Algebra 2 Polinomios Expressoes Radicais

51 Operations with Rational Expressions Student Outcomes Students perform addition and subtraction of rational expressions Summary In this lesson we extended addition and subtraction of rational numbers to addition and subtraction of rational expressions The process for adding or subtracting rational expressions can be summarized as follows Find a common multiple of the denominators to use as a common denominator Find equivalent rational expressions for each expression using the common denominator Add or subtract the numerators as indicated and simplify if needed Exercises 1 Write each sum or difference as a single rational expression a 7 8 3 5 b 5 10 2 6 2 c 4 𝑥 3 2𝑥 2 Write as a single rational expression a 1 𝑥 1 𝑥1 b 3𝑥 2𝑦 5𝑥 6𝑦 𝑥 3𝑦 c 𝑎𝑏 𝑎2 1 𝑎 d 1 𝑝2 1 𝑝2 e 1 𝑝2 1 2𝑝 f 1 𝑏1 𝑏 1𝑏 g 1 1 1𝑝 h 𝑝𝑞 𝑝𝑞 2 i 𝑟 𝑠𝑟 𝑠 𝑟𝑠 j 4 𝑥4 2 4𝑥 k 3𝑛 𝑛2 3 2𝑛 l 8𝑥 3𝑦2𝑥 10𝑦 2𝑥3𝑦 m 1 2𝑚4𝑛 1 2𝑚4𝑛 𝑚 𝑚24𝑛2 n 1 2𝑎𝑏𝑎𝑐 1 𝑏𝑐𝑏2𝑎 o 𝑏21 𝑏24 1 𝑏2 1 𝑏2 3 Write each rational expression as an equivalent rational expression in lowest terms a 1 𝑎 1 2𝑎 4 𝑎 b 5𝑥 2 1 5𝑥 4 1 5𝑥 c 1 5𝑥 3 𝑥2 1 1 𝑥 7 𝑥2 1 Extension 4 Suppose that 𝑥 0 and 𝑦 0 We know from our work in this section that 1 𝑥 1 𝑦 is equivalent to 1 𝑥𝑦 Is it also true that 1 𝑥 1 𝑦 is equivalent to 1 𝑥𝑦 Provide evidence to support your answer 5 Suppose that 𝑥 2𝑡 1𝑡2 and 𝑦 1𝑡2 1𝑡2 Show that the value of 𝑥2 𝑦2 does not depend on the value of 𝑡 6 Show that for any real numbers 𝑎 and 𝑏 and any integers 𝑥 and 𝑦 so that 𝑥 0 𝑦 0 𝑥 𝑦 and 𝑥 𝑦 𝑦 𝑥 𝑥 𝑦 𝑎𝑥𝑏𝑦 𝑥𝑦 𝑎𝑥𝑏𝑦 𝑥𝑦 2𝑎 𝑏 7 Suppose that 𝑛 is a positive integer a Rewrite the product in the form 𝑃 𝑄 for polynomials 𝑃 and 𝑄 3 1 𝑛 3 1 𝑛1 b Rewrite the product in the form 𝑃 𝑄 for polynomials 𝑃 and 𝑄 3 1 𝑛 3 1 𝑛1 3 1 𝑛2 c Rewrite the product in the form 𝑃 𝑄 for polynomials 𝑃 and 𝑄 3 1 𝑛 3 1 𝑛1 3 1 𝑛2 3 1 𝑛3 d If this pattern continues what is the product of 𝑛 of these factors 53 Rational Equations and Inequalities Student Outcomes Students solve rational equations monitoring for the creation of extraneous solutions Summary In this lesson we applied what we have learned in the past two lessons about addition subtraction multiplication and division of rational expressions to solve rational equations An extraneous solution is a solution to a transformed equation that is not a solution to the original equation For rational functions extraneous solutions come from the excluded values of the variable Rational equations can be solved one of two ways Write each side of the equation as an equivalent rational expression with the same denominator and equate the numerators Solve the resulting polynomial equation and check for extraneous solutions Multiply both sides of the equation by an expression that is the common denominator of all terms in the equation