1
$\begingroup$

I noticed that it's possible to factor out the greatest factor from an expression raised to power without having to resolve the power first.

For example,

$(3x^4 +15x^2)^2 = (3x^2)^2\cdot(x^2+5)^2 = 9x^8+90x^6+225x^4$

Which is equivalent to having resolved the power first

$(3x^4 + 15x^2)^2 =(3x^4 + 15x^2) \cdot (3x^4 + 15x^2) = 9x^8+90x^6+225x^4$

In general, it seems that

$(ca + cb)^n = c^n (a+b)^n$

Where $a, b, c$ are monomials and $n$ is a constant

Also, it would seem that if we have two factors both raised to the same power then we can distribute.

$c^n (a+b)^n = (ca + cb)^n$

Where $a, b, c$ are monomials and $n$ is a constant

I'm wondering why this appears to be true as in what properties does this observation follow from?

$\endgroup$
3
  • 2
    $\begingroup$ Are you referring to the property that $(a\cdot b)^n = a^n\cdot b^n$. $\endgroup$
    – amWhy
    Oct 9, 2020 at 20:22
  • 1
    $\begingroup$ I mean, $(ca + cb)^n = \Big(c(a + b)\Big)^n = c^n (a+b)^n$ by the very property I refer to in my first comment. Just tak $x=c, y= (a+b)$, so $(x\cdot y)^n = x^ny^n$. $\endgroup$
    – amWhy
    Oct 9, 2020 at 20:45
  • $\begingroup$ @amWhy Yes, that helps a lot. Thank you. $\endgroup$
    – Slecker
    Oct 10, 2020 at 2:52

1 Answer 1

2
$\begingroup$

This is rule 2 as explained here:

http://www.mclph.umn.edu/mathrefresh/exponents2.html

This is just rearranging the terms being multiplied in a product. For example:

$$(ab)^3 = (ab)*(ab)*(ab)$$

I can rearrange the a's and b's to bring the a's together and b's together. This follows from the associative and commutative properties of multiplication.

$$(ab)*(ab)*(ab) = (a*a*a)*(b*b*b) = a^3b^3$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .