SOLUTION: Factor w^6-729, Type N if the binomial is not factorable
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Question 324028: Factor w^6-729, Type N if the binomial is not factorable
Found 2 solutions by Fombitz, CharlesG2:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website!
Factor w^6-729, Type N if the binomial is not factorable
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see if it fits x^2 - a^2 which factors into (x + a)(x - a), the ax's cancel out
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sqrt(729) = 27
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w^6 - 729 = (w^3 + 27)(w^3 - 27)
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w^3 - 27 = (w - 3)(w^2 + 3w + 9)
check by distributing: w^3 + 3w^2 + 9w - 3w^2 - 9w - 27 = w^3 - 27
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w^6 - 729 = (w^3 + 27)(w - 3)(w^2 + 3w + 9)
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w^2 + 3w + 9 can not be factored, well not without going into complex numbers (those numbers that are in a+bi form, where i=sqrt(-1))
check by using discriminant b^2 - 4ac, where b=3 and a=1 and c=9
3^2 - 4*1*9 = 9 - 36 = -27
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w^3 + 27, see if there is a factor by dividing w + 3 into w^3 + 27
w^2 - 3w - 9
w + 3 --> w^3 + 0w^2 + 0w + 27
w^3 + 3w^2
- 3w^2 + 27
- 3w^2 - 9w
- 9w + 27
- 9w - 27
0
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w^6 - 729 = (w + 3)(w^2 - 3w - 9)(w - 3)(w^2 + 3w + 9)
the discriminant on w^2 - 3w - 9 = (-3)^2 - 4(1)(-9) = 9 - (-36) = 9 + 36 = 45
w12 = (-(-3) +- sqrt(45))/2 = (3 +- 3sqrt(5))/2
w1 = (3 + 3sqrt(5))/2
w2 = (3 - 3sqrt(5))/2
(w + (-3 - 3sqrt(5))/2)(w + (-3 + 3sqrt(5))/2)
w^2 + w(1/2)(-3 + 3sqrt(5)) + w(1/2)(-3 - 3sqrt(5) + (-3 - 3sqrt(5)(-3 + 3sqrt(5)(1/2)(1/2)
w^2 + (w/2)(-3 + 3sqrt(5)) + (w/2)(-3 - 3sqrt(5)) + (1/4)(9 - 45)
w^2 - (3/2)w + (3/2)sqrt(5)w - 3/2w - (3/2)sqrt(5)w - 36/4
w^2 - 3w - 9
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w^6 - 729 = (w + 3)(w^2 - 3w - 9)(w - 3)(w^2 + 3w + 9)
w^6 - 729 = (w + 3)(w + (-3 - 3sqrt(5))/2)(w + (-3 + 3sqrt(5))/2)(w - 3)(w^2 + 3w + 9)
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