(7x-4y)^5 binomial theorem
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Answer:
answer
Popular Problems Finite Math Expand using the Binomial Theorem (7x-4y)^2
(
7
x
−
4
y
)
2
Use the binomial expansion theorem to find each term. The binomial theorem states
(
a
+
b
)
n
=
n
∑
k
=
0
n
C
k
⋅
(
a
n
−
k
b
k
)
.
2
∑
k
=
0
2
!
(
2
−
k
)
!
k
!
⋅
(
7
x
)
2
−
k
⋅
(
−
4
y
)
k
Expand the summation.
2
!
(
2
−
0
)
!
0
!
⋅
(
7
x
)
2
−
0
⋅
(
−
4
y
)
0
+
2
!
(
2
−
1
)
!
1
!
⋅
(
7
x
)
2
−
1
⋅
(
−
4
y
)
+
2
!
(
2
−
2
)
!
2
!
⋅
(
7
x
)
2
−
2
⋅
(
−
4
y
)
2
Simplify the exponents for each term of the expansion.
1
⋅
(
7
x
)
2
⋅
(
−
4
y
)
0
+
2
⋅
(
7
x
)
⋅
(
−
4
y
)
+
1
⋅
(
7
x
)
0
⋅
(
−
4
y
)
2
Simplify each term.
Multiply
(
7
x
)
2
by
1
.
(
7
x
)
2
⋅
(
−
4
y
)
0
+
2
⋅
(
7
x
)
1
⋅
(
−
4
y
)
1
+
1
⋅
(
7
x
)
0
⋅
(
−
4
y
)
2
Apply the product rule to
7
x
.
7
2
x
2
⋅
(
−
4
y
)
0
+
2
⋅
(
7
x
)
1
⋅
(
−
4
y
)
1
+
1
⋅
(
7
x
)
0
⋅
(
−
4
y
)
2
Raise
7
to the power of
2
.
49
x
2
⋅
(
−
4
y
)
0
+
2
⋅
(
7
x
)
1
⋅
(
−
4
y
)
1
+
1
⋅
(
7
x
)
0
⋅
(
−
4
y
)
2
Apply the product rule to
−
4
y
.
49
x
2
⋅
(
(
−
4
)
0
y
0
)
+
2
⋅
(
7
x
)
1
⋅
(
−
4
y
)
1
+
1
⋅
(
7
x
)
0
⋅
(
−
4
y
)
2
Rewrite using the commutative property of multiplication.
49
⋅
(
−
4
)
0
(
x
2
y
0
)
+
2
⋅
(
7
x
)
1
⋅
(
−
4
y
)
1
+
1
⋅
(
7
x
)
0
⋅
(
−
4
y
)
2
Anything raised to
0
is
1
.
49
⋅
1
(
x
2
y
0
)
+
2
⋅
(
7
x
)
1
⋅
(
−
4
y
)
1
+
1
⋅
(
7
x
)
0
⋅
(
−
4
y
)
2
Multiply
49
by
1
.
49
(
x
2
y
0
)
+
2
⋅
(
7
x
)
1
⋅
(
−
4
y
)
1
+
1
⋅
(
7
x
)
0
⋅
(
−
4
y
)
2
Anything raised to
0
is
1
.
49
(
x
2
⋅
1
)
+
2
⋅
(
7
x
)
1
⋅
(
−
4
y
)
1
+
1
⋅
(
7
x
)
0
⋅
(
−
4
y
)
2
Multiply
x
2
by
1
.
49
x
2
+
2
⋅
(
7
x
)
1
⋅
(
−
4
y
)
1
+
1
⋅
(
7
x
)
0
⋅
(
−
4
y
)
2
Simplify.
49
x
2
+
2
⋅
(
7
x
)
⋅
(
−
4
y
)
1
+
1
⋅
(
7
x
)
0
⋅
(
−
4
y
)
2
Multiply
7
by
2
.
49
x
2
+
14
x
⋅
(
−
4
y
)
1
+
1
⋅
(
7
x
)
0
⋅
(
−
4
y
)
2
Simplify.
49
x
2
+
14
x
⋅
(
−
4
y
)
+
1
⋅
(
7
x
)
0
⋅
(
−
4
y
)
2
Rewrite using the commutative property of multiplication.
49
x
2
+
14
⋅
−
4
(
x
y
)
+
1
⋅
(
7
x
)
0
⋅
(
−
4
y
)
2
Multiply
14
by
−
4
.
49
x
2
−
56
(
x
y
)
+
1
⋅
(
7
x
)
0
⋅
(
−
4
y
)
2
Multiply
(
7
x
)
0
by
1
.
49
x
2
−
56
x
y
+
(
7
x
)
0
⋅
(
−
4
y
)
2
Apply the product rule to
7
x
.
49
x
2
−
56
x
y
+
7
0
x
0
⋅
(
−
4
y
)
2
Anything raised to
0
is
1
.
49
x
2
−
56
x
y
+
1
x
0
⋅
(
−
4
y
)
2
Multiply
x
0
by
1
.
49
x
2
−
56
x
y
+
x
0
⋅
(
−
4
y
)
2
Anything raised to
0
is
1
.
49
x
2
−
56
x
y
+
1
⋅
(
−
4
y
)
2
Multiply
(
−
4
y
)
2
by
1
.
49
x
2
−
56
x
y
+
(
−
4
y
)
2
Apply the product rule to
−
4
y
.
49
x
2
−
56
x
y
+
(
−
4
)
2
y
2
Raise
−
4
to the power of
2
.
49
x
2
−
56
x
y
+
16
y
2