Solve the resulting polynomial equation and check for extraneous solutions Exercises 1 Solve the following equations and check for extraneous solutions a 𝑥8 𝑥4 3 b 4𝑥8 𝑥2 4 c 𝑥4 𝑥3 1 d 4𝑥8 𝑥2 4 e 1 2𝑎 2 2𝑎3 0 f 3 2𝑥1 5 4𝑥3 g 5 𝑥5 2 5𝑥 2 𝑥 h 𝑦2 3𝑦2 𝑦 𝑦1 2 3 i 3 𝑥1 2 1𝑥 1 j 4 𝑥1 3 𝑥 3 0 k 𝑥1 𝑥3 𝑥5 𝑥2 17 6 l 𝑥7 4 𝑥1 2 5𝑥 3𝑥14 m 𝑏2𝑏6 𝑏2 2𝑏12 𝑏 𝑏39 2𝑏 n 1 𝑝𝑝4 1 𝑝6 𝑝 o 1 ℎ3 ℎ4 ℎ2 6 ℎ2 p 𝑚5 𝑚2𝑚 1 𝑚2𝑚 𝑚6 𝑚1 2 Create and solve a rational equation that has 0 as an extraneous solution 3 Create and solve a rational equation that has 3 as an extraneous solution Extension 4 Two lengths 𝑎 and 𝑏 where 𝑎 𝑏 are in golden ratio if the ratio of 𝑎 𝑏 is to 𝑎 is the same as 𝑎 is to 𝑏 Symbolically this is expressed as 𝑎 𝑏 𝑎𝑏 𝑎 We denote this common ratio by the Greek letter phi pronounced fee with symbol 𝜑 so that if 𝑎 and 𝑏 are in common ratio then 𝜑 𝑎 𝑏 𝑎𝑏 𝑎 By setting 𝑏 1 we find that 𝜑 𝑎 and 𝜑 is the positive number that satisfies the equation 𝜑 𝜑1 𝜑 Solve this equation to find the numerical value for 𝜑 5 Remember that if we use 𝑥 to represent an integer then the next integer can be represented by 𝑥 1 a Does there exist a pair of consecutive integers whose reciprocals sum to 5 5 Explain how you know b Does there exist a pair of consecutive integers whose reciprocals sum to 3 5 Explain how you know c Does there exist a pair of consecutive even integers whose reciprocals sum to 2 4 Explain how you know Does there exist a pair of consecutive even integers whose reciprocals sum to 5 7 Explain how you know 43 1 1 4a 4 a 44 a 4aa 16 a 4a2 2 x x 2 2 x x 2x 2 x2 2x 2 3 14 12 z 18 z2 14 12 z 14 18 z2 3 z 92 z2 4 5 x x3 10 5 x x3 5 10 x 5 x4 50 x 5 x 3 x 4 x x 4 x 3 x 3 4 x2 4 3 x 12 x2 x 12 6 3 z 1 3 z2 1 3 3 3 z2 3 z 3 z2 1 9 z3 3 z2 3 z 1 7 12 w 1 10 w 1 12 10 w w 12 w 10 w 1 120 w2 12 10 w 1 120 w2 2 w 1 1 5 8 6 w 3 w2 6 w w2 3 w2 6 w3 3 w2 9 15 s100 12 s200 0125 s 15 s100 12 s200 18 s 15 12 s100 s200 15 18 s100 s 152 s300 158 s101 10 2 q 1 2 q2 1 22 q2 2q 2 q2 1 4 q3 2 q2 2 q 1 11 x2 x 1 x 1 x2 x x2 x x 1 x x 1 x3 x2 x2 x x 1 x3 1 1 x2 1 1 x 1 x3 2 x2 2 x 1 12 4 x z 9 x y z 2 y z x y z 4 9 x x y z 4 x z2 2 x y z 2 y y z 2 y z z 36 x2 y z 4 x z2 2 x y z 2 y2 z 2 y z2 2 6 13 t 1 t t t2 1 t t t t 1 t2 1 t2 1 t2 1 t2 t2 t2 t2 1 t4 1 14 w 1 w4 w3 w2 w 1 w w4 w w3 w w2 w w w4 w3 w2 w 1 w5 w4 w3 w2 w w4 w3 w2 w 1 w5 1 1 w4 1 1 w3 1 1 w2 1 1 w 1 w5 0 w4 0 w3 0 w2 0 w 1 w5 1 15 z 2 z 1 3 z 2 z 2 3 z 2 2 2 z 3 z 2 z 6 z2 4 z 3 z 2 z 6 z2 z 2 6 z3 z2 2 z 3 16 x yy zz x x y x z y2 y zz x x y z x x y x y z x x z y2 z x y2 y z2 x y z2 x y2 z x2 y x2 z y2 z x y2 y z2 x y z 2 x y z x2 y x2 z x2 z y2 z x y2 y z2 17 x y4 14 x y 14 x 14 y 18 20 f10 10 f4 5 20 f10 10 f4 15 2015 f10 1015 f4 4 f10 2 f4 19 5 y y2 y 2 2 2 y3 5 y y2 5 y y 5 2 y 22 2y3 5 y3 5 y2 10 y 4 2 y3 5 2 y3 5 y2 