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
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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
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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.
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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:
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:
Rearranging the terms for factoring by grouping we get:
Factoring x from the "red" terms and from the "green" terms we get:
Factoring the "red" expression as a difference of squares we get:
Now we have (y-z) as a common factor of each term. We can factor it out:
Multiplying out the "red" expression we get:
Rearranging the terms for factoring by grouping we get:
Factoring z from the "red" terms and -x from the "green" terms we get:
Now we can factor the (x-y)'s:
With the Associative Property we can remove the extra parentheses giving:
And with the Commutative Property we have:
and we're done!