Simplify 2 square root of 32d^5-7/8 square root of 8d^5+d^2 square root of 98d

$2 \sqrt{32 d^{5}} - \frac{7}{8} \sqrt{8 d^{5}} + d^{2} \sqrt{98 d}$

Step 1

Simplify each term.

Rewrite $32 d^{5}$ as ${(4 d^{2})}^{2} \cdot (2 d)$ .

Factor $16$ out of $32$ .

$2 \sqrt{16 (2) d^{5}} - \frac{7}{8} \cdot \sqrt{8 d^{5}} + d^{2} \sqrt{98 d}$

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

$2 \sqrt{4^{2} \cdot 2 d^{5}} - \frac{7}{8} \cdot \sqrt{8 d^{5}} + d^{2} \sqrt{98 d}$

Factor out $d^{4}$ .

$2 \sqrt{4^{2} \cdot 2 (d^{4} d)} - \frac{7}{8} \cdot \sqrt{8 d^{5}} + d^{2} \sqrt{98 d}$

Rewrite $d^{4}$ as ${(d^{2})}^{2}$ .

$2 \sqrt{4^{2} \cdot 2 ({(d^{2})}^{2} d)} - \frac{7}{8} \cdot \sqrt{8 d^{5}} + d^{2} \sqrt{98 d}$

Move $2$ .

$2 \sqrt{4^{2} {(d^{2})}^{2} \cdot 2 d} - \frac{7}{8} \cdot \sqrt{8 d^{5}} + d^{2} \sqrt{98 d}$

Rewrite $4^{2} {(d^{2})}^{2}$ as ${(4 d^{2})}^{2}$ .

$2 \sqrt{{(4 d^{2})}^{2} \cdot 2 d} - \frac{7}{8} \cdot \sqrt{8 d^{5}} + d^{2} \sqrt{98 d}$

Add parentheses.

$2 \sqrt{{(4 d^{2})}^{2} \cdot (2 d)} - \frac{7}{8} \cdot \sqrt{8 d^{5}} + d^{2} \sqrt{98 d}$

Pull terms out from under the radical.

$2 (4 d^{2} \sqrt{2 d}) - \frac{7}{8} \cdot \sqrt{8 d^{5}} + d^{2} \sqrt{98 d}$

Multiply $4$ by $2$ .

$8 (d^{2} \sqrt{2 d}) - \frac{7}{8} \cdot \sqrt{8 d^{5}} + d^{2} \sqrt{98 d}$

Rewrite $8 d^{5}$ as ${(2 d^{2})}^{2} \cdot (2 d)$ .

Factor $4$ out of $8$ .

$8 d^{2} \sqrt{2 d} - \frac{7}{8} \cdot \sqrt{4 (2) d^{5}} + d^{2} \sqrt{98 d}$

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

$8 d^{2} \sqrt{2 d} - \frac{7}{8} \cdot \sqrt{2^{2} \cdot 2 d^{5}} + d^{2} \sqrt{98 d}$

Factor out $d^{4}$ .

$8 d^{2} \sqrt{2 d} - \frac{7}{8} \cdot \sqrt{2^{2} \cdot 2 (d^{4} d)} + d^{2} \sqrt{98 d}$

Rewrite $d^{4}$ as ${(d^{2})}^{2}$ .

$8 d^{2} \sqrt{2 d} - \frac{7}{8} \cdot \sqrt{2^{2} \cdot 2 ({(d^{2})}^{2} d)} + d^{2} \sqrt{98 d}$

Move $2$ .

$8 d^{2} \sqrt{2 d} - \frac{7}{8} \cdot \sqrt{2^{2} {(d^{2})}^{2} \cdot 2 d} + d^{2} \sqrt{98 d}$

Rewrite $2^{2} {(d^{2})}^{2}$ as ${(2 d^{2})}^{2}$ .

$8 d^{2} \sqrt{2 d} - \frac{7}{8} \cdot \sqrt{{(2 d^{2})}^{2} \cdot 2 d} + d^{2} \sqrt{98 d}$

Add parentheses.

$8 d^{2} \sqrt{2 d} - \frac{7}{8} \cdot \sqrt{{(2 d^{2})}^{2} \cdot (2 d)} + d^{2} \sqrt{98 d}$

Pull terms out from under the radical.

$8 d^{2} \sqrt{2 d} - \frac{7}{8} \cdot (2 d^{2} \sqrt{2 d}) + d^{2} \sqrt{98 d}$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{7}{8}$ into the numerator.

$8 d^{2} \sqrt{2 d} + \frac{- 7}{8} \cdot (2 d^{2} \sqrt{2 d}) + d^{2} \sqrt{98 d}$

Factor $2$ out of $8$ .

$8 d^{2} \sqrt{2 d} + \frac{- 7}{2 (4)} \cdot (2 d^{2} \sqrt{2 d}) + d^{2} \sqrt{98 d}$

Factor $2$ out of $2 d^{2} \sqrt{2 d}$ .

$8 d^{2} \sqrt{2 d} + \frac{- 7}{2 (4)} \cdot (2 (d^{2} \sqrt{2 d})) + d^{2} \sqrt{98 d}$

Cancel the common factor.

