SOLUTION: Dear Tutor, I would like to know how can I arrive/make: y*z*(z-y)+x*y*(y-x)+x*z*(x-z) into (x-y)*(y-z)*(z-x) I have tried multiplying all factors also tried

Algebra ->  Equations -> SOLUTION: Dear Tutor, I would like to know how can I arrive/make: y*z*(z-y)+x*y*(y-x)+x*z*(x-z) into (x-y)*(y-z)*(z-x) I have tried multiplying all factors also tried      Log On


   



Question 212101: Dear Tutor,

I would like to know how can I arrive/make:
y*z*(z-y)+x*y*(y-x)+x*z*(x-z)

into

(x-y)*(y-z)*(z-x)

I have tried multiplying all factors also tried to divide by factor:y and than factor:z but that's where I got stuck so I'm not sure if I'm even close to the right method.

Thank you for your time and effort,

Gabi

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
y%2Az%2A%28z-y%29%2Bx%2Ay%2A%28y-x%29%2Bx%2Az%2A%28x-z%29
This is a tough one. For one, you have to be good at factoring by grouping. Another key is recognizing that:
z-y = -1(y-z)
y-x = -1(x-y)
x-z = -1(z-x)
Substituting the first of these into the original we get:
-y%2Az%2A%28y-z%29%2Bx%2Ay%2A%28y-x%29%2Bx%2Az%2A%28x-z%29
Now we have one of the desired factors, (y-z), present. The rest is manipulations with the goal of having (y-z) as a common factor. First we'll multiply out the reset of the expression:
-y%2Az%2A%28y-z%29%2Bx%2Ay%5E2+-+x%5E2%2Ay+%2B+x%5E2%2Az+-+xz%5E2
Rearranging the terms for factoring by grouping we get:
-y%2Az%2A%28y-z%29%2B+red%28x%2Ay%5E2+-+xz%5E2%29+%2B+green%28-x%5E2%2Ay+%2B+x%5E2%2Az%29
Factoring x from the "red" terms and -x%5E2 from the "green" terms we get:
-y%2Az%2A%28y-z%29%2B+x%2Ared%28y%5E2+-+z%5E2%29+%2B+%28-x%5E2%29%2A%28y+-+z%29
Factoring the "red" expression as a difference of squares we get:
-y%2Az%2A%28y-z%29+%2B+x%2A%28y%2Bz%29%28y-z%29+%2B+%28-x%5E2%29%2A%28y+-+z%29
Now we have (y-z) as a common factor of each term. We can factor it out:
%28y-z%29%28-y%2Az+%2B+red%28x%2A%28y%2Bz%29%29+-+x%5E2%29
Multiplying out the "red" expression we get:
%28y-z%29%28-y%2Az+%2B+xy+%2B+xz+-+x%5E2%29
Rearranging the terms for factoring by grouping we get:
%28y-z%29%28red%28xz+-+y%2Az%29+%2B+green%28-x%5E2+%2B+xy%29%29
Factoring z from the "red" terms and -x from the "green" terms we get:
%28y-z%29%28z%28x+-+y%29+%2B+%28-x%29%28x+-+y%29%29
Now we can factor the (x-y)'s:
%28y-z%29%28%28x-y%29%28z-x%29%29
With the Associative Property we can remove the extra parentheses giving:
%28y-z%29%28x-y%29%28z-x%29
And with the Commutative Property we have:
%28x-y%29%28y-z%29%28z-x%29
and we're done!