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Let’s check out your problem:
Find the quadratic polynomial that completes the factorization.
\newline
w
3
+
8
=
(
w
+
2
)
(
_
_
_
_
_
)
w^3 + 8 = (w + 2)(\_\_\_\_\_)
w
3
+
8
=
(
w
+
2
)
(
_____
)
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Math Problems
Precalculus
Factor sums and differences of cubes
Full solution
Q.
Find the quadratic polynomial that completes the factorization.
\newline
w
3
+
8
=
(
w
+
2
)
(
_
_
_
_
_
)
w^3 + 8 = (w + 2)(\_\_\_\_\_)
w
3
+
8
=
(
w
+
2
)
(
_____
)
Factorization of sum of cubes:
We know that
w
3
+
8
w^3 + 8
w
3
+
8
is a sum of cubes, which
factors
as
(
w
+
2
)
(
w
2
−
2
w
+
4
)
(w + 2)(w^2 - 2w + 4)
(
w
+
2
)
(
w
2
−
2
w
+
4
)
.
Multiplication check:
Now we need to check if our factorization is correct by multiplying
(
w
+
2
)
(
w
2
−
2
w
+
4
)
(w + 2)(w^2 - 2w + 4)
(
w
+
2
)
(
w
2
−
2
w
+
4
)
.
Combining like terms:
w + \(2)(w^
2
2
2
-
2
2
2
w +
4
4
4
) = w^
3
3
3
-
2
2
2
w^
2
2
2
+
4
4
4
w +
2
2
2
w^
2
2
2
-
4
4
4
w +
8
8
8
\
Final verification:
Combine like terms:
w
3
+
0
w
2
+
0
w
+
8
=
w
3
+
8
w^3 + 0w^2 + 0w + 8 = w^3 + 8
w
3
+
0
w
2
+
0
w
+
8
=
w
3
+
8
.
Final verification:
Combine like terms:
w
3
+
0
w
2
+
0
w
+
8
=
w
3
+
8
w^3 + 0w^2 + 0w + 8 = w^3 + 8
w
3
+
0
w
2
+
0
w
+
8
=
w
3
+
8
.Since the original expression is
w
3
+
8
w^3 + 8
w
3
+
8
, our factorization is correct.
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\newline
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\newline
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Divide. If there is a remainder, include it as a simplified fraction.
\newline
(
24
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Question
Is the function
q
(
x
)
=
x
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q
(
x
)
=
x
6
−
9
even, odd, or neither?
\newline
Choices:
\newline
[[even][odd][neither]]
\text{[[even][odd][neither]]}
[[even][odd][neither]]
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Question
Find the product. Simplify your answer.
\newline
−
3
q
2
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−
3
q
2
+
q
)
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q
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3
q
2
+
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Question
Find the product. Simplify your answer.
\newline
(
r
+
3
)
(
4
r
+
2
)
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(
r
+
3
)
(
4
r
+
2
)
\newline
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Question
Find the roots of the factored polynomial.
\newline
(
x
+
7
)
(
x
+
4
)
(x + 7)(x + 4)
(
x
+
7
)
(
x
+
4
)
\newline
Write your answer as a list of values separated by commas.
\newline
x
=
x =
x
=
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Posted 8 months ago
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