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I am supposed to find the answer of: $x^2-2x-3$ when $x=3$. Can someone please help me? I know the answer is supposed to be 0, but I can not figure out why, or how you do it.

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    $\begingroup$ Do you mean, that you should calculate $3^2-2\cdot 3-3$? That is clearly zero... $\endgroup$ – mickep Sep 8 '15 at 18:36
  • $\begingroup$ Where do you get stuck? $\endgroup$ – lulu Sep 8 '15 at 18:36
  • $\begingroup$ Either you are asking a not so useful question or because there is calculus tag i may say your question is unclear $\endgroup$ – frunkad Sep 8 '15 at 18:41
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Calculating $x^2-2x-3$ when $x = 3$ simply means that you replace every instance of $x$ with $3$ in the first expression and evaluate it, i.e.

\begin{align} \color{red}{x}^2 - 2\color{red}{x} - 3 &= \color{red}{3}^2 -2\cdot \color{red}{3} - 3 \\&= 9 - 6 - 3 \\&= 0. \end{align}

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Wherever you see an $x$, put a $3$: \begin{align} x^2 - 2x - 3 &=3^2 - 2(3) - 3 \\ &=9 - 6 - 3 \\ &=0 \end{align}

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$$ x^{2}-2x-3$$ Then plug in $$x=3$$ for every x , so we have

$$ \left(3\right)^{2}-2\left(3\right)-3=9-6-3=3-3=0 $$

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Another way of doing it is by factorizing the expression.

$x^{2}-2x-3= (x-3)(x+1)$.

Putting $x=3$ in Right hand side of the above expression will give $0$. Thus, the answer is $0$.

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