10 y 4 3 y3 5 y2 10 y 4 4 20 a b ca b c17 sd 117 aa ab ac ab bb bc ac bc cc sd 117 a2 2 ab b2 c2 sd 117 a2 sd 2 ab sd b2 sd c2 sd 117 a2 sd 217 ab sd 117 b2 sd 117 c2 sd 21 2 x 9 5 x 2 2 29 x 52 x 12 22 92 x 59 29 x12 418 x 4518 x12 4 45 18 x 12 4918 12 x 491 182 x 4936 x 5 22 2 f3 2 f 1f2 f 2 2 f3 f2 2 f3 f 2 f3 2 2 f f2 2 f f 2 f 2 f2 f 2 2 f5 2 f4 4 f3 2 f3 2 f2 4 f f2 f 2 2 f5 2 f4 4 2 f3 2 1 f2 4 1 f 2 2 f5 2 f4 6 f3 3 f2 5 f 2 2 1 a b2 a ba b aa ab ab bb a2 2 ab b2 2 a 12 a 1a 1 aa a a 1 a2 2 a 1 3 4 b2 4 b4 b 44 4 b 4 b bb 16 4 4 b b2 16 8 b 16 4 3 12 3 13 1 33 3 3 1 9 3 3 1 16 6 5 x y z2 x y zx y z xx xy xz xy yy yz xz yz zz x2 2xy 2xz y2 2yz z2 6 x 1 z2 x 1 zx 1 z xx x xz x 1 z xz z zz x2 2x 2xz 2z z2 1 7 3 z2 3 z3 z 3x3 3z 3z zz 9 6z z2 z2 6z 9 8 p q3 p qp qp q pp pq pq qqp q p2 2pq q2p q p2p p2q 2pqq 2qqq p2 q2q p3 p2q 2p2q 2q2 p2 q3 p3 3p2q 3pq2 q3 7 9 p 13 p 1p 1p 1 p 1pp p p 11 p 1p2 2p 1 pp2 2pp p p2 2p 1 p3 2p2 p p2 2p 1 p3 3p2 3p 1 10 6 q3 6 q6 q6 q 6x6 6q 6q qq6 q 36 12q q26 q 36x6 36q 12x6q 1299 6q2 qq3 216 36q 72q 129 6q2 q3 q3 18q2 108q 216 3 1 s2 3s 1 s3 82 3s 3 s4 s2 3s 3 2 3s2 2s 1 3s3 s2 2s 2 3s3 s2 2s 2 3s3 3s2 6s 6 8 3 5s2 4s 1 5s3 s2 4s 4 ss3 s2 4s 4 s4 s3 4s2 4s 4 s 2s2 4s 1 s 2s3 s2 4s 1 s3 s2 4s s 2s3 2s2 8s 2 s4 s3 4s2 s 2s3 2s2 8s 2 s4 1 2s3 4 2s2 1 7s 2 s4 s3 2s2 6s 2 5 u 1u5 u4 u3 u2 u 1 u5 u4 u3 u2 u u u5 u4 u3 u2 u 1 u6 u5 u4 u3 u2 u u5 u4 u3 u2 u 1 u6 1 6 5 u 1u5 u4 u3 u2 u 1 5 u6 1 5 u6 5 7 u7 u3 1u 1u5 u4 u3 u2 u 1 u7 u3 1u6 1 u7 u6 u7 u3 u6 u3 u6 1 u13 u7 u9 u3 u6 1 u13 u9 u7 u6 u3 1 9 4 Yes all of the linked expressions are polynomials because sum of polynomials is a polynomial difference of polynomials is a polynomial and product of polynomials is a polynomial 45 1 1 25 x2 25 x 15 negative sign 52 x2 5x2 sqrt152 2 5 x sqrt15 10 sqrt15 x 25x 2 9 x2 y2 20 x y 8 sqrt82 32 x2 y2 3 x y2 2 3 x y sqrt8 6 sqrt8 x y 20 x y negative sign 3 50 y2 15 y 10 sqrt102 sqrt502 sqrt50 y2 2 sqrt50 y sqrt10 y 2 sqrt500 y 2 5 100 y 2 sqrt5 sqrt100 y 2 sqrt5 10 y 20 sqrt5 y 15 y 1 4 y8 y3 6 negative sign y42 sqrt62 2 y4 sqrt6 2 sqrt 6 y4 y3 5 x3 200 x3 sqrt3200 sqrt38253 sqrt38 sqrt3253 2 sqrt3253 x3 200 x 2 sqrt325x2 2 sqrt325 x 2 sqrt3252 x 2 sqrt325 x2 2 sqrt325 x 223 sqrt3522 x 2 sqrt325x2 2 sqrt325 x 4 sqrt354 x 2 sqrt325x2 2 sqrt325 x 4 5 sqrt35 x 2 sqrt325x2 2 sqrt325 20 sqrt35 6 2 x5 16 x 2 x x4 8 x22 sqrt82 sqrt232 2 sqrt22 2 x5 16 x 2 x2 2 sqrt2 x2 2 sqrt2 2 7 10x2 25y4 z6 52x2 5y4 z6 22 x2 2 x2 52y22z32 5 y2 z32 10x2 25 y4 z6 52 x 5 y2 z32 x 5 y2 z3 8 40 x6 y4 z2 25 x2 z10 5 x2 z2 8 x4 y4 5 z8 52 z42 5 z42 82 x22 y22 2 2 x2 y22 40 x6 y4 z2 25 x2 z10 5 x2 z2 2 2 x2 y2 5 z42 2 x2 y2 5 z4 9 5 x2 9 z2 sum of squares 52 x2 5 x2 10 x4 36 x2 6x2 6 x22 62 11 1 27 x9 1 3 x312 3 x3 3 x32 13 33 x33 3 x33 1 3 x31 3 x3 