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# Factorising into 3 brackets

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1. 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!
2. (Original post by mayjb)
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.
3. (Original post by mayjb)
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!
For the first one notice that is both parts- this means that everything in that expression is multiplied by , and you can put it outside of brackets like with any other common factor.
Rewrite it as:

and see if you can do it from that.
4. (Original post by mayjb)
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!
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.
5. 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
6. (Original post by RDKGames)
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.
Yes sorry I meant factorize
7. (Original post by HapaxOromenon3)
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.
Ah I see thank you!
8. (Original post by Uni12345678)
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
Thank you, this method was what I saw online but i didn't understand what it meant so thanks!

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