Section 5.3 Solutions
Section 5.3:
#1 – 10: Create a function with lead coefficient 1 that satisfies the conditions.
1) degree 3; zeros -4 and 6 + i
6 – i is also a zero as it is the conjugate pair to 6 + i
𝑓(𝑥) = (𝑥 − (6 − 𝑖))(𝑥 − (6 + 𝑖))(𝑥 − (−4))
=(𝑥 − 6 + 𝑖)(𝑥 − 6 − 𝑖)(𝑥 + 4)
=(𝑥 2 − 6𝑥 − 𝑖𝑥 − 6𝑥 + 36 + 6𝑖 + 𝑖𝑥 − 6𝑖 − 𝑖 2 )(𝑥 + 4)
=(𝑥 2 − 12𝑥 + 36 + 1)(𝑥 + 4)
=(𝑥 2 − 12𝑥 + 37)(𝑥 + 4)
=𝑥 3 + 4𝑥 2 − 12𝑥 2 − 48𝑥 + 37𝑥 + 148
Answer: 𝒇(𝒙) = 𝒙𝟑 − 𝟖𝒙𝟐 − 𝟏𝟏𝒙 + 𝟏𝟒𝟖
3) degree 3; zeros 2 and 3 - i
3 + i is also a zero
𝑓(𝑥) = (𝑥 − (3 − 𝑖))(𝑥 − (3 + 𝑖))(𝑥 − 2)
=(𝑥 − 3 + 𝑖)(𝑥 − 3 − 𝑖)(𝑥 − 2)
=(𝑥 2 − 3𝑥 − 𝑖𝑥 − 3𝑥 + 9 + 3𝑖 + 𝑖𝑥 − 3𝑖 − 𝑖 2 )(𝑥 − 2)
=(𝑥 2 − 6𝑥 + 9 − (−1))(𝑥 − 2)
=(𝑥 2 − 6𝑥 + 10)(𝑥 − 2)
=𝑥 3 − 2𝑥 2 − 6𝑥 2 + 12𝑥 + 10𝑥 − 20
Answer: f(x) = 𝒙𝟑 − 𝟖𝒙𝟐 + 𝟐𝟐𝒙 − 𝟐𝟎
5) degree 4; zeros 2i, and 5 – i
-2i and 5 + i are also zeros
𝑓(𝑥) = (𝑥 − (5 − 𝑖))(𝑥 − (5 + 𝑖))(𝑥 − 2𝑖)(𝑥 − (−2𝑖))
=(𝑥 − 5 + 𝑖)(𝑥 − 5 − 𝑖)(𝑥 − 2𝑖)(𝑥 + 2𝑖)
=(𝑥 2 − 5𝑥 − 𝑥𝑖 − 5𝑥 + 25 + 5𝑖 + 𝑖𝑥 − 5𝑖 − 𝑖 2 )(𝑥 2 − 4𝑖 2 )
=(𝑥 2 − 10𝑥 + 25 − (−1))(𝑥 2 + 4)
=(𝑥 2 − 10𝑥 + 26)(𝑥 2 + 4)
=𝑥 4 + 4𝑥 2 − 10𝑥 3 − 40𝑥 + 26𝑥 2 + 104
Answer: 𝒇(𝒙) = 𝒙𝟒 − 𝟏𝟎𝒙𝟑 + 𝟑𝟎𝒙𝟐 − 𝟒𝟎𝒙 + 𝟏𝟎𝟒
7) degree 4; zeros 3i, and 5i
Both -3i and -5i are also zeros
𝑓(𝑥) = (𝑥 − 5𝑖)(𝑥 − (−5𝑖))(𝑥 − 3𝑖)(𝑥 − (−3𝑖))
=(𝑥 − 5𝑖)(𝑥 + 5𝑖)(𝑥 − 3𝑖)(𝑥 + 3𝑖)
=(𝑥 2 − 25𝑖 2 )(𝑥 2 − 9𝑖 2 )
=(𝑥 2 + 25)(𝑥 2 + 9)
=𝑥 4 + 9𝑥 2 + 25𝑥 2 + 225
Answer: 𝒇(𝒙) = 𝒙𝟒 + 𝟑𝟒𝒙𝟐 + 𝟐𝟐𝟓
9) degree 4; zeros 2i, 3, - 4
-2i is another zero
𝑓(𝑥) = (𝑥 − 2𝑖)(𝑥 − (−2𝑖))(𝑥 − 3)(𝑥 − (−4))
=(𝑥 − 2𝑖)(𝑥 + 2𝑖)(𝑥 − 3)(𝑥 + 4)
=(𝑥 2 − 4𝑖 2 )(𝑥 2 + 4𝑥 − 3𝑥 − 12)
=(𝑥 2 + 4)(𝑥 2 + 𝑥 − 12)
=𝑥 4 + 𝑥 3 − 12𝑥 2 + 4𝑥 2 + 4𝑥 − 48
Answer: 𝒇(𝒙) = 𝒙𝟒 + 𝒙𝟑 − 𝟖𝒙𝟐 + 𝟒𝒙 − 𝟒𝟖
11) x3 – 4x2 + 4x – 16 = 0 (x = 2i is a solution)
I will do double synthetic division using 2i and -2i, much multiplication will be done via calculator, watch
the video or come visit me for calculator help
1
2i
1
1
-2i
1
-4
2i
-4+2i
4
-4-8i
-8i
-4+2i
-2i
-4
-8i
8i
0
we now know: x3 – 4x2 + 4x – 16 = (x-2i)(x- (-2i)(x-4)
