How do you simplify (-2a^7)(5a^-2)?

May 4, 2016

$- 10 {a}^{5}$

Explanation:

$\textcolor{red}{\text{How you would normally see the calculation}}$

variants on:

$\textcolor{red}{- \left(2 \times 5\right) \left({a}^{7 - 2}\right) = - 10 {a}^{5}}$
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$\textcolor{g r e e n}{\text{Very detailed solution split into 3 parts}}$

Part 1 $\to$ signs
Part 2 $\to$ numbers
Part 3 $\to$ the letters (variables)
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$\textcolor{b l u e}{\text{Part 1 "-> " signs}}$

Write the signs as -1 and +1$\text{ }$ ( this does work!)

The signs are not the same as each other so

$\left(- 1\right) \times \left(+ 1\right) = \textcolor{red}{\left(- 1\right)}$

$\textcolor{b l u e}{\text{So our answer is a negative value }}$
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$\textcolor{b l u e}{\text{Part 2 "-> " numbers}}$

$2 \times 5 = \textcolor{red}{10}$
$\textcolor{b l u e}{\text{The number part of the answer is 10}}$

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$\textcolor{b l u e}{\text{Part 3 "-> " letters (variables)}}$

We have:$\text{ } {a}^{7} \times {a}^{- 2}$

$\textcolor{b r o w n}{\text{Shortcut method}}$
Because the letters (variables) are the same we can do this:

${a}^{7} \times {a}^{- 2} \text{ "=" "a^(7-2)" } = \textcolor{red}{{a}^{5}}$
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$\textcolor{b r o w n}{\text{First principles method}}$

${a}^{- 2} \text{ is the same thing as } \frac{1}{a} ^ 2$

So::$\text{ "a^7xxa^(-2)" " =" } {a}^{7} \times \frac{1}{a} ^ 2$

Let ${a}^{7}$ be written as ${a}^{5} \times {a}^{2}$

Then we have: ${a}^{5} \times {a}^{2} \times \frac{1}{a} ^ 2$

This is the same as
$\text{ } {a}^{5} \times {a}^{2} / {a}^{2}$

But ${a}^{2} / {a}^{2} = 1$

$\text{ "a^5xxa^2/a^2" "=" } \textcolor{red}{{a}^{5}}$
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$\textcolor{b l u e}{\text{putting it all together}}$

$\textcolor{red}{\left(- 2 {a}^{7}\right) \left(5 {a}^{- 2}\right) = - 10 {a}^{5}}$