39.
f(x) 5 22x 3 2 3x 2 2 1
x-intercept: x ø 21.68
Local maximum: (0, 21)
Local minimum: (21, 22)
y
6
1
40.
x
7. f (x) 5 x5 2 x4 2 9
x
f (x)
f(x) 5 x4 1 3x3 2 x 2 2 8x 1 2
x-intercepts: x ø 0.25, x ø 1.34
22
0
21
1
257 211 29 29
8
Local maximum: (21.13, 7.06)
2
7
y
x
21
Local minimums: (22, 6), (0.88, 23.17)
Chapter Test for the chapter “Polynomials and
Polynomial Functions”
1. x3 p x2 p x24 5 x3 1 2 2 4 Product of powers property
Simplify exponents
5 x
2. (2x22y3)25
5 225(x22)25( y3)25
Power of a product property
5 225x10y215
Power of a power property
x10
2y
Negative exponent property
5 }
​  15  ​ 
x10
32y
Simplify and evaluate power.
1  2
Power of a quotient property
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
5 }
​  5 15  ​ 
(x24)22
x
  ​  ​ 5 }
​  2 22   
​
3.​ ​ }
2
y
(y )
24 22
8. (2x3 1 5x2 2 7x 1 4) 1 (x3 2 3x2 2 4x)
5 2x3 1 5x2 2 7x 1 4 1 x3 2 3x2 2 4x
5 3x3 1 2x2 2 11x 1 4
9. (3x3 2 4x2 1 3x 2 5) 2 (x2 1 4x 2 8)
5 3x3 2 4x2 1 3x 2 5 2 x2 2 4x 1 8
5 3x3 2 5x2 2 x 1 3
10. (3x 2 2)(x2 1 4x 2 7)
5 (3x 2 2)x2 1 (3x 2 2)4x 2 (3x 2 2)7
5 3x3 2 2x2 1 12x2 2 8x 2 21x 1 14
x8
5}
​  24  ​ 
y
Power of a power property
5 3x3 1 10x2 2 29x 1 14
5 x 8y 4
Negative exponent property
11. (3x 2 5)3 5 (3x)3 2 3(3x)2(5) 1 3(3x)(52) 2 53
3(xy)3
3x 3y 3
27x 2 5y
27x 2 5y
4.​ }
 
 ​ 
5}
​ 
 3 ​ 
3
Power of a product property
f (x)
22
21
0
8
1
0
5 27x3 2 15(9x2) 1 9x(25) 2 125
5 27x3 2 135x2 1 225x 2 125
12. 4
5. f (x) 5 2x3
x
1
2
21 28
3
16
222
12
28
32
22
8
10
3x3 2 14x2 1 16x 2 22
2
10
6. f (x) 5 x4 2 2x2 2 5x 1 1
22
21
0
19
5
1
2x2 1 2x 1 3
6x4 1 0x3 1   
7x2 1 4x 2 17 ​
13. 3x2 2 3x 1 2​qwww
x
21
f (x)
214
    
  ​5 3x2 2 2x 1 8 1 }
​ x 2  4 ​. 
So, ​ }}
x24
y
x
3
1
2
25 21
6x4 2 6x3 1 4x2
6x3 1 3x2 1 4x
6x3 2 6x2 1 4x
9x2 1 0x 2 17
9x2 2 9x 1 6
9x 2 23
Algebra 2
Worked-Out Solution Key
129
6x4 1 7x2 1 4x 2 17
3x 2 3x 1 2
9x 2 23
3x 2 3x 1 2
So, }}
​ 
  
  
  ​5 2x2 1 2x 1 3 1 }
​  2
  ​. 
2
14. 8x 1 27 5 (2x) 1 3 5 (2x 1 3)(4x 2 6x 1 9)
3
3
3
2
5 (x 2 1)(x 1 1)(x 1 6)
2
16. x 2 3x 2 4x 1 12 5 x2(x 2 3) 2 4(x 2 3)
5 (x 2 3)(x 2 4)
2
5 (x 2 3)(x 2 2)(x 1 2)
3
2
}
 
[x 2 (4 2 Ï
​ 10 ​ )]
}
5 [(x 2 1) 2 3i][(x 2 1) 1 3i[(x 2 4) 2 Ï
​ 10 ​  ]
}
[(x 2 4) 1 Ï
​ 10 ​] 
}
5 [(x 2 1)2 2 (3i)2][(x 2 4)2 2 (​Ï10 ​  )2]
5 (x2 2 2x 1 10)(x2 2 8x 1 6)
5 x4 2 10x3 1 32x2 2 92x 1 60
17. f(x) 5 x 1 x 2 22x 2 40
23. f (x) 5 x3 2 5x2 1 3x 1 4
Possible rational zeros: 61, 62, 64, 65, 68, 610, 620,
640
Test x 5 22: 22
1
1
1
222
240
22
2
40
21
220
0
f (x) 5 (x 1 2)(x2 2 x 2 20) 5 (x 1 2)(x 2 5)(x 1 4)
x-intercepts: x ø 20.6, x ø 1.6, x 5 4
Local maximum: (0.3, 4.5)
The real zeros are 22, 5, and 24.
Local minimum: (3, 25)
18. f (x) 5 4x4 2 8x3 2 19x2 1 23x 2 6
24. f (x) 5 x4 1 3x3 2 x2 2 6x 1 2
Possible rational zeros: 61, 62, 63, 66,
3
2
1
2
3
4
1
4
6​ } ​, 6​ } ​, 6​ } ​, 6​ } ​
Test x 5 22: 22
4
4
28
219
23
26
28
32
226
6
216
13
23
0
Local maximum: (20.88, 5.06)
f(x) 5 (x 1 2)(4x3 2 16x2 1 13x 2 3)
Test x 5 3: 3
4
216
13
23
12
212
3
24
1
0
4
x-intercepts: x ø 0.34, x ø 1.14
Local minimums: (22.16, 1.83), (0.79, 21.50)
25.
f(x) 5 (x 1 2)(x 2 3)(4x2 2 4x 1 1)
5 4(x 1 2)(x 2 3)​1 x 2 }
​ 2 ​ 2​
1 2
Since the points follow the shape of a parabola and there
is 1 turning point, the lowest degree of polynomial that the
data could represent is degree 2.
26.
1 1
The real zeros are 22, }
​ 2 ​, }
​ 2 ​, and 3.
19. f(x) 5 (x 1 1)(x 2 3)(x 2 4)
5 (x2 2 2x 2 3)(x 2 4)
5 x3 2 2x2 2 3x 2 4x2 1 8x 1 12
5 x3 2 6x2 1 5x 1 12
20. f(x) 5 (x 2 6)(x 2 2i)(x 1 2i)
5 (x 2 6)[x2 2 (2i)2]
5 (x 2 6)(x2 1 4)
5 x3 2 6x2 1 4x 2 24
Since the graph has 2 turning points, the lowest degree of
polynomial that the data could represent is degree 3.
}
}
​ 5 ​  )][x 2 (1 1 ​Ï 5 ​  )]
21. f(x) 5 (x 1 3)(x 1 1)[x 2 (1 2 Ï
130
}
}
5 (x2 1 4x 1 3)[(x 2 1) 1 Ï
​ 5 ​  ][(x 2 1) 2 Ï
​ 5 ​  ]
}
 
