help: factoring trinomials that =0

I

iliked0g5

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May 22, 2021
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Hello and first i want say thank you for taking your time to help me.
I am just starting and I don't remember how to start or how to do this

here is the question
2x^2+6x2x2+6x = 0
 
iliked0g5 said:
Hello and first i want say thank you for taking your time to help me.
I am just starting and I don't remember how to start or how to do this

here is the question
2x^2+6x2x2+6x = 0
Click to expand...

what, exactly, is that middle term …

6x2x2 ???
 
@Subhotosh Khan if i am wrong please tell me. I believe what you are saying is that the 2 and 3 make the 6 by multiplying. so here's what i think you did
(2x+2)+(x+3)
 
2*x^2 + 6*x = 2*x*x + 3*2*x

Look at the two terms (separated by the + sign). The common factors are 2 and x.

So, factorising them outside the brackets gives:

2*x*x + 3*2*x = 2*x*(x + 3)
 
iliked0g5 said:
@Subhotosh Khan if i am wrong please tell me. I believe what you are saying is that the 2 and 3 make the 6 by multiplying. so here's what i think you did
(2x+2)+(x+3)
Click to expand...
No, that's not what he did because it doesn't work: (2x+ 2)+ (x+ 3)= (2x+x)+ (2+ 3)= 3x+ 5, not \(\displaystyle 2x^2+ 6X\)! If you meant (2x+ 2)(x+ 3), multiplying instead of adding? That's also wrong: \(\displaystyle (2x+ 2)(x+ 3)= 2x(x+ 3)+ 2(x+ 3)= 2x^2+ 6x+ 2x+ 6= 2x^2+ 8x+ 6\).

I can't speak for Subhotosh Khan's thought processes but I suspect he saw that there was an "x" both terms so he could factor out an "x". And he saw that both "2" and "6" are even numbers so he could factor out a "2". \(\displaystyle 2x^2+ 6x= x(2x+ 6)= 2x(x+ 3) \)
 
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