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Will someone show me the steps to expanding $(\tan x - \sqrt{3})(3\tan x + \sqrt{3})$.

I have tried FOIL, but I get the wrong answer.

I keep getting $-3\sqrt{3}\tan x$ instead of $-2\sqrt{3}\tan x$.

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    Then post your steps, so that we know at which step you get it wrong. – 2017-01-23

3 Answers 3

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$$\begin{align}(\tan x - \sqrt{3})(3\tan x + \sqrt{3})&=\tan x\cdot3\tan x-\sqrt3\cdot3\tan x+\tan x\cdot\sqrt3-\sqrt3\cdot\sqrt3\\&=3\tan^2x-3\sqrt3\tan x+\sqrt3\tan x-3\\&=3\tan^2x-2\sqrt3\tan x-3\end{align}$$

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    an $x$ is missing – 2017-01-23
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    What happens to the second $3$ in the problem in the second phase? – 2017-01-23
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    @WilliamZlacki $-3\sqrt3\tan x+\sqrt3\tan x=-2\sqrt3\tan x$. Is this what you meant? – 2017-01-23
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    Oh I left it out when I did the addition, sorry. But yes, you're right, thanks. – 2017-01-23
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    @GoodDeeds, Looking back, now I seem, too, have misplaced my steps so called once or twice now I need help one more time on this since i didn't type it in up there. – 2017-06-19
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    @WilliamZlacki I couldn't understand that comment, sorry. Could you clarify please? – 2017-06-24
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    @GoodDeeds Nvm, figured it out. – 2017-06-24
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$$(tanx -\sqrt {3})(3tanx +\sqrt {3})$$ $$3tan^2 x+\sqrt {3} tanx -3\sqrt {3} tanx -3$$ $$3tan^2 x-2\sqrt {3} tanx-3$$ This is the expanded form..

.....--))

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it is $$3\tan(x)^2-3\sqrt{3}\tan(x)+\sqrt{3}\tan(x)-3=3\tan(x)^2-2\sqrt{3}\tan(x)-3$$

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    $$\sqrt{3}\tan(x)-3\sqrt{3}\tan(x)=-2\sqrt{3}\tan(x)$$ – 2017-01-23