anonymous
  • anonymous
factor (p+3q)-r(3q+p)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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terenzreignz
  • terenzreignz
A little (just a little) tricky, but you only need to remember two things: FIRST a + b = b + a And SECOND ac + bc = c(a + b)
terenzreignz
  • terenzreignz
So, the r(3q+p) may be written instead as r(p+3q) Turning this expression into (p+3q) - r(p+3q)
anonymous
  • anonymous
so now i have (p+3q)-rp-3qr ?

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terenzreignz
  • terenzreignz
No. While that is technically correct, you're being asked to factor, not distribute. Distributing is in fact a step backwards when it comes to factoring... Just remember ac + bc ----> c(a + b)
anonymous
  • anonymous
so what do i do ?
terenzreignz
  • terenzreignz
Do you not see anything that looks like this? ac + bc ? Maybbe ac - bc ? It works on the same principle ac - bc = c(a - b)
terenzreignz
  • terenzreignz
And you can be quite flexible when it comes to your choice of "c"
anonymous
  • anonymous
can you tell me the setup ? then ill solve it
terenzreignz
  • terenzreignz
Okay, why don't we let c = (p + 3q) So now, we have c - rc Does that look more agreeable to you? :)
anonymous
  • anonymous
do you know what the answer is ?
terenzreignz
  • terenzreignz
No ;)
anonymous
  • anonymous
can you just set up the problem for me ?
terenzreignz
  • terenzreignz
I am... We let c = (p + 3q) So that the expression becomes c - rc so use that second concept I told you, the one involving distribution...

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