Section 5.3 Solutions
1) degree 2; zeros 3i and -3i
First: Set each zero equal to x and then create factors.
x = 3i
x – 3i = 0
x = -3i
x + 3i = 0
Second: Multiply the factors to create the polynomial.
f(x) = (x – 3i)(x + 3i)
f(x) = x2 + 3xi – 3xi – 9i2
f(x) = x2 – 9(-1)
Answer #1: f(x) = x2 + 9
3) degree 3; zeros -4 and 5i and -5i
First: Set each zero equal to x and then create factors.
x = 5i
x – 5i = 0
x = -5i
x + 5i = 0
x = -4
x+4=0
Second: Multiply the factors to create the polynomial.
f(x) = (x – 5i)(x + 5i)(x+4)
f(x) = (x2 + 5xi – 5xi – 25i2)(x+4)
f(x) = (x2 – 25(-1))(x+4)
f(x) = (x2 + 25)(x + 4)
Answer #3: f(x) = x3 + 4x2 + 25x + 100
5) degree 3; zeros 2 and 3i
We know that -3i is also a zero
First: Set each zero equal to x and then create factors.
x = 3i
x – 3i = 0
x = -3i
x + 3i = 0
x=2
x-2=0
Second: Multiply the factors to create the polynomial.
f(x) = (x – 3i)(x + 3i)(x - 2)
f(x) = (x2 + 3xi – 3xi – 9i2)(x-2)
f(x) = (x2 – 9(-1))(x-2)
f(x) = (x2 + 9)(x -2)
Answer #5: f(x) = x3 - 2x2 + 9x - 18
7) degree 4; zeros 2i, and 6i
Both -2i and -6i are also zeros
First: Set each zero equal to x and then create factors.
x = 2i
x = -2i
x = 6i
x = -6i
x – 2i = 0
x + 2i = 0
x – 6i = 0
x + 6i = 0
Second: Multiply the factors to create the polynomial.
f(x) =(𝑥 − 2𝑖)(𝑥 + 2𝑖)(𝑥 − 6𝑖)(𝑥 + 6𝑖)
=(𝑥 2 − 4𝑖 2 )(𝑥 2 − 36𝑖 2 )
=(𝑥 2 + 4)(𝑥 2 + 36)
=𝑥 4 + 36𝑥 2 + 4𝑥 2 + 144
Answer #7: 𝒇(𝒙) = 𝒙𝟒 + 𝟒𝟎𝒙𝟐 + 𝟏𝟒𝟒
9) degree 2; zero 3 + i
3 – i is also a zero as it is the conjugate pair to 3 + i
First: Set each zero equal to x and then create factors.
x=3–i
x – (3 – i) = 0
x=3+i
x – (3 + i) = 0
Second: Multiply the factors to create the polynomial.
𝑓(𝑥) = (𝑥 − (3 − 𝑖))(𝑥 − (3 + 𝑖))
=(𝑥 − 3 + 𝑖)(𝑥 − 3 − 𝑖)
=(𝑥 2 − 3𝑥 − 𝑖𝑥 − 3𝑥 + 9 + 3𝑖 + 𝑖𝑥 − 3𝑖 − 𝑖 2 )
=(𝑥 2 − 6𝑥 + 9 + 1)
=(𝑥 2 − 6𝑥 + 10)
Answer #9: 𝒇(𝒙) = 𝒙𝟐 − 𝟔𝒙 + 𝟏𝟎
11) degree 2; zero 5 - i
5 + i is also a zero as it is the conjugate pair to 5 – i
First: Set each zero equal to x and then create factors.
x=5–i
x – (5 – i) = 0
x=5+i
x – (5 + i) = 0
Second: Multiply the factors to create the polynomial.
𝑓(𝑥) = (𝑥 − (5 − 𝑖))(𝑥 − (5 + 𝑖))
=(𝑥 − 5 + 𝑖)(𝑥 − 5 − 𝑖)
=(𝑥 2 − 5𝑥 − 𝑖𝑥 − 5𝑥 + 25 + 5𝑖 + 𝑖𝑥 − 5𝑖 − 𝑖 2 )
=(𝑥 2 − 10𝑥 + 25 + 1)
=(𝑥 2 − 10𝑥 + 26)
Answer #11: 𝒇(𝒙) = 𝒙𝟐 − 𝟔𝒙 + 𝟏𝟎
13) 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 #13: x = 2i, -2i, 4
you may write ±𝟐𝒊, 𝟒
-16
16
0
15) x3 – 4x2 + 9x – 36 = 0 (x = 3i is a solution)
I will do double synthetic division using 3i and -3i, much multiplication will be done via calculator, watch
the video or come visit me for calculator help
1
3i
1
1
-3i
1
-4
3i
-4+3i
9
-9-12i
-12i
-4+3i
-3i
-4
-12i
12i
0
we now know: x3 – 4x2 + 9x – 36 = (x-3i)(x- (-3i)(x-4)
now I can solve x3 – 4x2 + 4x – 16 = 0
(x-3i)(x- (-(3i))(x-4) = 0
(x-3i)(x+3i)(x-4) = 0
x-3i = 0
x+3i=0
x-4=0
Answer #15: x = 3i, -3i, 4
you may write ±𝟑𝒊, 𝟒
-36
36
0