 𝟑𝟑 𝟑𝟑𝟐𝟐𝟐𝟐𝟒𝟒 𝟐𝟐𝟐𝟐 𝟒𝟒𝟑𝟑 𝟑𝟑 𝟐𝟐𝟑𝟑 𝟐𝟐𝟐𝟐𝟑𝟑𝟑𝟑𝟑𝟑 𝟖𝟖 𝟑𝟑 𝟑𝟑 𝟐𝟐 𝟐𝟐𝟑𝟑 𝟐𝟐 𝟑𝟑 𝟑𝟑𝟑𝟑 𝟖𝟖𝟑𝟑 𝟒𝟒 𝟐𝟐𝟑𝟑𝟑𝟑𝟑𝟑 𝟐𝟐𝟐𝟐 𝟐𝟐𝟑𝟑 𝟐𝟐 𝟑𝟑 𝟔𝟔𝟔𝟔𝟑𝟑 𝟐𝟐 𝟑𝟑 Lesson 9: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015 𝟏𝟏 𝟐𝟐𝟑𝟑𝑎𝑎3 𝑎𝑎 𝑏𝑏which will allow me tosimplify the numeratorand denominatorseparately.𝟑𝟑 𝟐𝟐 𝟐𝟐𝟑𝟑 𝟐𝟐 𝟑𝟑 𝟑𝟑𝟐𝟐 𝟑𝟑3rewrite as 3 ,𝑏𝑏𝟑𝟑 𝟐𝟐𝟔𝟔Radicals and ConjugatesTo write the denominator as aninteger, I can multiply the numeratorand denominator by a commonfactor that makes the radicand in thedenominator a perfect cube.156M1-1A Story of Functions15Homework Helper
20ALGEBRA IISimplify Radical QuotientsSimplify each of the following quotients as far as possible.3.33 12 3𝟑𝟑3 3𝟑𝟑 𝟏𝟏𝟏𝟏 𝟑𝟑 𝟑𝟑 𝟑𝟑 𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏 𝟑𝟑 𝟑𝟑 𝟑𝟑 𝟑𝟑𝟑𝟑𝟑𝟑 𝟑𝟑I remember that an expression isin simplified radical form if it hasno terms with an exponentgreater than or equal to the indexof the radical and if it does nothave a radical in the denominator.𝟏𝟏𝟏𝟏 𝟑𝟑 𝟑𝟑 𝟑𝟑𝟑𝟑 𝟒𝟒 𝟏𝟏4.5 2 32 3 4 5𝟓𝟓 𝟐𝟐 𝟑𝟑 𝟐𝟐 𝟑𝟑 𝟒𝟒 𝟓𝟓 𝟓𝟓 𝟐𝟐 𝟑𝟑 𝟐𝟐 𝟑𝟑 𝟒𝟒 𝟓𝟓 𝟐𝟐 𝟑𝟑 𝟒𝟒 𝟓𝟓 𝟐𝟐 𝟑𝟑 𝟒𝟒 𝟓𝟓 𝟐𝟐 𝟑𝟑 𝟒𝟒 𝟓𝟓 𝟐𝟐 𝟑𝟑 𝟒𝟒 𝟓𝟓 𝟏𝟏𝟏𝟏 𝟔𝟔 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏 𝟐𝟐 𝟗𝟗 𝟒𝟒 𝟏𝟏𝟏𝟏(𝟐𝟐 𝟑𝟑)𝟐𝟐 (𝟒𝟒 𝟓𝟓)𝟐𝟐 𝟏𝟏𝟏𝟏 𝟔𝟔 𝟐𝟐𝟐𝟐 𝟏𝟏𝟏𝟏 𝟐𝟐(𝟑𝟑) 𝟒𝟒 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟖𝟖𝟖𝟖 𝟓𝟓 𝟔𝟔 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟑𝟑 𝟐𝟐 𝟏𝟏𝟏𝟏 𝟑𝟑𝟑𝟑𝟐𝟐 𝟓𝟓 𝟔𝟔 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟑𝟑 𝟐𝟐 𝟏𝟏𝟏𝟏 𝟔𝟔𝟔𝟔 I can convert thedenominator to an integerby multiplying both thenumerator and denominatorof the fraction by the radicalconjugate.Since 2 3 4 5 is theexpression, 2 3 4 5 is itsradical conjugate.𝟑𝟑 𝟐𝟐 𝟏𝟏𝟏𝟏 𝟓𝟓 𝟔𝟔 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏𝟑𝟑𝟑𝟑I know that for any positive numbers 𝑎𝑎 and 𝑏𝑏,22 𝑎𝑎 𝑏𝑏 𝑎𝑎 𝑏𝑏 𝑎𝑎 𝑏𝑏 . In this22case, 𝑎𝑎 2 3 12 and 𝑏𝑏 4 5 80.Lesson 9: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015Radicals and Conjugates166M1-1A Story of Functions15Homework Helper