9 x6 12 x4 y7 8 z3 23 3 2 z3 3 x4 3 y7 3 x 3 y73 x y2 3 x y 3 y3 x4 y7 8 z3 x y2 3 x y 2 zx y2 3 x y2 x y2 3 x y 2 z 2 z2 x y2 3 x y 2 zx2 y4 3 x2 y2 2 x y2 3 x y z 4 z2 2 1 No y4 4 y2 16 and 2 y2 4 8 y2 4 y2 y22 2 y23 43 y2 4y4 4 y2 16 4 y 2y 2y4 4 y2 16 3 y32 82 y3 8y3 8 y 2y2 2 y 4y 2y2 2 y 4 4 y6 64 y 2y 2y4 4 y2 16 y6 64 y 2y2 2 y 4y 2y2 2 y 4 y 2y 2y4 4 y2 16 y 2y2 2 y 4y 2y2 2 y 4 y4 4 y2 16 y2 2 y 4y2 2 y 4 5 If the factorization as difference of squares is applied first we get a product of a sum of cubes by a difference of cubes each of them can be factorized 3 1 169 π2 x10 y6 13 π x5 y313 π x5 y3 5 2 27x3 y12 z54 1 3xy4 z18 19x2 y8 z36 3xy4 z18 1 3 125x30 216x6 y216 5x10 6x2 y7225x20 30x12 y72 36x4 y144 57 Radical Equations and Functions Student Outcomes Students develop facility in solving radical equations Summary If 𝑎 𝑏 and 𝑛 is an integer then 𝑎𝑛 𝑏𝑛 However the converse is not necessarily true The statement 𝑎𝑛 𝑏𝑛 does not imply that 𝑎 𝑏 Therefore it is necessary to check for extraneous solutions when both sides of an equation are raised to an exponent Exercises Solve 1 2𝑥 6 𝑥 6 0 1 2𝑥 5 𝑥 7 0 2 𝑥 5 𝑥 6 2 3 2𝑥 5 𝑥 8 2 4 𝑥 4 2 𝑥 5 𝑥 5 3 𝑥 6 𝑥 3 5𝑥 6 3 7 2𝑥 6 𝑥 1 8 𝑥 12 𝑥 6 9 2𝑥 1 4𝑥 1 10 3𝑥 4𝑥 1 11 4𝑥 1 2 2𝑥 12 𝑥 3 4𝑥 2 13 2𝑥 8 3𝑥 12 0 14 𝑥 2𝑥 4 4 15 𝑥 2 9𝑥 36 16 Consider the right triangle 𝐴𝐵𝐶 shown to the right with 𝐴𝐵 8 and 𝐵𝐶 𝑥 a Write an expression for the length of the hypotenuse in terms of 𝑥 b Find the value of 𝑥 for which 𝐴𝐶 𝐴𝐵 9 17 Consider the triangle 𝐴𝐵𝐶 shown to the right where 𝐴𝐷 𝐷𝐶 and 𝐵𝐷 is the altitude of the triangle a If the length of 𝐵𝐷 is 𝑥 cm and the length of 𝐴𝐶 is 18 cm write an expression for the lengths of 𝐴𝐵 and 𝐵𝐶 in terms of 𝑥 b Write an expression for the perimeter of 𝐴𝐵𝐶 in terms of 𝑥 c Find the value of 𝑥 for which the perimeter of 𝐴𝐵𝐶 is equal to 38 cm 55 Expressions with Roots Student Outcomes Students understand that the sum of two square roots or two cube roots is not equal to the square root or cube root of their sum Students convert expressions to simplest radical form Students understand that the product of conjugate radicals can be viewed as the difference of two squares Summary For real numbers 𝑎 0 and 𝑏 0 where 𝑏 0 when 𝑏 is a denominator 𝑎𝑏 𝑎 𝑏 and 𝑎 𝑏 𝑎 𝑏 For real numbers 𝑎 0 and 𝑏 0 where 𝑏 0 when 𝑏 is a denominator 𝑎𝑏 3 𝑎 3 𝑏 3 and 𝑎 𝑏 3 𝑎 3 𝑏 3 Two binomials of the form 𝑎 𝑏 and 𝑎 𝑏 are called conjugate radicals 𝑎 𝑏 is the conjugate of 𝑎 𝑏 and 𝑎 𝑏 is the conjugate of 𝑎 𝑏 For example the conjugate of 2 3 is 2 3 To rewrite an expression with a denominator of the form 𝑎 𝑏 in simplest radical form multiply the numerator and denominator by the conjugate 𝑎 𝑏 and combine like terms Exercises 1 Express each of the following as a rational number or in simplest radical form Assume that the