My Math Forum Product of 4 factors

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 November 23rd, 2009, 12:57 AM #1 Member   Joined: Oct 2009 Posts: 59 Thanks: 0 Product of 4 factors Express as the product of four factors Is this correct? $a^6 - b^6= (a - b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^3)$
 November 23rd, 2009, 04:33 AM #2 Senior Member   Joined: Oct 2009 Posts: 105 Thanks: 0 Re: Product of 4 factors Try to distribute it out first. It's actually not too complicated, I worked it out in my head in just a few seconds.
November 23rd, 2009, 06:26 AM   #3
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Re: Product of 4 factors

Hello, Tartarus!

Quote:
 Express as the product of four factors. Is this correct? $a^6\,-\,b^6 \:=\: (a\,-\,b)(a^5\,+\,a^4b\,+\,a^3b^2\,+\,a^2b^3\,+\,ab^4\, +\,b^3)$

Obviously, you have only two factors . . .

We have the difference of two squares:

[color=beige]. . [/color]$a^6 \,-\, b^6 \:=\a^3)^2 \,-\, (b^3)^2 \;=\;(a^3\,-\,b^3)(a^3\,+\,b^3)" />

Then we have the difference of two cubes and the sum of two cubes:

[color=beige]. . [/color]$(a\,-\,b)(a^2\,+\,ab\,+\,b^2)(a\,+\,b)(a^2\,-\,ab\,+\,b^2)$

November 23rd, 2009, 12:47 PM   #4
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Re: Product of 4 factors

Quote:
Originally Posted by soroban
[qsize=120]Hello, Tartarus![/size]

Quote:
 Express as the product of four factors. Is this correct? $a^6\,-\,b^6 \:=\: (a\,-\,b)(a^5\,+\,a^4b\,+\,a^3b^2\,+\,a^2b^3\,+\,ab^4\, +\,b^3)$

Obviously, you have only two factors . . .

We have the difference of two squares:

[color=beige]. . [/color]$a^6 \,-\, b^6 \:=\a^3)^2 \,-\, (b^3)^2 \;=\;(a^3\,-\,b^3)(a^3\,+\,b^3)" />

Then we have the difference of two cubes and the sum of two cubes:

[color=beige]. . [/color]$(a\,-\,b)(a^2\,+\,ab\,+\,b^2)(a\,+\,b)(a^2\,-\,ab\,+\,b^2)$

Thanks
I tried another one:

Express as the product of three factors.

$a^4 - b^4$
$= (a^2)^2 - (b^2)^2$
$= (a^2 + b^2) (a^2 - b^2)$
$= (a^2 + b^2) (a - b) (a + b)$

 November 23rd, 2009, 06:35 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 Correct. Can you also do $^{y^8\,-\,16x^8}$?
November 24th, 2009, 01:47 PM   #6
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Re:

Quote:
 Originally Posted by skipjack Correct. Can you also do $^{y^8\,-\,16x^8}$?
Thanks
Yes, but not correctly. How do I know the amount of factors it should have?

$=(y^4)^2 - (4x^4)^2$
$= (y^4 + 4x^4)(y^4 - 4x^4)$

 November 26th, 2009, 11:31 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 Would knowing that help you find the factors?

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