SOLUTION: Factor each polynomial, if the polynomial is prime say so 3x^3y^2-3x^2y^2+3xy^2

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Question 162337This question is from textbook Elementary and Intermediate
: Factor each polynomial, if the polynomial is prime say so
3x^3y^2-3x^2y^2+3xy^2 This question is from textbook Elementary and Intermediate

Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
equation is 3%2Ax%5E3%2Ay%5E2-3%2Ax%5E2%2Ay%5E2%2B3%2Ax%2Ay%5E2
looks like you can factor out 3*x*y^2 to get
3%2Ax%2Ay%5E2%2A%28x%5E2+-+x+%2B+1%29
quadratic formula to find the roots of x%5E2+-+x+%2B+1 is
%28-b+%2B+sqrt%28b%5E2+-+4%2Aa%2Ac%29%29%2F%282%2Aa%29 and %28-b+-+sqrt%28b%5E2+-+4%2Aa%2Ac%29%29%2F%282%2Aa%29
looking at x%5E2+-+x+%2B+1 and using the quadratic formula, it looks like the solution to that quadratic equation is not real since
the part of the equation under the square root sign b%5E2+-+4%2Aa%2Ac is negative as shown below.
in the equation x%5E2+-+x+%2B+1
a = 1
b = -1
c = 1
4*a*c = 4
b^2 = 1
sqrt%28b%5E2+-+4%2Aa%2Ac%29 equals sqrt%281-4%29 = sqrt%28-3%29 which is negative so you don't have a real solution.
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the answer appears to be
3%2Ax%2Ay%5E2%2A%28x%5E2+-+x+%2B+1%29
x%5E2+-+x+%2B+1 appears to be prime because it can't be factored any further.