SOLUTION: Factor Completely. 6z^3-27z^2+12z I came up with a completely different answer than the options given and need to verify if this is correct. my answer is: 3z(z-4)(2z-1) th

Question 70832This question is from textbook Beginning Algebra
: Factor Completely. 6z^3-27z^2+12z
I came up with a completely different answer than the options given and need to verify if this is correct.
my answer is:
3z(z-4)(2z-1)
the options for this question is:
a)z(6z-1)(z-12)
b)3z(2z-1)(z-4)
c)z(6z+1)(z+12)
d)z(2z-1)(z-4)
Thanks verifying, I appreciate it. This question is from textbook Beginning Algebra

Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Your answer is the same as answer b. And it is correct.
.
I suspect that the confusion came because you didn't recognize that they were identical
because the two terms in parentheses appear in reverse order. That doesn't make a bit of
difference in multiplication. No matter what order they are in, if the three terms are there,
the product will be the same. [You could have had the 3z as the last multiplier and it
wouldn't have made your answer different from answer b.]
.
That was a good job of factoring, by the way.
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