1. 
Use long division to divide:
(3x3 + 11x2 + 1) by (x2 – 3) 2. 
Use synthetic division to divide: (x2 + 3) by (x + 3)
Factor f(x) = 2x3 + 7x2 – 33x – 18 given f(-6) = 0. 3. 
4. 
Solve f(x)= x3 + 11x2 – 150x – 1512
given that -14 is one zero. 5. 
Factor: a) 3x4 – x2 – 4
b) 3x4 – 7x3 – 24x + 56
Algebra II
1
Finding Real and
Rational Zeros
Algebra II
¡  To
solve, you factor the
problem, each factor must be
factored down to at most a
quadratic.
¡  To
solve, put each factor equal
to zero. Algebra II
3
1. 2x5 + 24x = 14x3
2x5 – 14x3 + 24x = 0
2x(x4 – 7x2 + 12) = 0
2x(x2 – 3)(x2 – 4) = 0
2x=0 x2– 3=0 x2– 4=0
x=0
x2=3
x2= 4
x=±√3 x = ±2 Algebra II
4
2. 2y5 – 18y = 0
2y(y4 – 9) = 0
2y(y2 – 3)(y2 + 3) = 0
2y=0
y2–3=0 y2+3=0
y=0
y2 = 3
y=±√3
Algebra II
y2 = –3 y=±i√3
5
3. 3x4 + 3x3 = 6x2 + 6x
3x4+3x3 – 6x2– 6x = 0
3x(x3+ x2– 2x – 2)= 0
3x[x2(x+1) – 2(x + 1)] = 0
3x(x +1)(x2 – 2) = 0
3x=0
x=0
x+1=0 x2–2=0
x= -1
x2 = 2
x=±√2
Algebra II
6
4. 3x7 = 81x4
3x7 – 81x4 = 0
3x4(x3 – 27) = 0
3x4(x–3)(x2+3x+9) = 0
3x4= 0 x–3=0 x2+3x+9 = 0
x=0
Algebra II
x=3
Not real!
7
5. 5x3 = 40
5x3 – 40 = 0
5(x3 – 8) = 0
5(x – 2)(x2 + 2x + 4)=0
5=0
NS
x – 2 = 0 x2 + 2x + 4= 0
x=2
Algebra II
2 imag solutions
8
6. 4x4 – 8x2 = 5
4x4 – 8x2 – 5 = 0
(2x2 + 1)(2x2 – 5) = 0
2x2+1=0 2x2-5=0 2x2 = -1
x2 = -1/2
2x2 = 5
x2 = 5/2
x =+i√2/2 x=+√10/2
Algebra II
9
1. 36r3 – r = 0
3. 3m2 = 75m4
2. 20x3 + 80x2 = -60x 4. -13y2 + 36 = -y4
Algebra II
10
5. 2x3 – x2 – 2x = -1
6. -20c2 + 50c = 8c3 – 125 (omit) Algebra II
11
7. f(x) = x4 – x3 – 12x2
Algebra II
12
8. f(x) = -4x3 + 12x2 – 9x
Algebra II
13
9. f(x) = x3 + 4x2 – 6x – 24
Algebra II
14
10. f(x) = x4 – 18x2 + 81
Algebra II
15
The revenue in thousands of dollars for a small business
can be modeled by R(t) = t3 – 8t2 + t + 82, where t is the
number of years since 1990. What was the revenue in
2002? In what year did the revenue reach $90,000? Algebra II
16
You are building a bin to hold cedar mulch for your garden.
The bin will hold 162 ft3 of mulch. the bin is x ft tall with
sides of 5x – 6 ft and 5x – 9 ft. What are the dimensions of
the bin?
Algebra II
17
Suppose a pyramid is 3x feet high and has a square base
measuring x – 4 ft on each side. if the volume is 128 cubic
feet, what are the dimensions of the pyramid?
(Hint: Use the formula V = ⅓Bh)
Algebra II
18
A catering company is designing a box. The volume of the
box is to be 54 cubic inches and the bottom of the box to
be a square. Suppose the bottom of the box has a width
that is 3 inches smaller than the height x of the box.
