(Another) simple maths question?

• (Another) simple maths question?
  Posted by Things on 16 Apr, 2013 08:03
• Exam next week, boo
  
  But working through some problems, I've come across something I'm really struggling to get my head around. I've looked back in my notes and can't really find where I covered this stuff in lectures.
  
  Basically, I'm trying to factorize (x+3)²+2(x+3), and supposedly meant to end up with (x+3)(x+5).
  
  The first thing I tried was expanding it, so I got:
  
  x² + 9 + 2x + 6, which can be simplified to x² + 2x + 15, a quadratic.
  
  Thus, (x+3)(x+5) makes sense, as 3x5 = 15, but isn't the sum of the 2 numbers (5 + 3) meant to add to "2x"?
  
  Probably something stupid I've done again - I'm good at that
  
  Dan
  
• #1 Reply
  Posted by Psi on 16 Apr, 2013 08:08
• The problem is in the expansion, is should be  x^2 + 8x + 15
  
  (Yeah, i cheated and expanded it with an online calculator. Sadly it doesn't show the working)
• #2 Reply
  Posted by Psi on 16 Apr, 2013 08:36
• Ok
  
  I think i have figured out the working for expanding it
  
  (x+3)²   +   2(x+3)=0
  
  
  Expand first part  (x + 3)²
  Apply rule (a+b)² = a² + 2ab + b²
  x² + (2* (x * 3)) + 3²
  x² + (2* (3x))  + 9
  x² + 6x + 9
  
  
  Expand second part   2(x+3)
  2x + 6
  
  Adding both parts together
  x² + (6x+2x) + (9+6)
  x² + 8x + 15
• #3 Reply
  Posted by Balaur on 16 Apr, 2013 08:40
• (x+3)^2 + 2 (x+3) = (x+3)(x+3) + 2(x+3) = (x+3) ((x+3)+2)) = (x+3)(x+5)
• #4 Reply
  Posted by Things on 16 Apr, 2013 09:20
• Quote from: Psi on 16 Apr, 2013 08:36

  Ok
  
  I think i have figured out the working for expanding it
  
  (x+3)²   +   2(x+3)=0
  
  
  Expand first part  (x + 3)²
  Apply rule (a+b)² = a² + 2ab + b²
  x² + (2* (x * 3)) + 3²
  x² + (2* (3x))  + 9
  x² + 6x + 9
  
  
  Expand second part   2(x+3)
  2x + 6
  
  Adding both parts together
  x² + (6x+2x) + (9+6)
  x² + 8x + 15
  

  
  Ah, of course! Thanks guys