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 May 21st, 2018, 11:18 AM #1 Newbie   Joined: May 2018 From: USA Posts: 1 Thanks: 0 Not understanding simple factorization I'm not understanding how this simple factorization works...I know how to solve it, but it is making no sense to me. 3p+3 -------- 3 I know the answer is p+1, but I don't understand how factoring 3p+3 works. My instructions say to rewrite as 3p + 1(3), then factor out the common term: 3(p+1). Why does this work when it's addition? Why wouldn't it be 6p+1? I'm probably missing something really simple but it makes no sense to me and I don't like to move forward with math unless I understand why it works that way. Thank you!
 May 21st, 2018, 11:48 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,098 Thanks: 1905 If you have 3 eggs and 6 slices of bread and wish to share this food equally between three people, you would give 1 egg and 2 slices of bread to each of those people. Thanks from Benit13 and dreamcatcher
 May 21st, 2018, 11:49 AM #3 Math Team   Joined: Jul 2011 From: Texas Posts: 2,805 Thanks: 1449 note the numerator has two terms with a common factor of $3$ ... $3p+3 = \color{red}{3}(p) + \color{red}{3}(1) = \color{red}{3}(p+1)$ ... it's just the reverse of the distributive property of multiplication over addition Thanks from dreamcatcher
May 21st, 2018, 02:11 PM   #4
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Quote:
 Originally Posted by dreamcatcher 3p+3 ------- 3
Well, to "see it", assign a value to p; let's have p=7:
unfactored:
(3p + 3) / 3 = (3*7 + 3) / 3 = (21 + 3) / 3 = 24 / 3 = 8
factored:
3*(p + 1) / 3 = p + 1 = 7 + 1 = 8

 May 22nd, 2018, 02:30 AM #5 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,133 Thanks: 719 Math Focus: Physics, mathematical modelling, numerical and computational solutions Multiplication tells us that we need to add up a number by itself a number of times equal to the other number. So, for example, if we have $\displaystyle 3 \times 4$ then the solution is to add 3 4s, like so $\displaystyle 3 \times 4 = 4 + 4 + 4 = 12$ We also know that $\displaystyle 4 = 1 + 3$ So, if we replace 4 in our original problem with 1 + 3, we get $\displaystyle 3 \times (1 + 3)$ and we need to evaluate this. Using the definition of multiplication, we expect $\displaystyle 3 \times (1 + 3) = 1 + 3 + 1 + 3 + 1 + 3$ We can also add up the items in any order, so let's swap some of them around to get all the similar numbers together in a group: $\displaystyle 3 \times (1 + 3) = 1 + 1 + 1 + 3 + 3 + 3$ Finally, we have 3 1s and 3 3s, so we can use the definition of multiplication to say that this is $\displaystyle 3 \times (1 + 3) = 3 \times 1 + 3 \times 3$ There's a pattern Thanks from greg1313

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