How do you list all possible roots and find all factors and zeroes of #3x^4-10x^3-24x^2-6x+5#?

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Answer 1 · 2016-07-14T13:14:20

#-1, -1, 1/3 and 5.#

Let #f(x)=3x^4-10x^3-24x^2-6x+5=0#

#f(-x)=3x^4+10x^3-24x^2+6x+5#

Sum of the coefficients = 0. So. -1 is a root of f(x) = 0.

# f'(-x)=12x^3-30x^2-48x-6#

#f'(-x)=-12x^3-30x^2+48x-6#

Here also, the sum of the coefficients = 0.. So, -1 is a double root.

Now #f(x) =(x+1)^2#(a quadratic in x.

In view of the the coefficient 3 of x^4 and the the constant being 5,

#f(x)=(x+1)^2(3x^2+kx+5)#.

Comparing coefficients of x^3, a+6=-19. So, # a= -16#.

Solving the quadratic #3x^2-16x+5=0, x=1/3 and 5#

The list is #-1, -1 , 1/3 and 5#