Section 5.2 solutions
#1- 10:
a) Perform the division using synthetic division.
b) if the remainder is 0 use the result to completely factor the dividend (this is the numerator or the
polynomial to the left of the division sign.)
1)
3𝑥 3 −17𝑥 2 +15𝑥−25
𝑥−5
a) I need to change the sign of the (-5) to positive for my synthetic division
3
-17
15
-2
5
3
Answer:
𝟑𝒙𝟑 −𝟏𝟕𝒙𝟐 +𝟏𝟓𝒙−𝟐𝟓
=
𝒙−𝟓
15
-10
5
-25
25
0
3x2 – 2x + 5 remainder 0
1b) since the remainder is 0 I need to factor the numerator. Synthetic division tells me this is true
3𝑥 3 − 17𝑥 2 + 15𝑥 − 25 = (x-5)(3x2 – 2x + 5)
I just need to factor a bit more
Answer: 𝟑𝒙𝟑 − 𝟏𝟕𝒙𝟐 + 𝟏𝟓𝒙 − 𝟐𝟓 = (x-5)(3x-5)(x+1)
3)
4𝑥 3 +8𝑥 2 −9𝑥−18
𝑥+2
3a) I need to change the sign of the 2 to negative for my synthetic division
4
-2
4
Answer:
𝟒𝒙𝟑 +𝟖𝒙𝟐 −𝟗𝒙−𝟏𝟖
=
𝒙+𝟐
8
-8
0
-9
0
-9
-18
18
0
4x2 – 9 remainder 0
3b) since the remainder is 0 I need to factor the numerator. Synthetic division tells me this is true
4𝑥 3 + 8𝑥 2 − 9𝑥 − 18= (x+2)(4x2 – 9) I just need to factor more
Answer: 𝟒𝒙𝟑 + 𝟖𝒙𝟐 − 𝟗𝒙 − 𝟏𝟖= (x+2)(2x+3)(2x-3)
5)
3𝑥 3 −16𝑥 2 −72
𝑥−6
5a) I need to change the sign of the (-6) to positive for my synthetic division. I need to think of the
numerator having the form 3x3 – 16x2 + 0x – 72.
3
-16
18
2
6
3
Answer:
𝟑𝒙𝟑 −𝟏𝟔𝒙𝟐 −𝟕𝟐
=
𝒙−𝟔
0
12
12
-72
72
0
3x2 + 2x + 12 remainder 0
5b) since the remainder is 0 I need to factor the numerator. Synthetic division tells me this is true
3x3 – 16x2 – 72 = (x-6)(3x2 + 2x + 12)
The 3x2 + 2x + 12 is prime, so I can’t factor more.
Answer: 3x3 – 16x2 – 72 = (x-6)(3x2 + 2x + 12)
7)
(5𝑥 3 + 6𝑥 + 8) ÷ (𝑥 + 2) this is the same as
5𝑥 3 +6𝑥+8
𝑥+2
a) I need to change the sign of the 2 to negative for my synthetic division . I need to insert a 0x2 term in
the numerator 5x3 + 0x2 + 6x + 8
5
-2
5
0
-10
-10
6
20
26
Answer: (𝟓𝒙𝟑 + 𝟔𝒙 + 𝟖) ÷ (𝒙 + 𝟐) = 5x2 – 10x + 26 remainder 44
7b) skip this part since the remainder is not 0.
8
-52
44
9) (𝑥 3 − 27) ÷ (𝑥 − 3)
this is the same as
𝑥 3 −27
𝑥−3
a) I need to change the sign of the (-3) to positive for my synthetic division
I need to insert a 0x2 and a 0x.
(x3 + 0x2 + 0x -27)÷(x-3)
1
3
1
0
3
3
0
9
9
-27
27
0
Answer: (𝒙𝟑 − 𝟐𝟕) ÷ (𝒙 − 𝟑) = x2 + 3x + 9 remainder of 0
9b) since the remainder is 0 I need to factor the numerator. Synthetic division tells me this is true
x3 – 27 = (x-3)(x2 + 3x + 9)
The x2 + 3x + 9 is prime
Answer: x3 – 27 = (x-3)(x2 + 3x + 9)
#11 – 20:
a) use your graphing calculator, or the rational root theorem to find a zero of the polynomial
i) you need to find one zero for a third degree polynomial
ii) you need to find two zeros for a fourth degree polynomial
b) use synthetic division to completely factor the polynomial (use “double” synthetic division for fourth
degree polynomials)
c) Use your answer to part b to solve f(x) = 0
11) f(x) = x3 + 2x2 – 5x – 6
here is a graph of f(x)
11a) Answer: I will use the numbers (-1) for my synthetic division, I could have also used 2.
11b) since x = -1 is a zero, I know (x+1) is a factor of f(x)
the synthetic division will get me the remaining factors.
