Download Week 1 Quiz: Review

Week 1 Quiz: Review
SGPE Summer School 2014
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).
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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
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