Algebra Quiz Answers

Week 1 Quiz: Review July 9, 2014 Rules for Exponents Question 1: Simplify the following expression using the various rules for manipulating exponents: x6 √ x3 (A) √ x4 (B) x3 √ (C) x9 (D) x5.5 (E) None of the above Answer: (C) 6 √x x3 = x6 3 x2 3 9 = x6− 2 = x4.5 = x 2 = √ x9 Question 2: Simplify the following expression using the various rules for manipulating exponents: √ x3 x 3 (A) x 2 (B) x7 √ (C) x4 √ (D) x7 (E) None of the above √ √ 1 1 7 Answer: (D) x3 x = x3 x 2 = x3+ 2 = x 2 = x7 Question 3: Simplify the following expression using the various rules for manipulating exponents:  (A) x7 y 11 (B) x12 y 24 x4 y8 (C) (x4 − y 8 )3 q 4 (D) 3 xy8 (E) None of the above 1 3 Answer: (B)  x4 y8 3 = x4∗3 y 8∗3 = x12 y 24 Question 4: Simplify the following expression using the various rules for manipulating exponents: 1 (x2 y 2 x3 y 3 )− 6 1 1 1 1 Answer: (x2 y 2 x3 y 3 )− 6 = (x2+3 y 2+3 )− 6 = (x5 y 5 )− 6 = (xy)5∗− 6 = 1 5 (xy) 6 Question 5: Simplify the following expression using the various rules for manipulating exponents: x−1 !2 1 y− 2  Answer: x−1 y 2 = −1 2 x−1∗2 y −1 2 ∗2 = x−2 y −1 = y x2 Question 6: Simplify the following expression using the various rules for manipulating exponents: 1 x− 4 y 3 !1 3 1 x2  Answer: 1 x− 4 y 3 1 1 x2 3 1 1 1 3 1 1 = (x− 4 − 2 y 3 ) 3 = (x− 4 y 3 ) 3 = x− 4 y = y 1 x4 Polynomials Question 7: Perform the indicated operation: (2x + 7)2 (A) 8x2 + 18x − 35 (B) 8x2 + 49 (C) 4x2 + 28x + 49 (D) 8x2 + 18x − 42 (E) None of the above Answer: (C) (2x + 7)2 = (2x + 7)(2x + 7) = 4x2 + 14x + 14x + 49 = 4x2 + 28x + 49 Question 8: Perform the indicated operation: (4x + 3y)(5x2 − 2xy + 6y 2 ) (A) 20x3 + 7x2 y + 18xy 2 + 18y 3 (B) 20x3 − 2xy + 18y 3 (C) 9x3 + 8x2 y + 6xy 2 + 9y 3 (D) 20x2 + 25xy + 18y 2 (E) None of the above Answer: (A) (4x+3y)(5x2 −2xy +6y 2 ) = 20x3 −8x2 y +24xy 2 +15x2 y −6xy 2 +18y 3 = 20x3 +7x2 y +18xy 2 +18y 3 2 Question 9: Perform the indicated operation: (7x2 + 3x)2 Answer: (7x2 + 3x)2 = (7x2 + 3x)(7x2 + 3x) = 49x4 + 21x3 + 21x3 + 9x2 = 49x4 + 42x3 + 9x2 Question 10: Perform the indicated operation: 1 (100x) 2 (100x2 + 80xy + 16y 2 ) 1 1 1 1 Answer: (100x) 2 (100x2 + 80xy + 16y 2 ) = 10x 2 (100x2 + 80xy + 16y 2 ) = 1000x + 800x 2 y + 160y 2 x 2 Factoring Question 11: Simplify by factoring out the greatest common factor: 55x8 y 9 − 22x6 y 4 − 99x5 y 7 (A) 11(5x8 y 9 − 2x6 y 4 − 9x5 y 7 ) (B) 11x4 (x4 y 9 − 22x2 y 4 − 99xy 7 ) (C) 11x4 y 4 (x4 y 5 − 22x2 − 99xy 3 ) (D) x5 y 4 (55x4 y 5 − 22x − 99y 3 ) (E) None of the above Answer: (E) 55x8 y 9 − 22x6 y 4 − 99x5 y 7 = 11x5 y 4 (5x3 y 5 − 2x − 9y 3 ) Question 12: Factor using integer coefficients: x2 + 13x + 36 (A) (x + 6)(x + 7) (B) (x + 12)(x + 1) (C) (x + 9)(x + 4) (D) (x + 5)(x + 8) (E) None of the above Answer: (C) Recall that factors c, d must satisfy two equations 1) cd = 36 and 2) c + d = 13. Candidates for cd = 36 are 1, 36; 2, 18; 3, 12; 4, 9; 6, 6 of these choices only 4, 9 satisfy c+d = 13. Thus x2 +13x+36 = (x+9)(x+4). Question 13: Factor using integer coefficients: x2 − 81 (A) (x + 9)2 (B) (x − 9)2 (C) (x + 9)(x − 9) (D) (x + 81)(x − 1) (E) None of the above Answer: (C) This is the important case of factoring the “difference of squares.” Recall that factors c, d must satisfy two equations 1) cd = −81 and 2) c + d = 0. Candidates for cd = −81 are 1, −81; 1, −81; −9, 9 of these choices only −9, 9 satisfy c + d = 0. Thus x2 − 81 = (x + 9)(x − 9). 