symbols 𝑎 𝑏 and 𝑥 represent positive numbers 1 30 2 82 3 28 4 98 5 28𝑥2 6 15 3 7 26𝑎 3 8 9𝑎2 9𝑏2 2 Express each of the following in simplest radical form combining terms where possible 1 25 55 20 2 33 3 4 1 3 3 54 3 8 3 7 1 4 3 4 5 8 3 40 3 8 9 3 3 Evaluate 𝑥2 𝑦2 when 𝑥 33 and 𝑦 15 4 Evaluate 𝑥2 𝑦2 when 𝑥 20 and 𝑦 10 5 Express each of the following as a rational expression or in simplest radical form Assume that the symbols 𝑥 and 𝑦 represent positive numbers 1 36 3 2 3 2 2 3 2 32 3 4 2 252 25 5 7 37 3 6 32 732 7 7 𝑥 3𝑥 3 8 2𝑥2 𝑦2𝑥2 𝑦 6 Simplify each of the following quotients as far as possible 1 21 3 3 2 5 4 5 1 3 3 2 32 5 4 25 3 35 42 7 If 𝑥 2 3 show that 𝑥 1 𝑥 has a rational value 8 Evaluate 5𝑥2 10𝑥 when the value of 𝑥 is 25 2 9 Write the factors of 𝑎4 𝑏4 Express 3 2 4 3 2 4 in a simpler form 10 The converse of the Pythagorean theorem is also a theorem If the square of one side of a triangle is equal to the sum of the squares of the other two sides then the triangle is a right triangle Use the converse of the Pythagorean theorem to show that for 𝐴 𝐵 𝐶 0 if 𝐴 𝐵 𝐶 then 𝐴 𝐵 𝐶 so that 𝐴 𝐵 𝐴 𝐵 41 Add and Subtract Polynomials Student Outcomes Students understand that the sum or difference of two polynomials produces another polynomial and relate polynomials to the system of integers students add and subtract polynomials Summary A monomial is a polynomial expression generated using only the multiplication operator Thus it does not contain or operators Monomials are written with numerical factors multiplied together and variable or other symbols each occurring one time using exponents to condense multiple instances of the same variable A polynomial is the sum or difference of monomials The degree of a monomial is the sum of the exponents of the variable symbols that appear in the monomial The degree of a polynomial is the degree of the monomial term with the highest degree Exercises 1 Celina says that each of the following expressions is actually a binomial in disguise 1 6𝑎𝑏𝑐 3𝑎2 7𝑎𝑏𝑐 2 7𝑥3 2𝑥2 10𝑥4 3𝑥5 3𝑥 2𝑥4 3 𝑡 22 6𝑡 4 6𝑎 1 10𝑎 1 100𝑎 1 5 2𝜋𝑟 𝜋𝑟2𝑟 2𝜋𝑟 𝜋𝑟2 2𝑟 For example she sees that the expression in i is algebraically equivalent to 11𝑎𝑏𝑐 2𝑎2 which is indeed a binomial She is happy to write this as 11𝑎𝑏𝑐 2𝑎2 if you prefer Is she right about the remaining four expressions 2 Janie writes a polynomial expression using only one variable 𝑥 with degree 4 Max writes a polynomial expression using only one variable 𝑥 with degree 8 1 What can you determine about the degree of the sum of Janies and Maxs polynomials 2 What can you determine about the degree of the difference of Janies and Maxs polynomials 3 Suppose Janie writes a polynomial expression using only one variable 𝑥 with degree of 6 and Max writes a polynomial expression using only one