1
2
-1
1
1
The result of my synthetic division gives me
-1
𝑥 3 +2𝑥 2 −5𝑥−6
𝑥+1
= 𝑥 2 + 𝑥 − 6 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 0
so now I can factor f(x)
f(x) = x3 + 2x2 – 5x – 6 = (x+1)(x2+x-6)
Answer #11b: f(x) = (x+1)(x-2)(x+3)
-5
-1
-6
-6
6
0
11c) Solve f(x) = 0
Just take answer to part b and set it equal to 0 and solve for x.
(x+1)(x-2)(x+3) = 0
x+1=0
x = -1
x–2=0
x=2
Answer #11c: x = -1, 2, -3
x+3=0
x = -3
13) f(x) = 2x3 – 13x2 + 24x – 9
here is a graph of f(x)
13a) I will use 3 is the value for my synthetic division
13b) since x = 3 is a zero, I know (x-3) is a factor of f(x)
the results of my synthetic division should help me get additional factors of f(x)
2
3
2
-13
6
-7
The result of the synthetic division tells me
Now I can factor f(x) = 2x3 – 13x2 + 24x – 9
24
-21
3
2𝑥 3 −13𝑥 2 +24𝑥−9
𝑥−2
= 2𝑥 2 − 7𝑥 + 3 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 0
= (x-3)(2x2 – 7x + 3)
Answer 13b: f(x) = (x-3)(x-3)(2x-1) or (x-3)2(2x-1)
-9
9
0
13c) Solve f(x) = 0
Just take answer to part b and set it equal to 0 and solve for x.
(x-3)(x-3)(2x-1) = 0
x-3=0
x=3
x–3=0
x=3
Answer 13c: x = 3, ½
2x – 1 = 0
2x = 1
x=½
15) f(x) =x3 – 5x2 – 4x - 20
here is a graph of f(x)
15a) I will use (-2), but I could have used any of the three x-intercepts.
15b) since x = (-2) is a zero I know (x+2) is a factor of f(x)
the results of my synthetic division should help me get more factors of f(x)
1
-2
1
-5
-2
-7
-4
14
10
20
-20
0
The result of my synthetic division tells me
f(x) = x3 – 5x2 – 4x - 20 = (x+2)(x2 -7x+10) (I just need to factor the second parenthesis to get my
answer)
Answer #15b: f(x) = (x+2)(x-2)(x-5)
15c) Solve f(x) = 0
Just take answer to part b and set it equal to 0 and solve for x.
(x+2)(x-2)(x-5) = 0
x+2=0
x–2=0
x–5=0
x = -2
x=2
x=5
Answer #15c: x = -2,2,5
17) f(x) = 2x4 + 17x3 + 35x2 – 9x – 45
here is a graph of f(x)
17a) I will use (-5) and (1) and perform “double synthetic division”
17b) since x = (-5) is a zero I know (x+5) is a factor of f(x)
since x = 1 is a zero I know (x-1) is a factor of f(x)
the results of my synthetic division should help me get more factors of f(x)
2
17
-10
7
-5
2
2
1
2
35
-35
0
7
2
9
The result of the double synthetic division tells me
f(x) = 2x4 + 17x3 + 35x2 – 9x – 45 = (x+5)(x-1)(2x2 + 9x + 9)
Answer: f(x) = (x+5)(x-1)(2x+3)(x+3)
-9
0
-9
0
9
9
-45
45
0
-9
9
0
17c) Solve f(x) = 0
Just take answer to part b and set it equal to 0 and solve for x.
(x+5)(x-1)(2x+3)(x+3) = 0
x+5=0
x = -5
x–1=0
x=1
−𝟑
Answer #17c: x = -5,1, 𝟐 ,-3
2x + 3 = 0
2x = -3
x = -3/2
x+3=0
x = -3
19) f(x) = x4 + 7x2 – 8
here is a graph of f(x)
19a) I will use (-1) and (1) and perform “double synthetic division”
since x = (-1) is a zero I know (x+1) is a factor of f(x)
since x = 1 is a zero I know (x-1) is a factor of f(x)
the results of my synthetic division should help me get more factors of f(x)
1
0
1
1
1
1
1
-1
1
7
1
8
1
-1
0
0
8
8
8
0
8
The result of the double synthetic division tells me
Answer: f(x) = x4 + 7x2 – 8 = (x+1)(x-1)(x2 + 8) this is the answer as the x2 + 8 is prime
-8
8
0
8
-8
0
19c) Solve f(x) = 0
Just take answer to part b and set it equal to 0 and solve for x.
(x+1)(x-1)(x2 + 8) = 0
x+1 = 0
x–1=0
x = -1
x=1
x2 + 8 = 0
x2 = -8
𝑥 = ±√−8 = ±2𝑖√2
Answer #19c: 𝒙 = 𝟏, −𝟏, ±𝟐√𝟐𝒊