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COLLEGE ALGEBRA for ENGINEERING (ENGALG1) REVIEWER Quiz Number: 1 I. Multiple Choice 1. Which of the following is NOT a polynomial? a. 2 + 5x2 – 3x b. 9x6 c. √𝑥 2 + 2𝑥 + 2 d. 8 𝑥2 1 + + 5x + 2 𝑥 2. The property of addition/multiplication in which the order of the addends/factors to be added/multiplied is not important is called _______? a. Associative Property b. Closure Property c. Commutative Property d. Distributive Property 3. Which of the following is in factored form? a. (5x + 3)x – 2(4x – 7) b. 5y(2y – 5) – (7y + 2)(2y + 7) c. (x – 2)(4x + 7)(5x + 3) d. (2y + 7)(7y – 2)2y + 5 4. The variable x in a. b. c. d. 𝒂𝒎 𝒂𝒏 = 𝒂𝒙 will be positive if: m>n m<n m≥n m≤n 5. The factors of 9x2 – 12x – 45 are: a. 3(x + 3)(3x + 5) b. 3(x + 3)(3x – 5) c. 3(x – 3)(3x + 5) d. 3(x – 3)(3x – 5) “We are built to build” II. 1. Subtract the product of x2m + 2xmym – y2m and xm – ym from the product of xm + ym and x2m – y2m. 2. Divide 3x3y + 6x4 – 25y2 + 5x2y by 3x2 – 5y. 3. Remove the grouping symbols then simplify: – 2xy {6x2y + [– 6x3 – 2x2y – 2{2xy – x[3x2y – (3x – 2y)] – 3x2 + 2x2y}] – 3x2} III. 1. 2. 3. 4. 5. Factor the algebraic expressions completely. 40x8zy2 + 625x2zy3m+2 9a2 + 25b2 – 4c2 – 9d2 – 30ab + 12cd 4x4 – 61x2y2 + 9y4 90x2 – 120xy + 40y2 – 21x + 14y – 12 (x + 3y)3 – x3 + 9xy2 Parts of the reviewer were obtained from ENGALG1 Quiz Number 1 of Ms. W. Baraoidan AY 2014 – 2015 Term 1 “We are built to build” ANSWERS & SOLUTIONS COLLEGE ALGEBRA for ENGINEERING (ENGALG1) REVIEWER Quiz Number: 1 I. Multiple Choice 8 1 1. d. 𝑥 2 + 𝑥 + 5x + 2 2. 3. 4. 5. c. Commutative Property c. (x – 2)(4x + 7)(5x + 3) a. m > n c. 3(x – 3)(3x + 5) II. 1. Subtract the product of x2m + 2xmym – y2m and xm – ym from the product of xm + ym and x2m – y2m. x2m + 2xmym – y2m xm – ym 𝑥 3𝑚 + 2𝑥 2𝑚 𝑦 𝑚 − 𝑥 𝑚 𝑦 2𝑚 − 𝑥 2𝑚 𝑦 𝑚 − 2𝑥 𝑚 𝑦 2𝑚 + 𝑦 3𝑚 𝑥 3𝑚 + 𝑥 2𝑚 𝑦 𝑚 − 3𝑥 𝑚 𝑦 2𝑚 + 𝑦 3𝑚 xm + ym x2m – y2m 𝑥 3𝑚 + 𝑥 2𝑚 𝑦 𝑚 𝑥 3𝑚 + 𝑥 2𝑚 𝑦 𝑚 − 𝑥 𝑚 𝑦 2𝑚 − 𝑦 3𝑚 − 𝑥 𝑚 𝑦 2𝑚 − 𝑦 3𝑚 (xm + ym) (x2m – y2m) = 𝑥 3𝑚 + 𝑥 2𝑚 𝑦 𝑚 − 𝑥 𝑚 𝑦 2𝑚 − 𝑦 3𝑚 − (x2m + 2xmym – y2m)(xm – ym) = 𝑥 3𝑚 + 𝑥 2𝑚 𝑦 𝑚 − 3𝑥 𝑚 𝑦 2𝑚 + 𝑦 3𝑚 = 2𝒙𝒎 𝒚𝟐𝒎 − 𝟐𝒚𝟑𝒎 2. Divide 3x3y + 6x4 – 25y2 + 5x2y by 3x2 – 5y. 