variable 𝑥 with degree of 5 1 What can you determine about the degree of the sum of Janies and Maxs polynomials 2 What can you determine about the degree of the difference of Janies and Maxs polynomials 4 Find each sum or difference by combining the parts that are alike 1 2𝑝 4 6𝑝 1 𝑝 7 2 7𝑥4 9𝑥 3𝑥4 13 3 6 𝑡 𝑡4 9𝑡 𝑡4 6 12𝑥 1 3𝑥 4 𝑥 15 7 13𝑥2 5𝑥 4𝑥2 1 8 9 𝑡 𝑡2 4 2 8𝑡 2𝑡2 4 5 𝑡2 7𝑡2 8 𝑡2 12 5 8𝑥3 5𝑥 4𝑥3 2 9 4𝑚 6 14𝑚 3 𝑚 2 10 15𝑥4 10𝑥 15𝑥4 4𝑥 43 Multiply and Divide Polynomials Student Outcomes Students understand that the product of two polynomials produces another polynomial students multiply polynomials Exercises 1 Use the distributive property to write each of the following expressions as the sum of monomials 1 4𝑎4 𝑎 2 𝑥𝑥 2 2 3 1 4 12𝑧 18𝑧2 4 5𝑥𝑥3 10 5 𝑥 3𝑥 4 6 3𝑧 13𝑧2 1 7 12𝑤 110𝑤 1 8 6𝑤 3𝑤2 9 15𝑠100 1 2 𝑠200 0125𝑠 10 2𝑞 12𝑞2 1 11 𝑥2 𝑥 1𝑥 1 12 4𝑥𝑧9𝑥𝑦 𝑧 2𝑦𝑧𝑥 𝑦 𝑧 13 𝑡 1𝑡 1𝑡2 1 14 𝑤 1𝑤4 𝑤3 𝑤2 𝑤 1 15 𝑧2𝑧 13𝑧 2 16 𝑥 𝑦𝑦 𝑧𝑧 𝑥 17 𝑥𝑦 4 18 20𝑓10 10𝑓4 5 19 5𝑦𝑦2 𝑦 2 22 𝑦3 20 𝑎𝑏𝑐𝑎𝑏𝑐 17 sd 21 2𝑥 9 5𝑥 2 2 22 2𝑓3 2𝑓 1𝑓2 𝑓 2 2 Use the distributive property and your wits to write each of the following expressions as a sum of monomials If the resulting polynomial is in one variable write the polynomial in standard form 1 𝑎 𝑏2 2 𝑎 12 3 4 𝑏2 4 3 12 5 𝑥 𝑦 𝑧2 6 𝑥 1 𝑧2 7 3 𝑧2 8 𝑝 𝑞3 9 𝑝 13 10 6 𝑞3 3 Use the distributive property and your wits to write each of the following expressions as a polynomial in standard form 1 𝑠2 3𝑠 1 2 3𝑠2 2𝑠 1 3 𝑠𝑠2 4𝑠 1 4 𝑠 2𝑠2 4𝑠 1 5 𝑢 1𝑢5 𝑢4 𝑢3 𝑢2 𝑢 1 6 5𝑢 1𝑢5 𝑢4 𝑢3 𝑢2 𝑢 1 7 𝑢7 𝑢3 1𝑢 1𝑢5 𝑢4 𝑢3 𝑢2 𝑢 1 4 Beatrice writes down every expression that appears in this problem set one after the other linking them with signs between them She is left with one very large expression on her page Is that expression a polynomial expression That is is it algebraically equivalent to a polynomial What if she wrote signs between the expressions instead What if she wrote signs between the expressions instead 45 Factoring Student Outcomes Students use the structure of polynomials to identify factors Sumary In this lesson we learned additional strategies for factoring polynomials The difference of squares identity 𝑎2 𝑏2 𝑎 𝑏𝑎 𝑏 can be used to factor more advanced binomials Trinomials can often be factored by looking for structure and then applying our previous factoring methods Sums and differences of cubes can be factored by the formulas 𝑥3 𝑎3 𝑥 𝑎𝑥2 𝑎𝑥 𝑎2 𝑥3 𝑎3 𝑥 𝑎𝑥2 𝑎𝑥 𝑎2 Exercises 1 If possible factor the following expressions using the techniques discussed in this lesson 1 25𝑥2 25𝑥 15 2 9𝑥2𝑦2 20𝑥𝑦 8 3 50𝑦2 15𝑦 10 4 𝑦8 𝑦3 6 5 𝑥3 200 6 2𝑥5 16𝑥 7 10𝑥2 25𝑦4𝑧6 8 40𝑥6𝑦4𝑧2 25𝑥2𝑧10 9 5𝑥2 9 10 𝑥4 36 11 1 27𝑥9 12 𝑥4𝑦7 8𝑧3 2 Consider the polynomial expression 𝑦4 4𝑦2 16 1 Is 𝑦4 4𝑦2 16 factorable using the methods we have seen so far 2 Factor 𝑦6 64 first as a