5 (x2 1 4x 1 3)[(x 2 1)2 2 (​Ï5 ​ )2]
5 (x2 1 4x 1 3)(x2 2 2x 2 4)
Algebra 2
Worked-Out Solution Key
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
3
}
2
5 x4 1 2x3 2 9x2 2 22x 2 12
​ 10 ​  )]
22. f (x) 5 [x 2 (1 1 3i)][x 2 (1 2 3i)][x 2 (4 1 Ï
15.x4 1 5x2 2 6 5 (x2 2 1)(x2 1 6)
GDP
27. Per Capita GDP 5 ​ }
   ​ 
Population
1.099 3 1013
  
​ 
5 ​ }
2.91 3 108
1013
10
1.099
35xy 22x24y 2
   
​ 
5 ​ }
5
Simplify.
2
22y   
​3 ​ }
  
​
5 ​ }
8
2.91
5 0.378 3 105
5 3.78 3 104
5. (4a5b22)23 5 423(a5)23(b22)23
Power of a product property
28. In order to use the Rational Zero Theorem, Mario would
have needed to factor x first and use the Theorem on the
resulting polynomial. He therefore left out x = 0 as a possible
(in fact, definite) zero.
29. Volume
Length
(Cubic
5 (inch)
inch)
1040 5 x
Width
Height
p (inch) p
(inch)
p (x 1 2) p (2x 2 3)
1040 5 2x3 1 x2 2 6x
3
0 5 2x 1 x 2 6x 2 1040
8
2
2
1
26
21040
16
136
1040
17
130
0
0 5 (x 2 8)(2x 1 17x 1 130)
2
5 423a215b6 Power of a power property
5}
​  15
   ​ 
b6
64a
2s 8
5 ​ }
  ​
r4
}
217 6 i​Ï 751 ​ 
 
 
​ 
5 }}
​ 
4
The only reasonable solution is x 5 8. So, the volume of
the prism can be modeled by V(x) 5 2x3 1 x2 2 6x. When
V 5 1040, the prism has a length of 8 in., width of 8 1 2
5 10 in., and a height of 2(8) 2 3 5 13 in.
x y
1. (3.4 3 103)(2.8 3 108) 5 (3.4 3 2.8)(103 3 108)
5 9.52 3 1011
)
5 5.8 3 (10
4
7x 4y21
Product of powers property
5 7x4 2 2y21 2 (23)
Quotient of powers property
x y
2 2
5 7x y x
23 22 21 0 1
y
81
16
1
2
3
0 1 16 81
The degree is even and the leading coefficient is positive.
So, f (x) → 1` as x → 1`.
y
1
ø 1.13 3 103 3 10224
ø 1.13 3 10221
1027
10
x
9. f (x) 5 x 3 1 x 1 4
)
5 1131.6496 3 10224
4.6
Simplify.
8. f (x) 5 x 4
26 4
4.6 3 1027
9.2 3 10
7x 3
y
10
For the chapter “Polynomials and Polynomial
Functions”
2. (5.8 3 10
Negative exponent property
5 }
​  2 23  
​
Extra Practice
26 4
Product of powers property
7.​ }
  
​• }
​  24  ​
2
}}
217 6 Ï
​ 172 2 4(2)(130) ​
  
Negative exponent property
6. (2r 3s3)(r27s5) 5 2r24s8
x 2 8 5 0 or 2x 1 17x 1 130 5 0
x 5 8 or x 5 }}
​ 
    
  ​
2(2)
Negative exponent property
5x 2
​ 
5 ​ }
4   
xy21
2
Test possible rational solutions:
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
22x23 2 1y5 2 3
214x23y5
4.​ }
   
​ 
5 }}
​ 
      
​ Quotient of powers property
3
5
x
23
y
226 26
22 21 0 1
2
2
3
4 6 14 34
The degree is odd and the leading coefficient is positive.
So, f (x) → 1` as x → 1` and f (x) → 2` as x → 2`.
y
3.​ }
 
 ​5 }
​ 9.2  ​3 }
​  29  
​
29
5 0.5 3 102
5 5 3 1021 3 102
5 5 3 101
10
1
x
Algebra 2
Worked-Out Solution Key
131