Question

M в квадрате(а+2)+(а+2) 2)p(a-5)-(5-a)

Answer 1

$1) m^2(a+2)+(a+2)=(a+2)(m^2+1) \\ \\ 2)p(a-5)-(5-a)=p(a-5)+(a-5)=(a-5)(p+1)$

Answer 2

To simplify the given expressions, let's take them one by one:

1. M^2(a + 2) + (a + 2)^2:

   First, we can factor out the common factor (a + 2):

   (a + 2)(M^2 + (a + 2))

2. p(a - 5) - (5 - a):

   Now, let's simplify the second expression:

   p(a - 5) - (5 - a)

   Distribute the negative sign inside the second parentheses:

   p(a - 5) - 5 + a

   Rearrange terms:

   a + a + p(a - 5) - 5

   Combine like terms:

   2a + p(a - 5) - 5

So, the simplified expressions are:

1. (a + 2)(M^2 + (a + 2))

2. 2a + p(a - 5) - 5