20ALGEBRA IILesson 10: The Power of Algebra—Finding Pythagorean TriplesUse the difference of squares identity 𝒙𝒙𝟐𝟐 𝒚𝒚𝟐𝟐 (𝒙𝒙 𝒚𝒚)(𝒙𝒙 𝒚𝒚) to simplify algebraic expressions.1. Use the difference of squares identity to find (𝑥𝑥 𝑦𝑦 1)(𝑥𝑥 𝑦𝑦 1).(𝒙𝒙 𝒚𝒚 𝟏𝟏)(𝒙𝒙 𝒚𝒚 𝟏𝟏) 𝒙𝒙 (𝒚𝒚 𝟏𝟏) 𝒙𝒙 (𝒚𝒚 𝟏𝟏) 𝒙𝒙𝟐𝟐 (𝒚𝒚 𝟏𝟏)𝟐𝟐 𝒙𝒙𝟐𝟐 𝒚𝒚𝟐𝟐 𝟐𝟐𝟐𝟐 𝟏𝟏 𝒙𝒙𝟐𝟐 𝒚𝒚𝟐𝟐 𝟐𝟐𝟐𝟐 𝟏𝟏I need to write the expression inthe form (𝑎𝑎 𝑏𝑏)(𝑎𝑎 𝑏𝑏).Since both the 𝑦𝑦-term and theconstant, 1, switch signsbetween the factors, I know thatthe 𝑏𝑏 component of the identityis 𝑦𝑦 1.Use the difference of squares identity to factor algebraic expressions.2. Use the difference of two squares identity to factor the expression (𝑥𝑥 𝑦𝑦)(𝑥𝑥 𝑦𝑦) 10𝑥𝑥 25.(𝒙𝒙 𝒚𝒚)(𝒙𝒙 𝒚𝒚) 𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 𝒙𝒙𝟐𝟐 𝒚𝒚𝟐𝟐 𝟏𝟏𝟏𝟏𝒙𝒙 𝟐𝟐𝟐𝟐 𝒙𝒙𝟐𝟐 𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐 𝒚𝒚𝟐𝟐 (𝒙𝒙 𝟓𝟓)𝟐𝟐 𝒚𝒚𝟐𝟐 (𝒙𝒙 𝟓𝟓) 𝒚𝒚 (𝒙𝒙 𝟓𝟓) 𝒚𝒚 I see that the constant term,25, is a perfect square andthat the coefficient of thelinear 𝑥𝑥 term is 10, which istwice the square root of 25.That reminds me of theidentity:(𝑥𝑥 𝑎𝑎)2 𝑥𝑥 2 2𝑎𝑎𝑎𝑎 𝑎𝑎2 .Lesson 10: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015The Power of Algebra—Finding Pythagorean Triples176M1-1A Story of Functions15Homework Helper
20ALGEBRA IIApply the difference of squares identity to find Pythagorean triples.3. Prove that the Pythagorean triple (15, 8, 17) can be found by choosing a pair of integers 𝑥𝑥 and 𝑦𝑦 with𝑥𝑥 𝑦𝑦 and computing (𝑥𝑥 2 𝑦𝑦 2 , 2𝑥𝑥𝑥𝑥, 𝑥𝑥 2 𝑦𝑦 2 ).We want 𝟐𝟐𝟐𝟐𝟐𝟐 𝟖𝟖, so 𝒙𝒙𝒙𝒙 𝟒𝟒, so we can set 𝒙𝒙 𝟒𝟒and 𝒚𝒚 𝟏𝟏.This means that 𝒙𝒙𝟐𝟐 𝒚𝒚𝟐𝟐 𝟒𝟒𝟐𝟐 𝟏𝟏𝟐𝟐 𝟏𝟏𝟏𝟏 and𝒙𝒙𝟐𝟐 𝒚𝒚𝟐𝟐 𝟒𝟒𝟐𝟐 𝟏𝟏𝟐𝟐 𝟏𝟏𝟏𝟏.I know 𝑥𝑥 4 and 𝑦𝑦 