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Math Problems
Precalculus
Find the roots of factored polynomials
Find (if possible) the rational zeros of the function.
\newline
(Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
\newline
f
(
x
)
=
4
x
3
−
34
x
2
+
80
x
−
32
f(x) = 4x^3 - 34x^2 + 80x - 32
f
(
x
)
=
4
x
3
−
34
x
2
+
80
x
−
32
Get tutor help
Perform the operation and reduce the answer fully. Make sure to express your answer as a simplified fraction.
\newline
1
10
÷
8
3
\frac{1}{10} \div \frac{8}{3}
10
1
÷
3
8
\newline
Answer:
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Perform the operation and reduce the answer fully. Make sure to express your answer as a simplified fraction.
\newline
2
3
÷
8
9
\frac{2}{3} \div \frac{8}{9}
3
2
÷
9
8
\newline
Answer:
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x
2
−
10
x
+
14
=
0
x^{2}-10x+14=0
x
2
−
10
x
+
14
=
0
\newline
One solution to the given equation can be written as
x
=
5
+
n
x=5+\sqrt{n}
x
=
5
+
n
, where
n
n
n
is a constant. What is the value of
n
n
n
?
Get tutor help
Find the sum of the first
7
7
7
terms of the following sequence. Round to the nearest hundredth if necessary.
\newline
125
,
100
,
80
,
…
125, \quad 100, \quad 80, \ldots
125
,
100
,
80
,
…
\newline
Sum of a finite geometric series:
\newline
S
n
=
a
1
−
a
1
r
n
1
−
r
S_{n}=\frac{a_{1}-a_{1} r^{n}}{1-r}
S
n
=
1
−
r
a
1
−
a
1
r
n
\newline
Answer:
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Find the sum of the first
7
7
7
terms of the following sequence. Round to the nearest hundredth if necessary.
\newline
27
,
−
54
,
108
,
…
27, \quad-54, \quad 108, \ldots
27
,
−
54
,
108
,
…
\newline
Sum of a finite geometric series:
\newline
S
n
=
a
1
−
a
1
r
n
1
−
r
S_{n}=\frac{a_{1}-a_{1} r^{n}}{1-r}
S
n
=
1
−
r
a
1
−
a
1
r
n
\newline
Answer:
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Evaluate:
\newline
∑
n
=
0
2
(
x
+
n
)
\sum_{n=0}^{2}(x+n)
n
=
0
∑
2
(
x
+
n
)
\newline
Answer:
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Find the numerical answer to the summation given below.
\newline
∑
n
=
0
92
(
6
n
+
3
)
\sum_{n=0}^{92}(6 n+3)
n
=
0
∑
92
(
6
n
+
3
)
\newline
Answer:
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Find the numerical answer to the summation given below.
\newline
∑
n
=
6
94
(
5
n
+
7
)
\sum_{n=6}^{94}(5 n+7)
n
=
6
∑
94
(
5
n
+
7
)
\newline
Answer:
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Find the numerical answer to the summation given below.
\newline
∑
n
=
0
95
(
5
n
+
7
)
\sum_{n=0}^{95}(5 n+7)
n
=
0
∑
95
(
5
n
+
7
)
\newline
Answer:
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Find the numerical answer to the summation given below.
\newline
∑
n
=
2
61
(
5
n
+
9
)
\sum_{n=2}^{61}(5 n+9)
n
=
2
∑
61
(
5
n
+
9
)
\newline
Answer:
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Find the numerical answer to the summation given below.
\newline
∑
n
=
0
93
(
3
n
+
10
)
\sum_{n=0}^{93}(3 n+10)
n
=
0
∑
93
(
3
n
+
10
)
\newline
Answer:
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Find the numerical answer to the summation given below.
\newline
∑
n
=
0
68
(
6
n
+
3
)
\sum_{n=0}^{68}(6 n+3)
n
=
0
∑
68
(
6
n
+
3
)
\newline
Answer:
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Find the numerical answer to the summation given below.
