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Perhaps something I should have learned during my math classes in high school, But why can't we multiply the exponent in $(x-1)^2$ by each term in the brackets so that we would get $x^2-1^2$ instead of $x^2-2x+1$.

I know that the former is incorrect and the latter is correct, and I also know how we come to the correct answer. I would just like if somebody could explain me why the first one is incorrect what do we exactly do wrong here, as my math teacher just said that the latter was the correct method but why is the first one not.

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    $\begingroup$ everything is just conventions and conventions are created to make life easier. Ask yourself following question: How often do you need to raise the exponents of each term and sum them together instead of multiplying the factors together. $\endgroup$ – achille hui Sep 16 '17 at 16:56
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The reason that the first "method" is incorrect is simply because it gives wrong answers. Exponentiation doesn't work like that. For example you can verify:

\begin{equation} (3-1)^2 = 4 \neq 8 = 3^2 - 1^2 \end{equation}

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  • $\begingroup$ Thanks! I just couldn't think about an example and started over complicating things $\endgroup$ – SOMI Sep 16 '17 at 16:54
  • $\begingroup$ No problem. If you feel that an answer is correct please mark it as accepted. $\endgroup$ – Mohit Sep 16 '17 at 16:55
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since we have $$(x-1)^2=(x-1)\cdot (x-1)=x^2-x-x+1$$

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  • $\begingroup$ Thanks I got that but why can we do $(3^2 + 6^9)^2 = 3^4+6^18$ and why we can't do that for $(x+1)^2$ $\endgroup$ – SOMI Sep 16 '17 at 16:50
  • $\begingroup$ You can't. With a calulator you can verify that $(3^2 + 6^9)^2 \neq 3^4+6^{18}$ $\endgroup$ – Mohit Sep 16 '17 at 16:53
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The fallacy is called the Freshman's dream, so named as it's a trap that most students fall for shortly after being introduced to exponents. (though more so with radicals) Here, we can see that

$$(x-1)^2 \equiv (x-1)(x-1)$$

After some thought, you can realise that, $$(x-1)(x-1) \equiv x(x-1) - 1(x-1)$$

Then using the distributive property,

$$x(x-1) - 1(x-1) \equiv x^2 - x - x + 1$$ $$\equiv x^2 - 2x + 1 \blacksquare$$

The only number that satisfies $(x-1)^2 = x^2 - 1$ is 1 that is,

$$(x-1)^2 = x^2 - 1$$ $$\implies x^2 - 2x + 1 = x^2 - 1$$ $$\implies -2x = -2$$ $$\implies x = 1$$

Many students think that the distributive property that applies to multiplication should also logically apply to exponentiation, but it does not.

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$a^2= a\times a$, so using this we are getting that $(x-1)^2= x^2-2x+1$.

Can you give me a reason why one should take $(x-1)^2= x^2-1^2$?

Instead of asking why former one is incorrect, think why it should be correct?

Can you arrive at your result using proper mathematical steps?

And by example, you can also verify that your result is wrong. As $(2-1)^2 \neq 2^2 - 1^2$.

Hope that helps you.

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