1because 𝑥𝑥𝑥𝑥 4 and 𝑥𝑥 𝑦𝑦.Therefore, the values 𝒙𝒙 𝟒𝟒 and 𝒚𝒚 𝟏𝟏 generate thePythagorean triple 𝒙𝒙𝟐𝟐 𝒚𝒚𝟐𝟐 , 𝟐𝟐𝟐𝟐𝟐𝟐, 𝒙𝒙𝟐𝟐 𝒚𝒚𝟐𝟐 (𝟏𝟏𝟏𝟏, 𝟖𝟖, 𝟏𝟏𝟏𝟏).4. Prove that the Pythagorean triple (15, 36, 39) cannot be found by choosing a pair of integers 𝑥𝑥 and 𝑦𝑦with 𝑥𝑥 𝑦𝑦 and computing (𝑥𝑥 2 𝑦𝑦 2 , 2𝑥𝑥𝑥𝑥, 𝑥𝑥 2 𝑦𝑦 2 ).Since 𝟑𝟑𝟑𝟑 is the only even number in(𝟏𝟏𝟏𝟏, 𝟑𝟑𝟑𝟑, 𝟑𝟑𝟑𝟑), we must have 𝟐𝟐𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑, so 𝒙𝒙𝒙𝒙 𝟏𝟏𝟏𝟏. I need to try all the cases where𝑥𝑥𝑥𝑥 18 and 𝑥𝑥 𝑦𝑦.If 𝒙𝒙 𝟏𝟏𝟏𝟏 and 𝒚𝒚 𝟏𝟏, then 𝒙𝒙𝟐𝟐 𝒚𝒚𝟐𝟐 𝟏𝟏𝟖𝟖𝟐𝟐 𝟏𝟏𝟐𝟐 𝟑𝟑𝟑𝟑𝟑𝟑 𝟏𝟏 𝟑𝟑𝟑𝟑𝟑𝟑. So this combination does notproduce a number in the triple.If 𝒙𝒙 𝟗𝟗 and 𝒚𝒚 𝟐𝟐, then 𝒙𝒙𝟐𝟐 𝒚𝒚𝟐𝟐 𝟗𝟗𝟐𝟐 𝟐𝟐𝟐𝟐 𝟖𝟖𝟖𝟖 𝟒𝟒 𝟕𝟕𝟕𝟕. So this combination does not producea number in the triple.If 𝒙𝒙 𝟔𝟔 and 𝒚𝒚 𝟑𝟑, then 𝒙𝒙𝟐𝟐 𝒚𝒚𝟐𝟐 𝟔𝟔𝟐𝟐 𝟑𝟑𝟐𝟐 𝟑𝟑𝟑𝟑 𝟗𝟗 𝟐𝟐𝟐𝟐. So this combination does not producea number in the triple.There are no other integer values of 𝒙𝒙 and 𝒚𝒚 that satisfy 𝒙𝒙𝒙𝒙 𝟏𝟏𝟏𝟏 and 𝒙𝒙 𝒚𝒚. Therefore, the triple(𝟏𝟏𝟏𝟏, 𝟑𝟑𝟑𝟑, 𝟑𝟑𝟑𝟑) cannot be found using the method described.Lesson 10: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015The Power of Algebra—Finding Pythagorean Triples186M1-1A Story of Functions15Homework Helper
20ALGEBRA IILesson 11: The Special Role of Zero in FactoringFind Solutions to Polynomial EquationsFind all solutions to the given equations.2 (3𝑥𝑥1. 2𝑥𝑥(𝑥𝑥 1)The exponent 2 means that the equation hasa repeated factor, which corresponds to arepeated solution. 4) 0𝟐𝟐𝟐𝟐 𝟎𝟎 or (𝒙𝒙 𝟏𝟏) 𝟎𝟎 or (𝟑𝟑𝟑𝟑 𝟒𝟒) 𝟎𝟎Solutions: 𝟎𝟎, 𝟏𝟏,𝟒𝟒𝟑𝟑To find all the solutions, I need toset each factor equal to zero andsolve the resulting equations.2. 