difference of cubes and then factor completely 𝑦23 43 3 Factor 𝑦6 64 first as a difference of squares and then factor completely 𝑦32 82 4 Explain how your answers to parts b and c provide a factorization of 𝑦4 4𝑦2 16 5 If a polynomial can be factored as either a difference of squares or a difference of cubes which formula should you apply first and why 3 Create expressions that have a structure that allows them to be factored using the specified identity Be creative and produce challenging problems 1 Difference of squares 2 Difference of cubes 3 Sum of cubes 47 Polynomial Equations Student Outcomes Students find solutions to polynomial equations where the polynomial expression is not factored into linear factors Students construct a polynomial function that has a specified set of zeros with stated multiplicity Sumary Given any two polynomial functions 𝑝 and 𝑞 the solution set of the equation 𝑝𝑥𝑞𝑥 0 can be quickly found by solving the two equations 𝑝𝑥 0 and 𝑞𝑥 0 and combining the solutions into one set The number 𝑎 is a zero of a polynomial function 𝑝 with multiplicity 𝑚 if the factored form of 𝑝 contains 𝑥 𝑎𝑚 Exercises For Problems 14 find all solutions to the given equations 1 𝑥 4𝑥 2 0 2 𝑥 6𝑥 2𝑥 3 0 3 3𝑥 4𝑥 5 0 4 2𝑥 53𝑥 1𝑥 1 0 5 Find four solutions to the equation 𝑥2 9𝑥5 16 0 6 Find the zeros with multiplicity for the function 𝑝𝑥 𝑥3 8𝑥5 4𝑥3 7 Find two different polynomial functions that have zeros at 1 3 and 5 of multiplicity 1 8 Find two different polynomial functions that have a zero at 2 of multiplicity 5 and a zero at 4 of multiplicity 3 9 Find three solutions to the equation 𝑥2 9𝑥3 8 0 10 Find two solutions to the equation 𝑥3 64𝑥5 1 0 11 If 𝑝 𝑞 𝑟 𝑠 are nonzero numbers find the solutions to the equation 𝑝𝑥 𝑞𝑟𝑥 𝑠 0 in terms of 𝑝 𝑞 𝑟 𝑠 Use the identity 𝑎2 𝑏2 𝑎 𝑏𝑎 𝑏 to solve the equations given in Problems 1213 12 3𝑥 22 5𝑥 12 13 𝑥 72 2𝑥 42 14 Consider the polynomial function 𝑃𝑥 𝑥3 2𝑥2 2𝑥 5 a Divide 𝑃 by the divisor 𝑥 1 and rewrite in the form 𝑃𝑥 divisorquotient remainder b Evaluate 𝑃1 15 Consider the polynomial function 𝑄𝑥 𝑥6 3𝑥5 4𝑥3 12𝑥2 𝑥 3 a Divide 𝑄 by the divisor 𝑥 3 and rewrite in the form 𝑄𝑥 divisorquotient remainder b Evaluate 𝑄3 16 Consider the polynomial function 𝑅𝑥 𝑥4 2𝑥3 2𝑥2 3𝑥 2 a Divide 𝑅 by the divisor 𝑥 2 and rewrite in the form 𝑅𝑥 divisorquotient remainder b Evaluate 𝑅2 17 Consider the polynomial function 𝑆𝑥 𝑥7 𝑥6 𝑥5 𝑥4 𝑥3 𝑥2 𝑥 1 a Divide 𝑆 by the divisor 𝑥 1 and rewrite in the form 𝑆𝑥 divisorquotient remainder b Evaluate 𝑆1 18 Make a conjecture based on the results of Problems 1417 47 1 x4x20 x40 or x20 x4 or x2 S24 2 x6x2x30 x60 or x20 or x30 x6 or x2 or x3 S3 2 6 3 3x4x50 3x40 or x50 3x4 or x5 x43 or x5 S5 43 4 2x53x1x10 2x50 or 3x10 or x10 2x5 or 3x1 or x1 x52 or x13 or x1 S13 1 52 5 x2 9x5 16 0 x3x3x5 16 0 x30 or x30 