\newline
∑
n
=
1
95
(
2
n
+
1
)
\sum_{n=1}^{95}(2 n+1)
n
=
1
∑
95
(
2
n
+
1
)
\newline
Answer:
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Find the numerical answer to the summation given below.
\newline
∑
n
=
1
93
(
4
n
+
3
)
\sum_{n=1}^{93}(4 n+3)
n
=
1
∑
93
(
4
n
+
3
)
\newline
Answer:
Get tutor help
If
a
1
=
2
,
a
2
=
0
a_{1}=2, a_{2}=0
a
1
=
2
,
a
2
=
0
and
a
n
=
a
n
−
1
−
a
n
−
2
a_{n}=a_{n-1}-a_{n-2}
a
n
=
a
n
−
1
−
a
n
−
2
then find the value of
a
4
a_{4}
a
4
.
\newline
Answer:
Get tutor help
If
a
1
=
2
,
a
2
=
0
a_{1}=2, a_{2}=0
a
1
=
2
,
a
2
=
0
and
a
n
=
3
a
n
−
1
+
3
a
n
−
2
a_{n}=3 a_{n-1}+3 a_{n-2}
a
n
=
3
a
n
−
1
+
3
a
n
−
2
then find the value of
a
4
a_{4}
a
4
.
\newline
Answer:
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If
a
1
=
3
,
a
2
=
2
a_{1}=3, a_{2}=2
a
1
=
3
,
a
2
=
2
and
a
n
=
a
n
−
1
+
2
a
n
−
2
a_{n}=a_{n-1}+2 a_{n-2}
a
n
=
a
n
−
1
+
2
a
n
−
2
then find the value of
a
5
a_{5}
a
5
.
\newline
Answer:
Get tutor help
If
a
1
=
2
,
a
2
=
0
a_{1}=2, a_{2}=0
a
1
=
2
,
a
2
=
0
and
a
n
=
3
a
n
−
1
−
3
a
n
−
2
a_{n}=3 a_{n-1}-3 a_{n-2}
a
n
=
3
a
n
−
1
−
3
a
n
−
2
then find the value of
a
5
a_{5}
a
5
.
\newline
Answer:
Get tutor help
If
a
1
=
3
,
a
2
=
5
a_{1}=3, a_{2}=5
a
1
=
3
,
a
2
=
5
and
a
n
=
2
a
n
−
1
+
2
a
n
−
2
a_{n}=2 a_{n-1}+2 a_{n-2}
a
n
=
2
a
n
−
1
+
2
a
n
−
2
then find the value of
a
5
a_{5}
a
5
.
\newline
Answer:
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If
a
1
=
0
,
a
2
=
1
a_{1}=0, a_{2}=1
a
1
=
0
,
a
2
=
1
and
a
n
=
3
a
n
−
1
+
3
a
n
−
2
a_{n}=3 a_{n-1}+3 a_{n-2}
a
n
=
3
a
n
−
1
+
3
a
n
−
2
then find the value of
a
6
a_{6}
a
6
.
\newline
Answer:
Get tutor help
If
a
1
=
3
,
a
2
=
0
a_{1}=3, a_{2}=0
a
1
=
3
,
a
2
=
0
and
a
n
=
3
a
n
−
1
+
3
a
n
−
2
a_{n}=3 a_{n-1}+3 a_{n-2}
a
n
=
3
a
n
−
1
+
3
a
n
−
2
then find the value of
a
5
a_{5}
a
5
.
\newline
Answer:
Get tutor help
If
a
1
=
1
,
a
2
=
3
a_{1}=1, a_{2}=3
a
1
=
1
,
a
2
=
3
and
a
n
=
a
n
−
1
+
a
n
−
2
a_{n}=a_{n-1}+a_{n-2}
a
n
=
a
n
−
1
+
a
n
−
2
then find the value of
a
4
a_{4}
a
4
.