$8 d^{2} \sqrt{2 d} + \frac{- 7}{2 \cdot 4} \cdot (2 (d^{2} \sqrt{2 d})) + d^{2} \sqrt{98 d}$

Rewrite the expression.

$8 d^{2} \sqrt{2 d} + \frac{- 7}{4} \cdot (d^{2} \sqrt{2 d}) + d^{2} \sqrt{98 d}$

Combine $d^{2}$ and $\frac{- 7}{4}$ .

$8 d^{2} \sqrt{2 d} + \frac{d^{2} \cdot - 7}{4} \cdot \sqrt{2 d} + d^{2} \sqrt{98 d}$

Combine $\frac{d^{2} \cdot - 7}{4}$ and $\sqrt{2 d}$ .

$8 d^{2} \sqrt{2 d} + \frac{d^{2} \cdot - 7 \sqrt{2 d}}{4} + d^{2} \sqrt{98 d}$

Move $- 7$ to the left of $d^{2}$ .

$8 d^{2} \sqrt{2 d} + \frac{- 7 \cdot d^{2} \sqrt{2 d}}{4} + d^{2} \sqrt{98 d}$

Move the negative in front of the fraction.

$8 d^{2} \sqrt{2 d} - \frac{(7) \cdot d^{2} \sqrt{2 d}}{4} + d^{2} \sqrt{98 d}$

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

Factor $49$ out of $98$ .

$8 d^{2} \sqrt{2 d} - \frac{7 d^{2} \sqrt{2 d}}{4} + d^{2} \sqrt{49 (2) d}$

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

$8 d^{2} \sqrt{2 d} - \frac{7 d^{2} \sqrt{2 d}}{4} + d^{2} \sqrt{7^{2} \cdot 2 d}$

Add parentheses.

$8 d^{2} \sqrt{2 d} - \frac{7 d^{2} \sqrt{2 d}}{4} + d^{2} \sqrt{7^{2} \cdot (2 d)}$

Pull terms out from under the radical.

$8 d^{2} \sqrt{2 d} - \frac{7 d^{2} \sqrt{2 d}}{4} + d^{2} (7 \sqrt{2 d})$

Rewrite using the commutative property of multiplication.

$8 d^{2} \sqrt{2 d} - \frac{7 d^{2} \sqrt{2 d}}{4} + 7 d^{2} \sqrt{2 d}$

Step 2

To write $8 d^{2} \sqrt{2 d}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$8 d^{2} \sqrt{2 d} \cdot \frac{4}{4} - \frac{7 d^{2} \sqrt{2 d}}{4} + 7 d^{2} \sqrt{2 d}$

Step 3

Simplify terms.

Combine $8 d^{2} \sqrt{2 d}$ and $\frac{4}{4}$ .

$\frac{8 d^{2} \sqrt{2 d} \cdot 4}{4} - \frac{7 d^{2} \sqrt{2 d}}{4} + 7 d^{2} \sqrt{2 d}$

Combine the numerators over the common denominator.

$\frac{8 d^{2} \sqrt{2 d} \cdot 4 - 7 d^{2} \sqrt{2 d}}{4} + 7 d^{2} \sqrt{2 d}$

Step 4

Simplify each term.

Simplify the numerator.

Factor $d^{2} \sqrt{2 d}$ out of $8 d^{2} \sqrt{2 d} \cdot 4 - 7 d^{2} \sqrt{2 d}$ .

Factor $d^{2} \sqrt{2 d}$ out of $8 d^{2} \sqrt{2 d} \cdot 4$ .

$\frac{d^{2} \sqrt{2 d} (8 \cdot 4) - 7 d^{2} \sqrt{2 d}}{4} + 7 d^{2} \sqrt{2 d}$

Factor $d^{2} \sqrt{2 d}$ out of $- 7 d^{2} \sqrt{2 d}$ .

$\frac{d^{2} \sqrt{2 d} (8 \cdot 4) + d^{2} \sqrt{2 d} (- 7)}{4} + 7 d^{2} \sqrt{2 d}$

Factor $d^{2} \sqrt{2 d}$ out of $d^{2} \sqrt{2 d} (8 \cdot 4) + d^{2} \sqrt{2 d} (- 7)$ .

$\frac{d^{2} \sqrt{2 d} (8 \cdot 4 - 7)}{4} + 7 d^{2} \sqrt{2 d}$

Multiply $8$ by $4$ .

$\frac{d^{2} \sqrt{2 d} (32 - 7)}{4} + 7 d^{2} \sqrt{2 d}$

Subtract $7$ from $32$ .

$\frac{d^{2} \sqrt{2 d} \cdot 25}{4} + 7 d^{2} \sqrt{2 d}$

Move $25$ to the left of $d^{2} \sqrt{2 d}$ .

$\frac{25 d^{2} \sqrt{2 d}}{4} + 7 d^{2} \sqrt{2 d}$

Simplify 2 square root of 32d^5-7/8 square root of 8d^5+d^2 square root of 98d

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