How do you multiply $2 y (y - 4)$ $2y(y-4)$ ?

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How do you multiply $2 y (y - 4)$ $2y(y-4)$ ?

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Answer 1 · 2017-04-28T05:26:37

$2 y^{2} - 8 y$ $2y^2 - 8y$

Explanation:

Use distribution $a (b + c) = a b + a c$ $a(b + c) = ab + ac$ and the exponent rules $x = x^{1} and x^{m} \cdot x^{n} = x^{m + n}$ $x = x^1 " and " x^m * x^n = x^(m + n)$ :

Given: $2 y (y - 4)$ $2y (y - 4)$

Distribute: $2 y (y - 4) = 2 y y + 2 y (- 4)$ $2y (y - 4) = 2y y + 2y(-4)$

Simplify using multiplication and the exponent rules:

$= 2 y^{2} - 8 y$ $= 2y^2 -8y$

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How do you multiply $2 y (y - 4)$ $2y(y-4)$ ?

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How do you multiply $2 y (y - 4)$ $2y(y-4)$ ?