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

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

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Answer 1 · 2015-07-28T09:47:32

You expand the paranthesis and add or subtract like terms.

Explanation:

Your starting expression is

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

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

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

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

Your expression becomes

$- 3 y^{2} - 4 y + 6 + 2 y^{2} + 14 y - 2$ $-3y^2 - 4y + 6 + 2y^2 + 14y - 2$

You can simplify this further by grouping like terms together to get

$(- 3 y^{2} + 2 y^{2}) + (14 y - 4 y) + (6 - 2)$ $(-3y^2 + 2y^2) +(14y - 4y) + (6-2)$

$- y^{2} + 10 y + 4$ $color(green)(-y^2 + 10y + 4)$

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

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Explanation:

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How do you multiply -3y^2-4y+6+2(y^2+7y-1)−3y2−4y+6+2(y2+7y−1)-3y^2-4y+6+2(y^2+7y-1)?

1 Answer

Explanation:

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