# How do you multiply -3y^2-4y+6+2(y^2+7y-1)?

Jul 28, 2015

You expand the paranthesis and add or subtract like terms.

#### Explanation:

$- 3 {y}^{2} - 4 y + 6 + 2 \left({y}^{2} + 7 y - 1\right)$

Notice that your expression contains the product between $2$ and a paranthesis that features three terms, ${y}^{2}$, $7 y$, and $- 1$.

To expand this paranthesis, multiply each of these terms by $2$ to get

$2 \left({y}^{2} + 7 y - 1\right) = 2 \cdot {y}^{2} + 2 \cdot 7 y + 2 \cdot \left(- 1\right) = 2 {y}^{2} + 14 y - 2$

$- 3 {y}^{2} - 4 y + 6 + 2 {y}^{2} + 14 y - 2$
$\left(- 3 {y}^{2} + 2 {y}^{2}\right) + \left(14 y - 4 y\right) + \left(6 - 2\right)$
$\textcolor{g r e e n}{- {y}^{2} + 10 y + 4}$