3 Question 14: Factor using integer coefficients: 7x2 − 39x − 18 (A) (x + 9)2 (B) (7x + 3)(x − 6) (C) (x + 9)(x − 9) (D) (x + 81)(x − 1) (E) None of the above Answer: (B) I solve these types of problems slightly differently than described in the Schaum’s guide. You can decide for yourself which way you prefer. First, convert 7x2 − 39x − 18 to x2 − 39x − 7 ∗ 18 = x2 − 39x − 126. Factor x2 − 39x − 126. Need to find c, d that satisfy cd = −126 and c + d = −39. Try -42 and 3. Now we know that x2 − 39x − 126 = (x + 3)(x − 42). Since we multiplied by 7 in the first step, we need to divide by 7: 3 42 (x + )(x − ) 7 7 Since 42 is divisible by 7, the second term becomes (x − 6). However, 3 is not divisible by 7, thus we “slide” the 7 in front of the x. Thus the first term becomes (7x + 3). Thus 7x2 − 39x − 18 can be factored as (7x + 3)(x − 6). N.B.: This rule works only for quadratics. Question 15: Factor the following polynomial: 16x2 − 49y 2 (A) (x + 7)(16x + 7y 2 ) (B) (4x − 7y 2 ) (C) (4x − 7y)(4x − 7y) (D) (4x − 7y)(4x + 7y) (E) None of the above Answer: (D) Answering this question requires a simply application of the “differences-of-square” rule for factoring polynomials. Best to just memorize this formula as it generally useful. Any polynomial that is expressed as a “difference-of-squares” (i.e., anything √ of the form a2 − b2 where p a and b are general expressions), factors according to (a − b)(a + b). In this case a = 16x2 = 4x and b = 49y 2 = 7y. Applying the rules yields answer D: (4x − 7y)(4x + 7y). Fractions Question 16: Multiply the following rational expressions involving quotients of binomials and reduce to lowest terms. x−5 x+2 . x+8 x−9 (A) (x−5)(x+2) (x+8)(x−9) (B) x2 −3x−10 x2 −x−72 (C) (x−5)(x+3) (x+8)(x−9) (D) x2 +3x−10 x2 −x+72 4 (E) None of the above Answer: (B) Work through the following algebra: x−5 x+2 (x − 5)(x + 2) x2 − 5x + 2x − 10 x2 − 3x − 10 . = = 2 = 2 x+8 x−9 (x + 8)(x − 9) x − 9x + 8x − 72 x − x − 72 Question 17: Divide the following expressions: 15ch4 3c4 h ÷ 2 3 5x yz 55y 2 z (A) 55h3 y c3 x2 z 2 (B) 55h2 y 2 c4 xz 2 (C) 55h3 y c3 xz 2 (D) 55h2 y c 3 x2 z 2 (E) None of the above Answer: (A) Work through the following algebra: 15ch4 3c4 h ÷ = 5x2 yz 3 55y 2 z 15ch4 5x2 yz 3 3c4 h 55y 2 z = 15ch4 55y 2 z 5h3 11y 55h3 y = = 5x2 yz 3 3c4 h x2 z 2 c3 c3 x2 z 2 Question 18: Add or subtract the following fractions: x2 (A) 12+7x x2 +x−42 (B) 12(x+7)+7x3 −49 (x2 −49)(x+7) (C) 12(x+7)+7x(x2 −49) x2 −49 (D) 12+7x(x−7) x2 +49 12 7x + − 49 x + 7 (E) None of the above Answer: (E) Work through the following algebra: 7x 12 7x 12 + 7x(x − 7) 12 + 7x2 − 49x 12 + = + = = x2 − 49 x + 7 (x − 7)(x + 7) x + 7 x2 − 49 x2 − 49 Question 19: Multiply the following rational expressions involving quotients of binomials and reduce to lowest terms. x−4 x+7 . x−1 x−6 Answer: x−4 x+7 x−1 . x−6 = (x−4)(x+7) (x−1)(x−6) = x2 −4x+7x−28 x2 −x−6x+6 = x2 +3x−28 x2 −7x+6 Question 20: Divide the following expressions: 11w2 x4 3wx5 ÷ 3 8yz 3 2y z Answer: 11w2 x4 8yz 3 ÷ 3wx5 2y 3 z = 11w2 x4 ∗2y 3 z 8yz 3 ∗3wx5 = 22w2 x4 y 3 z 24wx5 yz 3 = 11wy 2 12xz 2 5 Radicals Question 21: Simplify the following the radicals: p 81x8 y 6 (A) 9x4 y 3 p (B) 9 x8 y 6 (C) −9x4 y 3 (D) 9x4 y 3 and − 9x4 y 3 (E) None of the above Answer: (D) Work through the following algebra: p p p √ √ p 81x8 y 6 = 81 x8 y 6 = +/ − 9 (x4 )2 (y 3 )2 = +/ − 9(x4 )(y 3 )2 Question 22: Use the properties of radicals to solve for y: p 3y = 6x4 (A) y = 12x10 (B) y = 36x8 (C) y = 12x8 (D) y = 24x4 (E) None of the above Answer: (C) Work through the following algebra: p 3y =6x4 3y =(6x4 )2 3y =(36x8 ) y =12x8 Question 23: Simplify the following radical: p 450x4 y 5 z 6 p √ √ p √ √ √ √ √ 5 5 √ √ Answer: 450x4 y 5 z 6 = 450 x4 y 5 z 6 = 5 18x2 yy 2 z 3 or 5 18x2 y 2 z 3 = 3∗5 18x2 yy 2 z 3 or 3∗5 18x2 y 2 z 3 = √ √ 5 √ 15 18x2 yy 2 z 3 or 15 18x2 y 2 z 3 Question 24: Simplify the following radical: p 169x6 y 8 Answer: p √ p √ 169x6 y 8 = 169 x6 y 8 = +/ − 13x2 y 4 6