Draw an illustration for this situation.
b) 
Write a polynomial equation representing this situation.
c) 
Find the dimensions of the box.
d) 
What would the volume of the box be, in cubic inches, if
the the height were 10 in?
a) 
Algebra II
19
Between 1985 through 1995, the number of home
computers, in thousands, sold in Canada is estimated by
this equation c(t) = 0.92(t3 + 8t2 + 40t + 400), where t is
the number of years since 1985. How many computers
where there in 1990? In what year did home computer
sales reach 1.5 million?
Algebra II
20
¡  Possible
Rational Solutions: ± p/q
¡  p
à factors of the a0 (constant term)
¡  q
à factors of the an (leading coefficient)
¡  Remember
that these are possible
zeros, not actual zeros. Algebra II
21
1. f(x) = x3 + 14x2 + 41x – 56
p = -56
q = 1 factors:
1 ∙ 56
1 ∙ 1
2∙ 28
4 ∙ 14
7 ∙ 8
Put all factors of p over all
factors of q to get possible
solutions: ± 1, 2, 4, 7, 8, 14, 28, 56
Algebra II
2. f(x) = 2x3 + 7x2 – 7x + 30
p = 30
q =2 factors:
1 ∙ 30
1 ∙ 2
2 ∙ 15
3 ∙ 10
5 ∙ 6
Put all factors of p over all
factors of q to get possible
solutions: ± 1, 2, 3, 5, 6, 10, 15, 30, ½,
3/2, 5/2, 15/2
22
3. f(x) = 4x3 + 10x2 + 4x – 8
p =-8
q =4 4. f(x) = 3x3 + 8x2 – 2x + 10
factors:
p = 10
q = 3 factors:
1 ∙ 8
1 ∙ 4
2∙ 4
2 ∙ 2
Put all factors of p over all
factors of q to get
possible solutions: 1 ∙ 10
1 ∙ 3
2 ∙ 5
Put all factors of p over all
factors of q to get
possible solutions: ± 1, 2, 4, 8, ½, 1/4
± 1, 2, 5, 10, 1/3, 2/3, 5/3,
10/3
Algebra II
23
¡  To
find which possible solutions
work, try them with condensed
synthetic division until you get
one with a remainder of zero,
then factor the depressed
polynomial.
Algebra II
24
1. f(x) = x3 + 2x2 – 11x – 12 possible: ± 1, 2, 3, 4, 6, 12
1
-1
1
1
1
2
3
1
-11
-8
-12
-12
-20 0 solution
x = -1 is one solution, factor x2 + x – 12 to get the
other solutions. (x + 4)(x – 3) are the factors, so x = -4 and x = 3 are
the other zeros. Algebra II
25
2. f(x) = x3 – 4x2 – 11x + 30 possible: ± 1, 2, 3, 5, 6, 10, 15, 30
1
-4
-11
30
1
1
-3
-14
16 -1
1
-5
-6
36
2
1 -2
-15
0 solution
x = 2 is one solution, factor x2 – 2x – 15 to get the
other solutions.
(x – 5)(x + 3) are the factors, so x = 5 and x = -3 are
the other zeros. Algebra II
26
3. f(x) = 2x3 + 9x2 + 7x – 6 possible: ± 1, 2, 3, 6, ½ , 3/2
2
9
7
-6
1 2 11 18
12 -1 2
7
0
-6
2 2 13 33
60 -2 2 5
-3
0
solution
x = -2 is one solution, factor 2x2 + 5x – 3 to get the
other solutions.
(2x – 1)(x + 3) are the factors, so x = ½ and x = -3
are the other zeros. Algebra II
27
4. f(x) = x3 – 7x2 + 10x + 6 possible: ± 1, 2, 3, 6
1 -7 10
6
1 1 -6
4
10 -1 1 -8
18 -12
2 1 -5
0
6
-2 1 -9 28 -50
3 1 -4 -2
0 solution
x = 3 is one solution, factor x2 – 4x – 2 to get the other
solutions. x2 – 4x – 2 is prime so use the quadratic formula.
x = 2 ± √6 and x = 3 are the zeros. Algebra II
28
1. 4, -4, and 1
x = 4 x = -4
2. 1, -4, 5
x = 1
(x – 4)(x + 4)(x – 1)
(x2
– 16)(x – 1)
f(x) = x3–x2–16x+16
Algebra II
x = 1 x = -4 x = 5
(x – 1)(x + 4)(x – 5) (x2 + 3x – 4)(x – 5) f(x)=x3–5x2+3x2–15x–4x+20
f(x) = x3 – 2x2 – 19x + 20
29
Algebra II
30