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

  • Thread starter Thread starter confused88
  • Start date Start date
AI Thread Summary
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.
confused88
Messages
22
Reaction score
0
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?
 
Mathematics news on Phys.org
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!
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Back
Top