Grouping terms to start factoring.

[bergausstein]

how would i go about properly grouping terms to start my factoring?

1. $12x^2y^3-24y^3z-6x^6+30y^4-15x^4y+4x^2y+12x^4z+10y^2-8yz$

thanks!

 

[MarkFL]

I suspect there is a small typo in the problem, because it almost factors, but not quite. I can get it to:

\(\displaystyle (5y-4z)\left(2y\left(3y^2+1 \right)-3x^4 \right)+2x^2\left(2y\left(3y+1 \right)-3x^4 \right)\)

 

[bergausstein]

I suspect there is a small typo in the problem, because it almost factors, but not quite. I can get it to:

\(\displaystyle (5y-4z)\left(2y\left(3y^2+1 \right)-3x^4 \right)+2x^2\left(2y\left(3y+1 \right)-3x^4 \right)\)

yes, the first term is $12x^2y^3$

 

[bergausstein]

$(5y-4z+2x^2)(6y^3-3x^4+2y)$ this should be the completely factored form.

but how did you quickly determine the proper grouping is this $\displaystyle (5y-4z)\left(2y\left(3y^2+1 \right)-3x^4 \right)+2x^2\left(2y\left(3y+1 \right)-3x^4 \right)$

can you show me the steps how did you arrive at this.

 

[MarkFL]

yes, the first term is $12x^2y^3$

Okay, so we have:

\(\displaystyle 12x^2y^3-24y^3z-6x^6+30y^4-15x^4y+4x^2y+12x^4z+10y^2-8yz\)

If we observe that we have 3 groups of 3 coefficients in the ratio $2:5:-4$, we can group as follows:

\(\displaystyle \left(12x^2y^3+30y^4-24y^3z \right)+\left(4x^2y+10y^2-8yz \right)+\left(-6x^6-15x^4y+12x^4z \right)\)

Next, pull the greater factor common to each term in each group:

\(\displaystyle 6y^3\left(2x^2+5y-4z \right)+2y\left(2x^2+5y-4z \right)-3x^4\left(2x^2+5y-4z \right)\)

And now we see we have a factor common to all 3 groups:

\(\displaystyle \left(2x^2+5y-4z \right)\left(6y^3+2y-3x^4 \right)\)

 

[bergausstein]

okay, i have another problem similar to problem above.

$4xy+32x^3-12x^2+y^3+8x^2y^2-3xy^2-3xy^2-2x^2y-16x^4+6x^3$

following your method i got

$(4xy+32x^3-12x^2)+(8x^2y^2-16x^4+6x^3)+(y^3-3xy^2-2x^2y)$

$4x(y+8x^2-3x)+2x^2(4y^2-8x^2+3x)+y(y^2-3xy-2x^2)$

i can't see a common factor to each term. can you tell why is that?

 

[MarkFL]

Okay, we have:

\(\displaystyle 4xy+32x^3-12x^2+y^3+8x^2y^2-3xy^2-2x^2y-16x^4+6x^3\)

I see a common ratio of $-2:4:1$:

\(\displaystyle \left(-2x^2y+4xy+y^3 \right)-\left(-6x^3+12x^2+3xy^2\right)+\left(-16x^4+32x^3+8x^2y^2 \right)\)

I will let you continue...

 

[bergausstein]

this is what i have

$(4x+y^2-2x^2)(y+8x^2-3x)$

but how do you determine that ratio thing?

 

[MarkFL]

this is what i have

$(4x+y^2-2x^2)(y+8x^2-3x)$

but how do you determine that ratio thing?

Looks good.

You just have to look for it...the more experience you have, the better you will be at more quickly spotting it, it it exists.

 

[bergausstein]

uhm it sounds difficult. but will that method always works in this type of problem.

 

[MarkFL]

uhm it sounds difficult. but will that method always works in this type of problem.

If it is factorable it will work. I would hesitate to say there is an "always will work" method when it comes to factoring. Factoring is a skill that requires a great deal of trial and error at times, but as you develop experience, you become better at it.

Since there were 9 terms in the polynomial, then we should look at making 3 groups of 3 terms. Then you look for a common ratio among the numeric coefficients. I noticed a $4:1:-2$ ratio, but as we see from the factored form, there is also $1:8:-3$ ratio too.

This worked for this polynomial. Other polynomials might require other techniques. :D

 

[bergausstein]

why do you call it common ratio?

 

[MarkFL]

why do you call it common ratio?

Because it is a ratio common to all groups.