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

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Answer 1 · 2015-07-28T09:47:32

You expand the paranthesis and add or subtract like terms.

Your starting expression is

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

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

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

$2(y^2 + 7y - 1) = 2* y^2 + 2 * 7y + 2 * (-1)=2y^2 + 14y - 2$

Your expression becomes

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

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

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

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

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