Math, asked by rayupureddy, 4 months ago

(7x-4y)^5 binomial theorem​

Answers

Answered by gauripagade20
1

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

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