now I can solve x3 – 4x2 + 4x – 16 = 0
(x-2i)(x- (-2i)(x-4) = 0
(x-2i)(x+2i)(x-4) = 0
x-2i = 0
x+2i=0
x-4=0
Answer: x = 2i, -2i, 4 you may write ±𝟐𝒊, 𝟒
-16
16
0
13) 2x4 + 5x3 +5x2 +20x-12 = 0 (2i is a solution)
I will do double synthetic division using 2i and -2i, much multiplication will be done via calculator, watch
the video or come visit me for calculator help
2
5
4i
5+4i
2i
2
5
-8+10i
-3+10i
20
-20-6i
-6i
2
5+4i
-3+10i
-4i
-10i
2
5
-3
4
3
2
2
We now know that 2x + 5x +5x +20x-12 =(x-2i)(x-(-2i)(2x + 5x – 3)
-2i
-12
12
0
-6i
6i
0
I can now solve 2x4 + 5x3 +5x2 +20x-12 = 0
(x-2i)(x-(-2i)(2x2 + 5x – 3)= 0
(x-2i)(x+2i)(2x-1)(x+3) = 0
x-2i = 0
x + 2i = 0
2x -1 = 0
x+3=0
𝟏
𝟐
Answer: x = ±𝟐𝒊, , −𝟑
15) x3 -7x2 – x + 87 = 0 (5 + 2i is a zero)
5 – 2i is also a zero. I will do double synthetic division with 5 +2i and 5 – 2i
1
-7
5+2i
-2+2i
5+2i
1
-1
-14+6i
-15+6i
1
-2+2i
5-2i
1
3
We now know that x3 -7x2 – x + 87 = (x- (5+2i))(x-(5-2i)(x+3)
5-2i
So I can solve x3 -7x2 – x + 87 = 0
(x- (5+2i))(x-(5-2i))(x+3) = 0
x – (5+2i) = 0
Answer: x = 5 + 2i
x – (5-2i) = 0
x = 5 – 2i
x+3=0
x = -3
87
-87
0
-15+6i
15-6i
0
17) x4 +6x3 + 2x2 – 26x + 17 = 0 (-4+i is a solution)
I will perform double synthetic division with -4 + i and -4 – i
1
6
-4+i
2+i
-4+i
1
1
-4-i
1
2
-9-2i
-7-2i
2+i
-4-i
-2
-26
30+i
4+i
-7-2i
8+2i
1
we can now factor x4 +6x3 + 2x2 – 26x + 17 = (x – (-4+i)(x-(-4-i))(x2 – 2x + 1)
x4 +6x3 + 2x2 – 26x + 17 = 0
(x – (-4+i))(x-(-4-i))(x2 -2x+1) = 0
x-(-4+i) = 0
x – (-4-i) = 0
Answer: x = -4+i, x = -4 – i, x = 1
x2 – 2x+ 1 = 0
(x-1)(x-1) = 0
17
-17
0
4+i
-4-i
0
19) x3 -8x2 +25x-26 = 0 (3+2i is a solution)
I will perform double synthetic division with 3 + 2i and 3 – 2i
1
-8
3+2i
-5+2i
3+2i
1
1
3-2i
1
25
-19-4i
6-4i
-5+2i
3-2i
-2
Thus we can factor x3 -8x2 +25x-26 = (x- (3+2i))(x-(3-2i)(x-2)
x3 -8x2 +25x-26 = 0
(x – (3 + 2i)(x – (3 – 2i)(x-2) = 0
x- (3-2i) = 0
x – (3-2i) = 0
answer: x = 3+2i, 3-2i, 2
x–2=0
-26
26
0
6-4i
-6+4i
0