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Simplify $4 m + 5 n 2 + 5 m + 4 n 2$ .

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41 m 2 + 41 n 2 + 80 m n

41 m 2 + 41 n 2 − 80 m n

41 m 2 − 41 n 2 + 80 m n

− 41 m 2 + 41 n 2 + 80 m n

answer is A.

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Detailed Solution

It is given

4 m + 5 n 2 + 5 m + 4 n 2

.
We know the identity

a + b 2 = a 2 + b 2 + 2 a b

.
Expanding the whole square terms by the use of identity,
Let us expand

4 m + 5 n 2

first,

⇒ 4 m + 5 n 2 = 4 m 2 + 5 n 2 + 2 × 4 m × 5 n ⇒ 4 m + 5 n 2 = 16 m 2 + 25 n 2 + 40 m n

Now let us expand

5 m + 4 n 2

⇒ 5 m + 4 n 2 = 5 m 2 + 4 n 2 + 2 × 5 m × 4 n ⇒ 5 m + 4 n 2 = 25 m 2 + 16 n 2 + 40 m n

On substituting these values in

4 m + 5 n 2 + 5 m + 4 n 2

we get,

⇒ 4 m + 5 n 2 + 5 m + 4 n 2 = 16 m 2 + 25 n 2 + 40 m n + 25 m 2 + 16 n 2 + 40 m n

Adding all

m 2

terms together and all

n 2

terms together,

⇒ 4 m + 5 n 2 + 5 m + 4 n 2 = 16 + 25 m 2 + 25 + 16 n 2 + 40 + 40 m n

⇒ 4 m + 5 n 2 + 5 m + 4 n 2 = 41 m 2 + 41 n 2 + 80 m n

Therefore,

4 m + 5 n 2 + 5 m + 4 n 2 = 41 m 2 + 41 n 2 + 80 m n .

Hence, option (1) is correct.

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