Simplify square root of 49(x+2y)^10

$\sqrt{49 {(x + 2 y)}^{10}}$

Step 1

Rewrite $49 {(x + 2 y)}^{10}$ as ${(7 {(x + 2 y)}^{5})}^{2}$ .

$\sqrt{{(7 {(x + 2 y)}^{5})}^{2}}$

Step 2

Pull terms out from under the radical, assuming positive real numbers.

$7 {(x + 2 y)}^{5}$

Step 3

Use the Binomial Theorem.

$7 (x^{5} + 5 x^{4} (2 y) + 10 x^{3} {(2 y)}^{2} + 10 x^{2} {(2 y)}^{3} + 5 x {(2 y)}^{4} + {(2 y)}^{5})$

Step 4

Simplify terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$7 (x^{5} + 5 \cdot 2 x^{4} y + 10 x^{3} {(2 y)}^{2} + 10 x^{2} {(2 y)}^{3} + 5 x {(2 y)}^{4} + {(2 y)}^{5})$

Multiply $5$ by $2$ .

$7 (x^{5} + 10 x^{4} y + 10 x^{3} {(2 y)}^{2} + 10 x^{2} {(2 y)}^{3} + 5 x {(2 y)}^{4} + {(2 y)}^{5})$

Apply the product rule to $2 y$ .

$7 (x^{5} + 10 x^{4} y + 10 x^{3} (2^{2} y^{2}) + 10 x^{2} {(2 y)}^{3} + 5 x {(2 y)}^{4} + {(2 y)}^{5})$

Rewrite using the commutative property of multiplication.

$7 (x^{5} + 10 x^{4} y + 10 \cdot 2^{2} x^{3} y^{2} + 10 x^{2} {(2 y)}^{3} + 5 x {(2 y)}^{4} + {(2 y)}^{5})$

Raise $2$ to the power of $2$ .

$7 (x^{5} + 10 x^{4} y + 10 \cdot 4 x^{3} y^{2} + 10 x^{2} {(2 y)}^{3} + 5 x {(2 y)}^{4} + {(2 y)}^{5})$

Multiply $10$ by $4$ .

$7 (x^{5} + 10 x^{4} y + 40 x^{3} y^{2} + 10 x^{2} {(2 y)}^{3} + 5 x {(2 y)}^{4} + {(2 y)}^{5})$

Apply the product rule to $2 y$ .

$7 (x^{5} + 10 x^{4} y + 40 x^{3} y^{2} + 10 x^{2} (2^{3} y^{3}) + 5 x {(2 y)}^{4} + {(2 y)}^{5})$

Rewrite using the commutative property of multiplication.

$7 (x^{5} + 10 x^{4} y + 40 x^{3} y^{2} + 10 \cdot 2^{3} x^{2} y^{3} + 5 x {(2 y)}^{4} + {(2 y)}^{5})$

Raise $2$ to the power of $3$ .

$7 (x^{5} + 10 x^{4} y + 40 x^{3} y^{2} + 10 \cdot 8 x^{2} y^{3} + 5 x {(2 y)}^{4} + {(2 y)}^{5})$

Multiply $10$ by $8$ .

$7 (x^{5} + 10 x^{4} y + 40 x^{3} y^{2} + 80 x^{2} y^{3} + 5 x {(2 y)}^{4} + {(2 y)}^{5})$

Apply the product rule to $2 y$ .

$7 (x^{5} + 10 x^{4} y + 40 x^{3} y^{2} + 80 x^{2} y^{3} + 5 x (2^{4} y^{4}) + {(2 y)}^{5})$

Rewrite using the commutative property of multiplication.

$7 (x^{5} + 10 x^{4} y + 40 x^{3} y^{2} + 80 x^{2} y^{3} + 5 \cdot 2^{4} x y^{4} + {(2 y)}^{5})$

Raise $2$ to the power of $4$ .

$7 (x^{5} + 10 x^{4} y + 40 x^{3} y^{2} + 80 x^{2} y^{3} + 5 \cdot 16 x y^{4} + {(2 y)}^{5})$

Multiply $5$ by $16$ .

$7 (x^{5} + 10 x^{4} y + 40 x^{3} y^{2} + 80 x^{2} y^{3} + 80 x y^{4} + {(2 y)}^{5})$

Apply the product rule to $2 y$ .

$7 (x^{5} + 10 x^{4} y + 40 x^{3} y^{2} + 80 x^{2} y^{3} + 80 x y^{4} + 2^{5} y^{5})$

Raise $2$ to the power of $5$ .

$7 (x^{5} + 10 x^{4} y + 40 x^{3} y^{2} + 80 x^{2} y^{3} + 80 x y^{4} + 32 y^{5})$

Apply the distributive property.

$7 x^{5} + 7 (10 x^{4} y) + 7 (40 x^{3} y^{2}) + 7 (80 x^{2} y^{3}) + 7 (80 x y^{4}) + 7 (32 y^{5})$

Step 5

Simplify.

Multiply $10$ by $7$ .

$7 x^{5} + 70 (x^{4} y) + 7 (40 x^{3} y^{2}) + 7 (80 x^{2} y^{3}) + 7 (80 x y^{4}) + 7 (32 y^{5})$

Multiply $40$ by $7$ .

$7 x^{5} + 70 (x^{4} y) + 280 (x^{3} y^{2}) + 7 (80 x^{2} y^{3}) + 7 (80 x y^{4}) + 7 (32 y^{5})$

Multiply $80$ by $7$ .

$7 x^{5} + 70 (x^{4} y) + 280 (x^{3} y^{2}) + 560 (x^{2} y^{3}) + 7 (80 x y^{4}) + 7 (32 y^{5})$

Multiply $80$ by $7$ .

$7 x^{5} + 70 (x^{4} y) + 280 (x^{3} y^{2}) + 560 (x^{2} y^{3}) + 560 (x y^{4}) + 7 (32 y^{5})$

Multiply $32$ by $7$ .

$7 x^{5} + 70 (x^{4} y) + 280 (x^{3} y^{2}) + 560 (x^{2} y^{3}) + 560 (x y^{4}) + 224 y^{5}$

Step 6

Remove parentheses.

$7 x^{5} + 70 x^{4} y + 280 x^{3} y^{2} + 560 x^{2} y^{3} + 560 x y^{4} + 224 y^{5}$

Simplify square root of 49(x+2y)^10

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