(Another) simple maths question?

[Things]

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

[Psi]

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)

[Psi]

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

[Balaur]

(x+3)^2 + 2 (x+3) = (x+3)(x+3) + 2(x+3) = (x+3) ((x+3)+2)) = (x+3)(x+5)

[Things]

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