Asked by julez
Solve the following equation in the real number system. Please show all of your work.
2x^3-3x^2-14x+15=0
2x^3-3x^2-14x+15=0
Answers
Answered by
Damon
I can see right off that x = 1 works
2-3-14+15 = 0 sure enough
so divide by (x-1)--> 2x^2-x-1
(2x+1)(x-1)
so x = 1 again and x = -1/2
2-3-14+15 = 0 sure enough
so divide by (x-1)--> 2x^2-x-1
(2x+1)(x-1)
so x = 1 again and x = -1/2
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