Find all complex zeros of this Polynomial function. Give exact values. List multiple zeros as necessary. f(x)=3x^3 - 9x^2 - 31x + 5
by 
Anonymous
ANSWERS: 1
-
Go here: http://www.1728.com/cubic.htm and find the answer. Or. I saw in another answer that all the rational solutions are all of the form p/q where p is a factor of 5, the last number and q is a factor of 3, the first number. That leaves few possibilities for p/q: 1,1/3,5/3,5 or -1,-1/3,-5/3,-5 x=+5 works: 3*125-9*25-31*5+5 = 0, so (x-5) is a factor. dividing this out: 3x^2 * (x-5) = 3x^3 - 15x^2 too few x^2, need to add 6x^2 6x * (x-5) = 6x^2 - 30 x need to subtract one more x: -1 * (x-5) = -x + 5 add the two sides of the three equations: (3x^2 + 6x - 1)(x-5) = 3x^3 - 9x^2 - 31 x + 5 (3x^2 + 6x - 1) = 3(x^2 + 2x - (1/3)) = 3((x + 1)^2 - (4/3)) = 3((x + 1 + 2sqrt(3)/3)(x + 1 - 2sqrt(3)/3) the other two roots are x = -1 +/- 2sqrt(3)/3
RELATED QUESTIONS
Copyright 2023, Wired Ivy, LLC