𝟓𝒙𝒚𝟐 6x4 + 3x3y + 5x2y – 25y2 ÷ 3x2 – 5y = 2x2 + xy + 5y + 𝟑𝒙𝟐 – 𝟓𝐲 “We are built to build” 3. Remove the grouping symbols then simplify: – 2xy {6x2y + [– 6x3y – 2x2y – 2{2xy – x[3x2y – (3x – 2y)] – 3x2 + 2x2y}] – 3x2} = = = = = = = = = III. – 2xy {6x2y + [– 6x3y – 2x2y – 2{2xy – x[3x2y – (3x – 2y)] – 3x2 + 2x2y}] – 3x2} – 2xy {6x2y + [– 6x3y – 2x2y – 2{2xy – x[3x2y – 3x + 2y] – 3x2 + 2x2y}] – 3x2} – 2xy {6x2y + [– 6x3y – 2x2y – 2{2xy – 3x3y + 3x2 – 2xy – 3x2 + 2x2y}] – 3x2} – 2xy {6x2y + [– 6x3y – 2x2y – 2{– 3x3y + 2x2y}] – 3x2} – 2xy {6x2y + [– 6x3y – 2x2y + 6x3y – 4x2y] – 3x2} – 2xy {6x2y + [– 6x3y – 6x2y + 6x3y] – 3x2} – 2xy {6x2y + [– 6x2y] – 3x2} – 2xy {6x2y – 6x2y – 3x2} – 2xy {– 3x2} 6x3y Factor the algebraic expressions completely. 1. 40x8zy2 + 625x2zy3m+2 = 5x2zy2 (8x6z + 125y3m) = 5x2zy2 (2x2z + 5ym)(4x4z – 10x2zym + 25y2m) 2. 9a2 + 25b2 – 4c2 – 9d2 – 30ab + 12cd = 9a2 – 30ab + 25b2 – 4c2 + 12cd – 9d2 = (9a2 – 30ab + 25b2) – (4c2 – 12cd + 9d2) = (3a – 5b)2 – (2c – 3d)2 = [3a – 5b + (2c – 3d)][3a – 5b – (2c – 3d)] = (3a – 5b + 2c – 3d)(3a – 5b – 2c + 3d) 3. 4x4 – 61x2y2 + 9y4 4x4 = (2x2)2 9y4 = (3y2)2 (2x2)( 3y2) = 12x2y2 = 4x4 – 61x2y2 + 49x2y2+ 9y4 – 49x2y2 = (4x4 – 12x2y2 + 9y4) – 49x2y2 = (2x2 – 3y2)2 – 49x2y2 = (2x2 – 3y2)2 – (7xy)2 = (2x2 – 3y2 + 7xy) (2x2 – 3y2 – 7xy) = (2x2 + 7xy – 3y2) (2x2 – 7xy – 3y2) Take out common factors Pattern: (x3 + y3) = (x + y)(x2 – xy + y2) Factoring by Grouping Perfect Square Trinomial Pattern: a2 – b2 = (a +b)(a – b) Middle Term of Resultant Perfect Square Binomial Add and subtract a perfect square monomial Pattern: a2 – b2 = (a +b)(a – b) “We are built to build” 4. 90x2 – 120xy + 40y2 – 21x + 14y – 12 = (90x2 – 120xy + 40y2) – (21x – 12y) – 12 = 10(9x2 – 12xy + 4y2) – 7(3x – 2y) – 12 = 10(3x – 2y)2 – 7(3x – 2y) – 12 = [2(3x – 2y) – 3] [5(3x – 2y) + 4] = (6x – 4y – 3)(15x – 10y + 4) Factoring by Grouping Take out common factors Perfect Square Trinomial 5. (x + 3y)3 – x3 + 9xy2 (Instruction: Do not expand) 3 3 2 = (x + 3y) – (x – 9xy ) Factoring by Grouping 3 2 2 = (x + 3y) – x(x – 9y ) Pattern: a2 – b2 = (a +b)(a – b) = (x + 3y)3 – x(x + 3y)(x – 3y) Take out common factors 2 = (x + 3y)[(x + 3y) – x(x – 3y)] = (x + 3y)[x2 + 6xy + 9y2 – x2 + 3xy] = (x + 3y)[9xy + 9y2] = 9y(x + y)(x + 3y) Prepared by: April Oilec Deanielle B. Pacayra Academics Officer Civil Engineering Society “We are built to build”