\newline
Answer:
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If
f
(
1
)
=
0
,
f
(
2
)
=
3
f(1)=0, f(2)=3
f
(
1
)
=
0
,
f
(
2
)
=
3
and
f
(
n
)
=
3
f
(
n
−
1
)
−
2
f
(
n
−
2
)
f(n)=3 f(n-1)-2 f(n-2)
f
(
n
)
=
3
f
(
n
−
1
)
−
2
f
(
n
−
2
)
then find the value of
f
(
5
)
f(5)
f
(
5
)
.
\newline
Answer:
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If
f
(
1
)
=
1
,
f
(
2
)
=
0
f(1)=1, f(2)=0
f
(
1
)
=
1
,
f
(
2
)
=
0
and
f
(
n
)
=
f
(
n
−
1
)
+
2
f
(
n
−
2
)
f(n)=f(n-1)+2 f(n-2)
f
(
n
)
=
f
(
n
−
1
)
+
2
f
(
n
−
2
)
then find the value of
f
(
4
)
f(4)
f
(
4
)
.
\newline
Answer:
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If
f
(
1
)
=
1
,
f
(
2
)
=
3
f(1)=1, f(2)=3
f
(
1
)
=
1
,
f
(
2
)
=
3
and
f
(
n
)
=
3
f
(
n
−
1
)
−
2
f
(
n
−
2
)
f(n)=3 f(n-1)-2 f(n-2)
f
(
n
)
=
3
f
(
n
−
1
)
−
2
f
(
n
−
2
)
then find the value of
f
(
5
)
f(5)
f
(
5
)
.
\newline
Answer:
Get tutor help
If
f
(
1
)
=
5
,
f
(
2
)
=
0
f(1)=5, f(2)=0
f
(
1
)
=
5
,
f
(
2
)
=
0
and
f
(
n
)
=
3
f
(
n
−
1
)
−
3
f
(
n
−
2
)
f(n)=3 f(n-1)-3 f(n-2)
f
(
n
)
=
3
f
(
n
−
1
)
−
3
f
(
n
−
2
)
then find the value of
f
(
4
)
f(4)
f
(
4
)
.
\newline
Answer:
Get tutor help
If
f
(
1
)
=
2
,
f
(
2
)
=
1
f(1)=2, f(2)=1
f
(
1
)
=
2
,
f
(
2
)
=
1
and
f
(
n
)
=
3
f
(
n
−
1
)
+
f
(
n
−
2
)
f(n)=3 f(n-1)+f(n-2)
f
(
n
)
=
3
f
(
n
−
1
)
+
f
(
n
−
2
)
then find the value of
f
(
6
)
f(6)
f
(
6
)
.
\newline
Answer:
Get tutor help
If
f
(
1
)
=
5
,
f
(
2
)
=
2
f(1)=5, f(2)=2
f
(
1
)
=
5
,
f
(
2
)
=
2
and
f
(
n
)
=
3
f
(
n
−
1
)
−
3
f
(
n
−
2
)
f(n)=3 f(n-1)-3 f(n-2)
f
(
n
)
=
3
f
(
n
−
1
)
−
3
f
(
n
−
2
)
then find the value of
f
(
6
)
f(6)
f
(
6
)
.
\newline
Answer:
Get tutor help
If
f
(
1
)
=
0
,
f
(
2
)
=
1
f(1)=0, f(2)=1
f
(
1
)
=
0
,
f
(
2
)
=
1
and
f
(
n
)
=
3
f
(
n
−
1
)
+
3
f
(
n
−
2
)
f(n)=3 f(n-1)+3 f(n-2)
f
(
n
)
=
3
f
(
n
−
1
)
+
3
f
(
n
−
2
)
then find the value of
f
(
6
)
f(6)
f
(
6
)
.