𝑥𝑥(𝑥𝑥 2 25)(𝑥𝑥 2 1) 0𝒙𝒙(𝒙𝒙 𝟓𝟓)(𝒙𝒙 𝟓𝟓)(𝒙𝒙 𝟏𝟏)(𝒙𝒙 𝟏𝟏) 𝟎𝟎𝒙𝒙 𝟎𝟎 or (𝒙𝒙 𝟓𝟓) 𝟎𝟎 or (𝒙𝒙 𝟓𝟓) 𝟎𝟎 or (𝒙𝒙 𝟏𝟏) 𝟎𝟎 or (𝒙𝒙 𝟏𝟏) 𝟎𝟎Solutions: 𝟎𝟎, 𝟓𝟓, 𝟓𝟓, 𝟏𝟏, 𝟏𝟏I recognize the expression on the leftside of the equation as the differenceof two perfect squares. I know that3. (𝒙𝒙 𝟑𝟑)𝟐𝟐 (𝟑𝟑𝒙𝒙 6)𝟐𝟐(𝒙𝒙 𝟑𝟑)𝟐𝟐 (𝟑𝟑𝟑𝟑 𝟔𝟔)𝟐𝟐 𝟎𝟎𝑎𝑎2 𝑏𝑏 2 (𝑎𝑎 𝑏𝑏)(𝑎𝑎 𝑏𝑏). In thiscase, 𝑎𝑎 (𝑥𝑥 3) and 𝑏𝑏 (3𝑥𝑥 6). (𝒙𝒙 𝟑𝟑) (𝟑𝟑𝟑𝟑 𝟔𝟔) (𝒙𝒙 𝟑𝟑) (𝟑𝟑𝟑𝟑 𝟔𝟔) 𝟎𝟎(𝟒𝟒𝟒𝟒 𝟑𝟑)( 𝟐𝟐𝟐𝟐 𝟗𝟗) 𝟎𝟎𝟒𝟒𝟒𝟒 𝟑𝟑 𝟎𝟎 or 𝟐𝟐𝟐𝟐 𝟗𝟗 𝟎𝟎I collected like terms to rewrite thefactors.𝟑𝟑 𝟗𝟗𝟒𝟒 𝟐𝟐Solutions: ,Determine Zeros and Multiplicity for Polynomial Functions4. Find the zeros with multiplicity for the function 𝑝𝑝(𝑥𝑥) (𝑥𝑥 3 1)(𝑥𝑥 4 9𝑥𝑥 2 ).The number of times a solution appears as a factor in a polynomial function is its multiplicity.𝟎𝟎 (𝒙𝒙 𝟏𝟏)(𝒙𝒙𝟐𝟐 𝒙𝒙 𝟏𝟏)𝒙𝒙𝟐𝟐 (𝒙𝒙𝟐𝟐 𝟗𝟗)𝟎𝟎 (𝒙𝒙 𝟏𝟏)(𝒙𝒙𝟐𝟐 𝒙𝒙 𝟏𝟏)𝒙𝒙𝟐𝟐 (𝒙𝒙 𝟑𝟑)(𝒙𝒙 𝟑𝟑)I factored the expression (𝑥𝑥 3 1) using thedifference of cubes pattern.Since (𝑥𝑥 2 𝑥𝑥 1) does not factor, it does notcontribute any zeros of the function 𝑝𝑝.Then 𝟏𝟏 is a zero of multiplicity 𝟏𝟏, 𝟎𝟎 is a zero of multiplicity 𝟐𝟐, 𝟑𝟑 is a zero of multiplicity 𝟏𝟏, and 𝟑𝟑 is azero of multiplicity 𝟏𝟏.Lesson 11: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015The Special Role of Zero in Factoring196M1-1A Story of Functions15Homework Helper
20ALGEBRA IIConstruct a Polynomial Function That Has a Specified Set of Zeros with Stated Multiplicity5. Find two different polynomial functions that have a zero at 1 of multiplicity 3 and a zero at 2 ofmultiplicity 2.𝒑𝒑(𝒙𝒙) (𝒙𝒙 𝟏𝟏)𝟑𝟑 (𝒙𝒙 𝟐𝟐)𝟐𝟐I need to include the linear factor (𝑥𝑥 1)three times and the linear factor (𝑥𝑥 2) twicein the statement of each function.𝒒𝒒(𝒙𝒙) 𝟓𝟓(𝒙𝒙 𝟏𝟏)𝟑𝟑 (𝒙𝒙 𝟐𝟐)𝟐𝟐Compare the Remainder of 𝒑𝒑(𝒙𝒙) (𝒙𝒙 𝒂𝒂) with 𝒑𝒑(𝒂𝒂)6. Consider the polynomial function 𝑝𝑝(𝑥𝑥) 𝑥𝑥 3 𝑥𝑥 2 𝑥𝑥 6.a.Divide 𝑝𝑝 by the divisor (𝑥𝑥 2), and rewrite in the form 𝑝𝑝(𝑥𝑥) (divisor)(quotient) remainder.Either the division algorithm or the reverse tabular method can be used to find the quotient andremainder.