Factorising into 3 brackets

I have these two questions:

x(x+1)(2x+7) + 6(x+1)(x+3)

and

N^2(N+1)^2 + 4(N+1) ^3

and just having started C1 am unsure how to solve them. The answers say they should become three brackets but I don't know how to form them without expanding everything, then I don't know where to go from there. Thanks for any help!

I have these two questions:

x(x+1)(2x+7) + 6(x+1)(x+3)

and

N^2(N+1)^2 + 4(N+1) ^3

and just having started C1 am unsure how to solve them. The answers say they should become three brackets but I don't know how to form them without expanding everything, then I don't know where to go from there. Thanks for any help!

I have these two questions:

x(x+1)(2x+7) + 6(x+1)(x+3)

and

N^2(N+1)^2 + 4(N+1) ^3

and just having started C1 am unsure how to solve them. The answers say they should become three brackets but I don't know how to form them without expanding everything, then I don't know where to go from there. Thanks for any help!

You can't solve them because they're not equations, that's not what you're doing most likely. Look to get them into a form where they're products of two factors. This means factor out common terms out of the whole expression.

In the first expression, there is a common factor of x+1, so take it out to give:
(x+1)(x(2x+7) + 6(x+3)) = (x+1)(2x^2+13x+18) = (x+1)(x+2)(2x+9).

In the second expression, there is a common factor of (N+1)^2, so take it out to give:
(N+1)^2(N^2 + 4(N+1)) = (N+1)^2(N^2+4N+4) = (N+1)^2(N+2)^2.

Yes expand everything, then put in values of x between -3 and 3 (call this a) and if you find one that makes the equation equal zero, then (x-a) is a factor. Taking one factor out means you will be left with just a quadratic to factorise and that should leave three brackets! If you need any help I can do the working and put it on here if you wish