Help Me Factorise: 3(n+r)(n+r-1) + (n+r) to (n+r)(3n+3r-2)

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To factor the expression 3(n+r)(n+r-1) + (n+r) into (n+r)(3n+3r-2), start by substituting n+r with x, simplifying to 3x(x-1) + x. Factor out x to get x(3(x-1) + 1). Expand 3(x-1) to collect like terms, leading to x(3x - 3 + 1). Finally, substituting back n+r for x results in the desired factorization.
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Can someone please help me with very simple math.. and explain slowly how they got from
3(n+r)(n+r-1) + (n+r)
to (n+r)(3n+3r-2)?

I forgot how to factorise, so can someone help me do this please?
 
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The n+r is already factored out. Just write everything in terms of that and you'll see.

(It's actually easier to understand by doing, in this case, than to have it explained!)
 
Explain slowly you say? Maybe that can be expressed through example.

ab+ac=a(b+c)
xy+x=x(y+1)
pqr+r=r(pq+1)

If you understand how to do the opposite of factorising which is expanding, then these results would be easy to confirm.

So for 3(n+r)(n+r-1) + (n+r)
let n+r=x
So we now have 3x(x-1)+x
Factorising x out: x(3(x-1)+1)
Expand 3(x-1) and collect like terms, then substitute n+r=x back into the result.
 
Oh wow i get it now. Thank you!
 
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