\newline
Answer:
Get tutor help
If
f
(
1
)
=
5
,
f
(
2
)
=
2
f(1)=5, f(2)=2
f
(
1
)
=
5
,
f
(
2
)
=
2
and
f
(
n
)
=
3
f
(
n
−
1
)
+
3
f
(
n
−
2
)
f(n)=3 f(n-1)+3 f(n-2)
f
(
n
)
=
3
f
(
n
−
1
)
+
3
f
(
n
−
2
)
then find the value of
f
(
4
)
f(4)
f
(
4
)
.
\newline
Answer:
Get tutor help
If
f
(
1
)
=
1
f(1)=1
f
(
1
)
=
1
and
f
(
n
+
1
)
=
f
(
n
)
2
−
2
f(n+1)=f(n)^{2}-2
f
(
n
+
1
)
=
f
(
n
)
2
−
2
then find the value of
f
(
4
)
f(4)
f
(
4
)
.
\newline
Answer:
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If
f
(
1
)
=
2
f(1)=2
f
(
1
)
=
2
and
f
(
n
+
1
)
=
f
(
n
)
2
−
2
f(n+1)=f(n)^{2}-2
f
(
n
+
1
)
=
f
(
n
)
2
−
2
then find the value of
f
(
3
)
f(3)
f
(
3
)
.
\newline
Answer:
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If
f
(
1
)
=
2
f(1)=2
f
(
1
)
=
2
and
f
(
n
+
1
)
=
f
(
n
)
2
−
5
f(n+1)=f(n)^{2}-5
f
(
n
+
1
)
=
f
(
n
)
2
−
5
then find the value of
f
(
4
)
f(4)
f
(
4
)
.
\newline
Answer:
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If
f
(
1
)
=
2
f(1)=2
f
(
1
)
=
2
and
f
(
n
+
1
)
=
f
(
n
)
2
−
3
f(n+1)=f(n)^{2}-3
f
(
n
+
1
)
=
f
(
n
)
2
−
3
then find the value of
f
(
3
)
f(3)
f
(
3
)
.
\newline
Answer:
Get tutor help
If
f
(
1
)
=
2
f(1)=2
f
(
1
)
=
2
and
f
(
n
+
1
)
=
f
(
n
)
2
−
3
f(n+1)=f(n)^{2}-3
f
(
n
+
1
)
=
f
(
n
)
2
−
3
then find the value of
f
(
4
)
f(4)
f
(
4
)
.
\newline
Answer:
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If
f
(
1
)
=
1
,
f
(
2
)
=
0
f(1)=1, f(2)=0
f
(
1
)
=
1
,
f
(
2
)
=
0
and
f
(
n
)
=
2
f
(
n
−
1
)
−
3
f
(
n
−
2
)
f(n)=2 f(n-1)-3 f(n-2)
f
(
n
)
=
2
f
(
n
−
1
)
−
3
f
(
n
−
2
)
then find the value of
f
(
4
)
f(4)
f
(
4
)
.
\newline
Answer:
Get tutor help
If
f
(
1
)
=
1
,
f
(
2
)
=
3
f(1)=1, f(2)=3
f
(
1
)
=
1
,
f
(
2
)
=
3
and
f
(
n
)
=
f
(
n
−
1
)
+
3
f
(
n
−
2
)
f(n)=f(n-1)+3 f(n-2)
f
(
n
)
=
f
(
n
−
1
)
+
3
f
(
n
−
2
)
then find the value of
f
(
4
)
f(4)
f
(
4
)
.
\newline
Answer:
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Given that
f
(
x
)
=
x
+
3
,
g
(
x
)
=
−
4
x
f(x)=x+3, \quad g(x)=-4 x
f
(
x
)
=
x
+
3
,
g
(
x
)
=
−
4
x
and
h
(
x
)
=
−
3
f
(
x
)
+
3
g
(
x
+
1
)
h(x)=-3 f(x)+3 g(x+1)
h
(
x
)
=
−
3
f
(
x
)
+
3
g
(
x
+
1
)
, then what is the value of
h
(
6
)
h(6)
h
(
6
)
?
