1. Simplify each.
a ) 5− 3 4 − 62 = 5− 3(−32) = 5+ 96 = 101
(
)
b) 49 + 74 = 123
c) −5− 6 + 4 (−2) = −5− 6 − 8 = −19 = 19
d ) − 32 + 33 + 3−2 = −9 + 27 + 1 = 18 1
9
9
2. Evaluate each if w = 4, x = 2, y = −3, and z = 2 .
3
2
a ) w 2 + y 2 = 42 + (−3) = 16 + 9 = 25 = 5
b)
xy
z
2(−3)
=
= −6 ÷ 2 = −6 ⋅ 3 = −9
3
2
3
2
c) w − yz = 4 − (−3) 2 = 4 + 2 = 6 = 6
3
d ) y + xz
−1
= −3+ 2
( 23 ) = −3+ 3 = 0
!
2$
e ) w 3− 2x − y 2 = 4 # 3− 2 ( 2) − (−3) & = 4 (3− 4 − 9) = 4 (−10) = −40
"
%
(
)
3. Use the rules of exponents to simplify each. Leave no negative exponents.
10 6
a ) 5x 6 y 3 z −5 −2x 4 y 3 z 4 = −10x10 y 6 z −1 = −10x
(
b) 7x −2 y8
(
c) −6x4
7 2
y
9x y
−3
)
3
)
y
z
= 343x −6 y 24 = 343y6
24
x
= −2x
3 5
y
3
d ) 25x8 y16
(
12
)
= 25x8 y16 = 5x 4 y8
4. Simplify each.
a ) 3x 2 − 5x +11 + 5x 2 + 6x − 23 = 8x 2 + x −13
(
b) (−x
2
)(
− 7x + 8) − (3x
2
)
2
− 6x + 3 = −4x 2 − x + 5
)
2
c) (3x − 8) = (3x ) − 2 (3x ) (8) + 82 = 9x 2 − 48x + 64
d ) (5x −1) (7x + 4) = 35x 2 + 20x − 7x − 4 = 35x 2 +13x − 4
2
e ) (8x − 5) (8x + 5) = (8x ) − 52 = 64x 2 − 25
f ) ( 2x + 3) x 2 − 4x + 5 = 2x 3 + 3x 2 − 8x 2 −12x +10x +15 = 2x 3 − 5x 2 − 2x +15
(
)
g ) − 5x 3 9x 2 +10x = −45x5 − 50x 4
(
)
h ) 3 250 + 7 40 = 3 25⋅10 + 7 4 ⋅10 = 15 10 +14 10 = 29 10
i ) 6 3 ⋅ 7 12 = 42 48 = 42 16 ⋅ 3 = 168 3
j ) 12
(
)
6 +2 8 = 2 3
(
)
6 + 4 2 = 2 18 + 8 6 = 6 3 + 8 6
2
( ) = 49 −14 13 +13 = 62 −14 13
l ) ( 17 − 4) ( 17 + 4) = 17 −16 = 1
m) (3 5 + 4 2 ) ( 5 − 3 2 ) = 3⋅ 5− 9 10 + 4 10 −12 ⋅ 2 = 15− 5 10 − 24 = −9 − 5 10
k ) 7 − 13
6 ⋅ 3
3
3
n)
72
50
o)
=6
3
3
36
25
=
=6
5 3 ⋅ 3+ 5
3− 5 3+ 5
p)
=2 3
5
= 15
3+5 15
9−5
= 15
3+5 15
4
q) (7 − 3i ) + ( 4 − 9i ) = 11−12i
r ) (11+ 2i ) − (7 + 5i ) = 4 − 3i
s) ( 4 + 9i ) ( 4 − 9i ) = 16 + 81 = 97
2
t ) ( 4 − 9i ) = 16 − 72i + 81i 2 = −65− 72i
u ) (5− 2i ) (6 + i ) = 30 + 5i −12i − 2i 2 = 30 − 7i + 2 = 32 − 7i
2
v ) (5+ 2i ) = 25+ 20i + 4i 2 = 21+ 20i
w) 3i ( 28 −13i ) − i (5i ) +12i = 84i − 39i 2 − 5i 2 +12i = 44 + 96i
x)
y)
z)
3 ⋅ x−4 + 7 ⋅ x+7
x+7 x−4 x−4 x+7
5x+3
( x+5)( x−5)
−
=
5 ⋅ x−5
x−5 x−5
3x − x+5 + 4
2x−1 2x−1 2x−1
3x−12 + 7x+49
x+7
( )( x−4) ( x+7)( x−4)
=
5x+3
−
5x−25
cc)
x +3
2
2x− 3
x
⋅ 2x =
2x
x 2 +6x
4x 2 −6
10x+37
x+7
( )( x−4)
=
28
( x+5)( x−5) ( x+5)( x−5) ( x+5)( x−5)
3x−( x+5)+4 2x−1
=
=
=1
2x−1
2x−1
( x−3)( x+2) ⋅ 6x 2 = 2x
5
3x( x+2) 5( x−3)
(3x−4)(3x+4) ⋅ (2x+1)( x+3) = (3x−4)(2x+1)
bb)
(3x+4)( x+1) ( x+3)( x−1) ( x+1)( x−1)
aa )
=
5. Write an equation for each line.
a) y = − 4 x − 23
b) m =
3
7−4
5−(−1)
= 3 = 1 ; y − 7 = 1 ( x − 5) ; y − 4 = 1 ( x +1) .
6
2
2
c) x = 9
e) m =
2
d) y = 4 7
−A
B
=
−5 = 5 ;
−3 3
b = −π ; y = 5 x − π .
3
f ) slope of the perpendicular line is m = 4; y + 3 = 4 ( x − 3) .
2
6. Given the quadratic function f ( x ) = −2 ( x + 7) + 90, find each.
2
2
e) − 2 ( x + 7) + 90 = 0
a) f (−7) = −2 (−7 + 7) + 90 = −2 (0) + 90 = 90;
2
2
− 2 ( x + 7) = −90
b) f (0) = −2 (0 + 7) + 90 = −2 ( 49) + 90 = −8;
c) The vertex is ( h, k ) ⇒ (−7, 90) ;
( x + 7)
d ) The axis of symmetry is x = h ⇒ x = −7;
2
= 45
x + 7 = 45
f ) f (0) = −8 so the y-intercept is − 8;
x + 7 = ±3 5
g) The domain is (−∞, ∞) ;
x = −7 ± 3 5
h) The range is (−∞, 90#$.
7. Given the points (7, − 8) and (−3, 6) , find each.
a) m =
b) d =
−8−6
7−(−3)
= −14 = − 7 ;
10
5
2
(7 − (−3)) + (−8 − 6)
" 7+(−3) −8+6 %
c) M = $
,
=
2 '&
# 2
2
2
= 102 + (−14) = 100 +196 = 296 = 4 ⋅ 74 = 2 74;
( 42 , −22 ) = (2, −1).
8. Solve each equation.
a) 6x + 8 = 32
6x = 24
x=4
b) − 5 = 3 ( x −1)
c) − 4 ( x − 3) = 6 ( x + 5)
d ) 12 = 5+x
− 40 = 3( x −1)
− 4x +12 = 6x + 30
−10x +12 = 30
−10x = 18
12 ( x − 3) = 8 (5+ x )
8
− 40 = 3x − 3
− 37 = 3x
x = −37
3
x =−9
5
8
x−3
12x − 36 = 40 + 8x
4x − 36 = 40
4x = 76
x = 19