How to calculate LCM for rational equations ?

[kupid]

To calculate an LCM for a rational function, follow these steps:
1. Factor all denominator polynomials completely.
2. Make a list that contains one copy of each factor, all multiplied together.
3. The power of each factor in that list should be the highest power that factor is
raised to in any denominator.
4. The list of factors and powers you generated is the LCM.

View attachment 6872

4x = 2.2.x , x² = x.x , 2x² =2.x.x

So LCM = 2.2.x.x = 4x²

I still don't understand this clearly ...

 

[I like Serena]

Hey kupid,

Let's run through the steps.

To calculate an LCM for a rational function, follow these steps:
1. Factor all denominator polynomials completely.

$4x=2^2\cdot x,\quad x^2,\quad 2x^2 = 2\cdot x^2$

2. Make a list that contains one copy of each factor, all multiplied together.

$2\cdot x$

3. The power of each factor in that list should be the highest power that factor is
raised to in any denominator.

$2^2\cdot x^2$

4. The list of factors and powers you generated is the LCM.

$LCM=2^2\cdot x^2 = 4x^2$

 

[kupid]

Is there any mistake in the above post , because i am a bit confused .

To calculate an LCM
for a rational function, follow these steps:
1. Factor all denominator polynomials completely.

4x = 2.2.x = 2².x , x² = x.x , 2x² =2.x.x

2. Make a list that contains one copy of each factor from the "pairs of factors ", all multiplied together.

2.x

3. The power of each factor in that list should be the highest power that factor is
raised to in any denominator.

2².x²

4. The list of factors and powers you generated is the LCM

LCM = 2².x² = 4x²

 

[I like Serena]

Is there any mistake in the above post , because i am a bit confused .

No mistake. That's fine. (Nod)

It's just that when we factor completely, there's little point in writing out a power as a product.
We can keep it as a power.
So instead of writing $7x^8=7\cdot x\cdot x\cdot x\cdot x\cdot x\cdot x\cdot x\cdot x$, we can just leave it as $7x^8$.

 

[kupid]

OK , Thanks a lot .