\newline
Answer:
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Given that
f
(
x
)
=
x
−
4
,
g
(
x
)
=
−
3
x
f(x)=x-4, \quad g(x)=-3 x
f
(
x
)
=
x
−
4
,
g
(
x
)
=
−
3
x
and
h
(
x
)
=
−
f
(
x
)
+
2
g
(
x
−
1
)
h(x)=-f(x)+2 g(x-1)
h
(
x
)
=
−
f
(
x
)
+
2
g
(
x
−
1
)
, then what is the value of
h
(
7
)
h(7)
h
(
7
)
?
\newline
Answer:
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Given that
f
(
x
)
=
x
−
1
,
g
(
x
)
=
−
2
x
f(x)=x-1, \quad g(x)=-2 x
f
(
x
)
=
x
−
1
,
g
(
x
)
=
−
2
x
and
h
(
x
)
=
−
3
f
(
x
)
+
g
(
x
−
3
)
h(x)=-3 f(x)+g(x-3)
h
(
x
)
=
−
3
f
(
x
)
+
g
(
x
−
3
)
, then what is the value of
h
(
5
)
h(5)
h
(
5
)
?
\newline
Answer:
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Find the
5
th
5^{\text {th }}
5
th
term of the arithmetic sequence
−
4
x
−
8
,
2
x
−
13
,
8
x
−
18
,
…
-4 x-8,2 x-13,8 x-18, \ldots
−
4
x
−
8
,
2
x
−
13
,
8
x
−
18
,
…
\newline
Answer:
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Find the real zeros of the quadratic function using any method you wish. What are the x-intercepts, if any, of the graph of the function?
\newline
f
(
x
)
=
x
2
−
50
f(x)=x^{2}-50
f
(
x
)
=
x
2
−
50
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Solve the equation by factoring:
\newline
2
x
3
−
6
x
2
−
20
x
=
0
2 x^{3}-6 x^{2}-20 x=0
2
x
3
−
6
x
2
−
20
x
=
0
\newline
Answer:
x
=
x=
x
=
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Solve the equation by factoring:
\newline
36
x
2
−
160
x
−
2
x
3
=
0
36 x^{2}-160 x-2 x^{3}=0
36
x
2
−
160
x
−
2
x
3
=
0
\newline
Answer:
x
=
x=
x
=
Get tutor help
Solve the equation by factoring:
\newline
16
x
2
−
30
x
−
2
x
3
=
0
16 x^{2}-30 x-2 x^{3}=0
16
x
2
−
30
x
−
2
x
3
=
0
\newline
Answer:
x
=
x=
x
=
Get tutor help
Solve the equation by factoring:
\newline
2
x
3
+
12
x
2
−
32
x
=
0
2 x^{3}+12 x^{2}-32 x=0
2
x
3
+
12
x
2
−
32
x
=
0
\newline
Answer:
x
=
x=
x
=
Get tutor help
Solve the equation by factoring:
\newline
2
x
3
+
14
x
2
−
36
x
=
0
2 x^{3}+14 x^{2}-36 x=0
2
x
3
+
14
x
2
−
36
x
=
0
\newline
Answer:
x
=
x=
x
=
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Solve the equation by factoring:
\newline
60
x
−
24
x
2
−
3
x
3
=
0
60 x-24 x^{2}-3 x^{3}=0
60
x
−
24
x
2
−
3
x
3
=
0
\newline
Answer:
x
=
x=
x
=
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Solve for
x
x
x
.
\newline
1
5
−
5
x
=
−
8
5
x
\frac{1}{5}-\frac{5}{x}=\frac{-8}{5 x}
5
1
−
x
5
=
5
x
−
8
\newline
Answer:
x
=
x=
x
=
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