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Grouping terms to start factoring.

bergausstein

Active member
Jul 30, 2013
191
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!
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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

Active member
Jul 30, 2013
191
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

Active member
Jul 30, 2013
191
$(5y-4z+2x^2)(6y^3-3x^4+2y)$ 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

Administrator
Staff member
Feb 24, 2012
13,775
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

Active member
Jul 30, 2013
191
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?
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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

Active member
Jul 30, 2013
191
this is what i have

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

but how do you determine that ratio thing?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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

Active member
Jul 30, 2013
191
uhm it sounds difficult. but will that method always works in this type of problem.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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

Active member
Jul 30, 2013
191
why do you call it common ratio?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775