𝒙𝒙 𝟐𝟐𝒙𝒙𝟑𝟑 (𝒙𝒙𝟑𝟑𝒙𝒙𝟐𝟐 𝒙𝒙𝟐𝟐 𝟐𝟐𝒙𝒙𝟐𝟐 )𝟑𝟑𝒙𝒙𝟐𝟐 (𝟑𝟑𝒙𝒙𝟐𝟐 𝟑𝟑𝟑𝟑 𝒙𝒙b. 𝟔𝟔 𝒙𝒙 𝟔𝟔𝟔𝟔)𝟕𝟕𝟕𝟕 (𝟕𝟕𝟕𝟕𝒑𝒑(𝒙𝒙) (𝒙𝒙 𝟐𝟐) 𝒙𝒙𝟐𝟐 𝟑𝟑𝟑𝟑 𝟕𝟕 𝟖𝟖 𝟕𝟕 𝟔𝟔 𝟏𝟏𝟏𝟏)I used the division algorithmwe learned in Lesson 4 tofind the quotient andremainder.𝟖𝟖Evaluate 𝑝𝑝(2).𝒑𝒑(𝒙𝒙) (𝒙𝒙 𝟐𝟐) 𝒙𝒙𝟐𝟐 𝟑𝟑𝟑𝟑 𝟕𝟕 𝟖𝟖𝒑𝒑(𝟐𝟐) (𝟐𝟐 𝟐𝟐) 𝟐𝟐𝟐𝟐 𝟑𝟑 𝟐𝟐 𝟕𝟕 𝟖𝟖𝒑𝒑(𝟐𝟐) 𝟎𝟎 𝟖𝟖𝒑𝒑(𝟐𝟐) 𝟖𝟖Lesson 11: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015The Special Role of Zero in Factoring206M1-1A Story of Functions15Homework Helper
20ALGEBRA IILesson 12: Overcoming Obstacles In FactoringSolving Problems by Completing the SquareSolve each of the following equations by completing the square.1. 𝑥𝑥 2 8𝑥𝑥 3 0𝒙𝒙𝟐𝟐 𝟖𝟖𝟖𝟖 𝟑𝟑𝒙𝒙𝟐𝟐 𝟖𝟖𝟖𝟖 ( 𝟒𝟒)𝟐𝟐 𝟑𝟑 ( 𝟒𝟒)𝟐𝟐𝒙𝒙𝟐𝟐 𝟖𝟖𝒙𝒙 𝟏𝟏𝟏𝟏 𝟑𝟑 𝟏𝟏𝟏𝟏(𝒙𝒙 𝟒𝟒)𝟐𝟐 𝟏𝟏𝟏𝟏𝒙𝒙 𝟒𝟒 𝟏𝟏𝟏𝟏The solutions are 𝟒𝟒 𝟏𝟏𝟏𝟏 and 𝟒𝟒 𝟏𝟏𝟏𝟏.2. 2𝑥𝑥 2 6𝑥𝑥 1𝟒𝟒𝒙𝒙𝟐𝟐 𝟏𝟏𝟏𝟏𝒙𝒙 𝟐𝟐𝟒𝟒𝒙𝒙 𝟏𝟏𝟏𝟏𝒙𝒙 𝟗𝟗 𝟐𝟐 𝟗𝟗(𝟐𝟐𝒙𝒙 𝟑𝟑)𝟐𝟐 𝟕𝟕𝟐𝟐𝒙𝒙 𝟑𝟑 𝟕𝟕𝟐𝟐The solutions are𝟑𝟑 𝟕𝟕𝟑𝟑 𝟕𝟕and 𝟐𝟐 .𝟐𝟐I squared half the linear coefficientand added the result to both sides ofthe equation.I recognize that 𝑥𝑥 2 8𝑥𝑥 16 fits the pattern for thesquare of a difference: 𝑥𝑥 2 2𝑎𝑎𝑎𝑎 𝑎𝑎2 (𝑥𝑥 𝑎𝑎)2I multiplied both sidesof the equation by 2 sothe quadratic termwould be a perfectsquare. Then I cancomplete the squareusing the reversetabular method, shownto the right.𝟐𝟐𝒙𝒙 𝟑𝟑𝟐𝟐𝟐𝟐𝟒𝟒𝒙𝒙𝟐𝟐 𝟔𝟔𝟔𝟔 𝟑𝟑 𝟔𝟔𝟔𝟔𝟗𝟗3. 𝑥𝑥 4 6𝑥𝑥 2 4 0𝒙𝒙𝟒𝟒 𝟔𝟔𝒙𝒙𝟐𝟐 𝟒𝟒𝒙𝒙𝟒𝟒 𝟔𝟔𝒙𝒙𝟐𝟐 𝟗𝟗 𝟒𝟒 𝟗𝟗(𝒙𝒙𝟐𝟐 𝟑𝟑)𝟐𝟐 𝟓𝟓𝒙𝒙𝟐𝟐 𝟑𝟑 𝟓𝟓𝒙𝒙𝟐𝟐 𝟑𝟑 𝟓𝟓The solutions are 𝟑𝟑 𝟓𝟓, 𝟑𝟑 𝟓𝟓, 𝟑𝟑 𝟓𝟓, and 𝟑𝟑 𝟓𝟓.Lesson 12: 2015 Great Minds eureka-math.orgALG II-M1-HWH-1.3.0-08.2015Overcoming Obstacles In Factoring216M1-1A Story of Functions15Homework Helper