or x5 160 x3 or x3 or x5 16 x3 or x3 or x 516 Theres three real solutions actually S3 516 3 6 px x3 8x5 4x3 x 2x2 2x 4 x3 x2 4 x 2x2 2x 4 x3 x 2x 2 x3 x 2x 22 x2 2x 4 0 is a zero of p with multiplicity 3 2 is a zero of p with multiplicity 1 2 is a zero of p with multiplicity 2 7 p1x x 1x 3x 5 x2 3x x 3 x 5 x2 4x 3x 5 x2 x 5x2 4xx 45x 3x 35 x3 5x2 4x2 20x 3x 15 x3 9x2 23x 15 p2x 2x1x3x5 2x3 9x2 23x 15 2x3 18x2 46x 30 8 p1x x 25 x 43 x 25 x 43 p2x 10 x 25 x 43 9 x29x38 0 x3x 3 x 2x2 2x 4 0 Δ 22 414 4 16 12 0 so there is no x R such x2 2x 4 0 x 3 0 or x 3 0 or x 2 0 x 3 or x 3 or x 2 10 x3 64x5 1 0 x 4 x2 4x 16 x5 1 0 Δ 42 4116 16 64 48 0 so there is no x R that x2 4x 16 0 x 4 0 or x5 1 0 x 4 or x 51 1 11 px qrx s 0 px q 0 or rx s 0 px q or rx s x qp or x sr 12 3x 22 5x 12 3x 22 5x 12 0 3x 2 5x 13x 2 5x 1 0 3x 2 5x 18x 1 0 2x 38x 1 0 2x 3 0 or 8x 1 0 2x 3 or 8x 1 x 32 or x 18 13 x 72 2x 42 x 72 2x 42 0 x 7 2x 4x 7 2x 4 0 x 7 2x 43x 11 0 x 33x 11 0 6 x 3 0 or 3x 11 0 x 3 or x 113 14 a x3 2x2 2x 5 x 1 x2 3x 5 x3 x2 3x2 2x 5 3x2 3x 5x 5 5x 5 0 Px x 1x2 3x 5 0 b P1 1 112 3 1 5 0 0 7 15 a x6 3x5 0x4 4x3 12x2 x 3 x 3 x5 4x2 1 x6 3x5 0 4x3 12x2 x 3 4x3 12x2 0 x 3 x 3 0 Qx x 3x5 4x2 1 0 b Q3 3 335 4 32 1 0 16 a x4 2x3 2x2 3x 2 x 2 x3 2x 1 x4 2x3 0 2x2 3x 2 2x2 4x x 2 x 2 0 Rx x 2x3 2x 1 0 8 b R2 2 2 23 22 1 0 17 a x7 x6 x5 x4 x3 x2 x 1 x 1 x6 x4 x2 1 x7 x6 0 x5 x4 x3 x2 x 1 x5 x4 0 x3 x2 x 1 x3 x2 0 x 1 x 1 0 sx x 1 x6 x4 x2 1 0 b S1 1 1 16 14 12 1 0 9 4 1 1 1 6abc 3a2 7abc 3a2 6 7 abc 3a2 13abc Its a binomial 2 7x3 2x2 10x4 3x5 3x 2 x4 7 2 x32 10x4 3x5 3 2 x14 14x5 10x4 3x5 6 x5 14 3 6 x5 10x4 11x5 10 x4 Its a binomial 3 t 22 6t t2 2 t 2 22 6t t2 4t 4 6t t2 4 6 t 4 t2 2 t 4 It isnt a binomial 1 4 6a1 10a1 100a1 610100a1 96 a1 96a 96 Its a binomial 5 2πr πr2r 2πr πr2 2r 2πr πr2r 2r 2πr πr2r 2πrr πr2r 2πr2 πr3 Its a binomial 2 1 As Janies polynomial degree is 4 the highest degree of its terms is 4 while Maxs is 8 So when adding the polynomials the numerical factor of the term with degree 8 dont change Besides that when adding polynomials just the terms coefficients are changed and the only way of changing the degree is by adding polynomials of same degree whose higher terms coefficients add to 0 Therefore the degree of the sum of 2 Maxs and Janies polynomials is 8 2 As difference is a sum in disguise the degreell still be 8 3 1 The degree will be 6 2 The degree will be 6 4 1 2p 4 6p 1 p 7 2p 4 6p 1 p 7 2 6 1p 4 1 7 7p 4 2 7x4 9x 3x4 13 7x4 9x 3x4 313 7 3x4 9x 39 4x4 9x 39 3 7 13x2 5x 4x2 1 13x2 5x 4x2 4 13 4x2 5x 4 9x2 5x 4 8 9 t t2 42 8t 2t2 9 t t2 28t 2t2 9 t t2 28t 22t2 9 t t2 16t 4t2 1 4t2 1 16t 9 5t2 17t 9 9 4m 6 14m 3 m 2 4m 6 14m 143 m 2 4 14 1m 6 42 2 9m 50 10 15x4 10x 15x4 4x 15x4 10x 15x4 154x 15x4 10x 15x4 60x 15 15x4 10 60x 0x4 50x 50x