Find the Product -1/2y(2y^3-8)

$- \frac{1}{2} y (2 y^{3} - 8)$

Step 1

Simplify by multiplying through.

Apply the distributive property.

$- \frac{1}{2} \cdot (y (2 y^{3}) + y \cdot - 8)$

Reorder.

Rewrite using the commutative property of multiplication.

$- \frac{1}{2} \cdot (2 y \cdot y^{3} + y \cdot - 8)$

Move $- 8$ to the left of $y$ .

$- \frac{1}{2} \cdot (2 y \cdot y^{3} - 8 \cdot y)$

Step 2

Multiply $y$ by $y^{3}$ by adding the exponents.

Move $y^{3}$ .

$- \frac{1}{2} \cdot (2 (y^{3} y) - 8 \cdot y)$

Multiply $y^{3}$ by $y$ .

Raise $y$ to the power of $1$ .

$- \frac{1}{2} \cdot (2 (y^{3} y^{1}) - 8 \cdot y)$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- \frac{1}{2} \cdot (2 y^{3 + 1} - 8 \cdot y)$

Add $3$ and $1$ .

$- \frac{1}{2} \cdot (2 y^{4} - 8 \cdot y)$

$- \frac{1}{2} \cdot (2 y^{4} - 8 y)$

Step 3

Simplify terms.

Apply the distributive property.

$- \frac{1}{2} (2 y^{4}) - \frac{1}{2} (- 8 y)$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$\frac{- 1}{2} (2 y^{4}) - \frac{1}{2} (- 8 y)$

Factor $2$ out of $2 y^{4}$ .

$\frac{- 1}{2} (2 (y^{4})) - \frac{1}{2} (- 8 y)$

Cancel the common factor.

$\frac{- 1}{2} (2 y^{4}) - \frac{1}{2} (- 8 y)$

Rewrite the expression.

$- 1 y^{4} - \frac{1}{2} (- 8 y)$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$- 1 y^{4} + \frac{- 1}{2} (- 8 y)$

Factor $2$ out of $- 8 y$ .

$- 1 y^{4} + \frac{- 1}{2} (2 (- 4 y))$

Cancel the common factor.

$- 1 y^{4} + \frac{- 1}{2} (2 (- 4 y))$

Rewrite the expression.

$- 1 y^{4} - 1 (- 4 y)$

Simplify the expression.

Multiply $- 4$ by $- 1$ .

$- 1 y^{4} + 4 y$

Rewrite $- 1 y^{4}$ as $- y^{4}$ .

$- y^{4} + 4 y$

Find the Product -1/2y(2y^3-8)

Related Questions

Support LCMGCF.com by disabling your adblocker.