20ALGEBRA IISolving Problems by Using the Quadratic FormulaSolve the equation by using the quadratic formula.4. (𝑥𝑥 2 7𝑥𝑥 3)(𝑥𝑥 2 5𝑥𝑥 2) 0𝒙𝒙 𝟕𝟕 ( 𝟕𝟕)𝟐𝟐 𝟒𝟒(𝟏𝟏)(𝟑𝟑)𝟐𝟐(𝟏𝟏)The solutions areor 𝒙𝒙 I can find the solutions bysetting each factor equal tozero. I can solve the resultingquadratic equations using thequadratic formula:𝟓𝟓 ( 𝟓𝟓)𝟐𝟐 𝟕 𝟑𝟑𝟑𝟑 𝟕𝟕 𝟑𝟑𝟑𝟑 𝟓𝟓 𝟏𝟏𝟏𝟏𝟓𝟓 𝟏𝟏𝟏𝟏, 𝟐𝟐 , 𝟐𝟐 , and 𝟐𝟐 .𝟐𝟐If 𝑎𝑎𝑥𝑥 2 𝑏𝑏𝑥𝑥 𝑐𝑐 0, then𝑥𝑥 Solving Polynomial Equations by FactoringUse factoring to solve each equation.5. 6𝑥𝑥 2 7𝑥𝑥 3 0I split the linear term into two termswhose coefficients multiply to theproduct of the leading coefficient andthe constant (9 2) (6 3).Then I can factor the resultingquadratic expression by grouping.𝟔𝟔𝒙𝒙𝟐𝟐 𝟗𝟗𝒙𝒙 𝟐𝟐𝟐𝟐 𝟑𝟑 𝟎𝟎𝟑𝟑𝟑𝟑(𝟐𝟐𝟐𝟐 𝟑𝟑) 𝟏𝟏(𝟐𝟐𝟐𝟐 𝟑𝟑) 𝟎𝟎(𝟑𝟑𝟑𝟑 𝟏𝟏)(𝟐𝟐𝟐𝟐 𝟑𝟑) 𝟎𝟎𝟑𝟑𝟐𝟐𝟏𝟏𝟑𝟑The solutions are and .6. (2𝑥𝑥 1)(𝑥𝑥 4) (2𝑥𝑥 1)(𝑥𝑥 3)(𝟐𝟐𝒙𝒙 𝟏𝟏)(𝒙𝒙 𝟒𝟒) (𝟐𝟐𝒙𝒙 𝟏𝟏)(𝒙𝒙 𝟑𝟑) 𝟎𝟎(𝟐𝟐𝒙𝒙 𝟏𝟏) (𝒙𝒙 𝟒𝟒) (𝒙𝒙 𝟑𝟑) 𝟎𝟎(𝟐𝟐𝒙𝒙 𝟏𝟏)( 𝟕𝟕) 𝟎𝟎 𝑏𝑏 𝑏𝑏2 4𝑎𝑎𝑎𝑎.2𝑎𝑎I factored (2𝑥𝑥 1) from both expressions.𝟏𝟏𝟐𝟐The only solution is .7. (𝑥𝑥 2 5)2 3 0 𝒙𝒙𝟐𝟐 𝟓𝟓 𝟑𝟑 𝒙𝒙𝟐𝟐 𝟓𝟓 𝟑𝟑 𝟎𝟎 𝒙𝒙𝟐𝟐 𝟓𝟓 𝟑𝟑 𝟎𝟎 or 𝒙𝒙𝟐𝟐 𝟓𝟓 𝟑𝟑 𝟎𝟎𝒙𝒙𝟐𝟐 𝟓𝟓 𝟑𝟑 or 𝒙𝒙𝟐𝟐 𝟓𝟓 𝟑𝟑I factored the left side of theequation as a difference of twoperfect
2015-16. M1. ALGEBRA II. Lesson 1: Successive Differences in Polynomials . I found the second differences to be 10, and I know that the secon
10 9 8 7 6 5 4 3 2 1 Eureka Math Grade K, Module 4 Student File_A Contains copy-ready classwork and homework as well as templates (including cut outs) . 2 3 1 This work is derived from Eureka Math and licensed by Great Minds. 2015 Great Minds. eureka-math.org A
10 9 8 7 6 5 4 3 2 1 Eureka Math Algebra II, Module 1 Student File_B Contains Exit Ticket, and Assessment Materials Published by the non-profit Great Minds. . This work is derived from Eureka Math and licensed by Great Minds. 2015 Great Minds. eureka- math.org ALG II-M1-ETP-1.3.-05.2015 1. Lesson 2 M1 ALGEBRA II
