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3 years ago

Factor 2x^3 -6x^2 -80x completely

1 answer:

lisabon 2012 [21]3 years ago

4 0

$2x^3 -6x^2 -80x \\ \\ 2x( \frac{2x^3}{2x} + \frac{-6x^2}{2x} + \frac{-80x}{2x} ) \\ \\ 2x(x^2 - 3x - 40) \\ \\ 2x(x - 8)(x + 5) \\ \\$

The answer is: 2x(x - 8)(x + 5).

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Step-by-step explanation:

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3 years ago

Is −4[3(x−7)] and 6(14−2x) equiviant

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3 years ago

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4 years ago

Given sin theta= 6/11 and sec theta < 0, find cos theta and tan theta.

bekas [8.4K]

Answer: option a.

Step-by-step explanation:

By definition, we know that:

$cos^2(\theta)=1-sen^2(\theta)\\\\tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

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