Simplify 5 square root of 32x+4 square root of 98x

$5 \sqrt{32 x} + 4 \sqrt{98 x}$

Step 1

Simplify each term.

Rewrite $32 x$ as ${(2^{2})}^{2} \cdot (2 x)$ .

Factor $16$ out of $32$ .

$5 \sqrt{16 (2) x} + 4 \sqrt{98 x}$

Rewrite $16$ as $4^{2}$ .

$5 \sqrt{4^{2} \cdot 2 x} + 4 \sqrt{98 x}$

Rewrite $4$ as $2^{2}$ .

$5 \sqrt{{(2^{2})}^{2} \cdot 2 x} + 4 \sqrt{98 x}$

Add parentheses.

$5 \sqrt{{(2^{2})}^{2} \cdot (2 x)} + 4 \sqrt{98 x}$

Pull terms out from under the radical.

$5 (2^{2} \sqrt{2 x}) + 4 \sqrt{98 x}$

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

$5 (4 \sqrt{2 x}) + 4 \sqrt{98 x}$

Multiply $4$ by $5$ .

$20 \sqrt{2 x} + 4 \sqrt{98 x}$

Rewrite $98 x$ as $7^{2} \cdot (2 x)$ .

Factor $49$ out of $98$ .

$20 \sqrt{2 x} + 4 \sqrt{49 (2) x}$

Rewrite $49$ as $7^{2}$ .

$20 \sqrt{2 x} + 4 \sqrt{7^{2} \cdot 2 x}$

Add parentheses.

$20 \sqrt{2 x} + 4 \sqrt{7^{2} \cdot (2 x)}$

Pull terms out from under the radical.

$20 \sqrt{2 x} + 4 (7 \sqrt{2 x})$

Multiply $7$ by $4$ .

$20 \sqrt{2 x} + 28 \sqrt{2 x}$

Step 2

Add $20 \sqrt{2 x}$ and $28 \sqrt{2 x}$ .

$48 \sqrt{2 x}$