2015-16 Lesson 1 : Solve word problems with three addends, two of which make ten. 1 2 Homework Helper G1-M2-Lesson 1 Read the math story. Make a simple math drawing with labels. Circle 10 and solve. Maddy goes
Grade 2 Eureka Math, A Story of Units Module 1 Read on to learn a little bit about Eureka Math, the creators of A Story of Units: Eureka Math is a complete, PreK–12 curriculum and professional development platform. It follows the focus and coherence of the Common Core State Standards (CCSS) and carefully sequences the
LEAP 2025 Algebra I State Assessment, Review for Algebra I Exam, and Algebra I Final Exam Eureka Module Eureka Module 2 Topics A and B Eureka Module 2 Topic C and D Eureka Module 5 Topic A and B Eureka Modules 1-4 Topics A- D or E Suggested # of Days 8 days 11/5/19- 11/15/19 5 days 11/18/19-11/22/19 5 days 12/2/19 -12/6/19 10 days
10 9 8 7 6 5 4 3 2 1 Eureka Math Grade 1, Module 4 Student File_A Contains copy-ready classwork and homework as well as templates (including cut outs) . This work is derived from Eureka Math and licensed by Great Minds. 2015 Great Minds. eureka-math.org . A STORY OF UNITS. 1. G1-M4-SE-1.3.0-05.2015. . bond and place value chart. 23 .
Grade 4 Module 2 Lessons 1–5 Eureka Math . A Story of Units 4 2 Lesson 1 : Express metric length measurements in terms of a smaller unit; . Know and relate metric units to place value units in order to express measurements in different units. G4-M2-HWH-1.3.0-09.2015. 2015-16
10 9 8 7 6 5 4 3 2 1 Eureka Math Algebra II, Module 2 Student_A Contains copy-ready classwork and homework as well as templates (including cut outs) Published by the non-profit Great Minds. . This work is derived from Eureka Math and licensed by Great Minds. A STORY OF FUNCTIONS 2015 Great Minds. eureka- math.org. ALG II-M2-SE-1.3. .
