factoring an expression

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yet2spark

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\(\displaystyle x=\dfrac{(a+2b+2c+d)a}{d-a}\)

It is known that x is an integer.

From this I think that the division must have remainder 0

I tried doing long division with this but am hopeless at it :(

I didn't know what to do with the 'a' factor in the numerator so multiplied it out but it didn't help anything.

Can anyone give me a hand please
 
yet2spark said:
\(\displaystyle x=\dfrac{(a+2b+2c+d)a}{d-a}\)

It is known that x is an integer.

From this I think that the division must have remainder 0

I tried doing long division with this but am hopeless at it :(

I didn't know what to do with the 'a' factor in the numerator so multiplied it out but it didn't help anything.

Can anyone give me a hand please
Click to expand...

Your expression is already in 'factored' form. What was the original question?
 
It's not a homework question, it's from a project I'm working on.

I was trying to use long division to divide the (d-a) into the numerator, knowing that it would leave a remainder because of the b and c but also knowing that this remainder could be equated to zero, which would be useful.

i.e. say the remainder was a+2b-c (I know it isn't) then I could say that a+2b=c
 
yet2spark said:
It's not a homework question, it's from a project I'm working on.

I was trying to use long division to divide the (d-a) into the numerator, knowing that it would leave a remainder because of the b and c but also knowing that this remainder could be equated to zero, which would be useful.

i.e. say the remainder was a+2b-c (I know it isn't) then I could say that a+2b=c
Click to expand...

Two of those conditions would be:

d = 0

or

a +2b + 2c + d = 0
 
How so?

Valid values for the equation are

a=13, b=6, c=9, d=20

we then have

\(\displaystyle x=\dfrac{(a+2b+2c+d)a}{d-a}\)

\(\displaystyle x=\dfrac{(13+2*6+2*9+20)(13)}{20-13}\)

which gives x=117
 
JeffM said:
... a computer program that solves this by brute force may be the simplest solution.
Click to expand...

Thanks Jeff and Denis. I already have brute-force at work, I was just hoping I had missed something in this equation. Oh well, back to letting the poot do what poots do best, number-crunching. Although I very much doubt brute-force will come up with a solution.
 
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