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 $-1, -1 , 1/3 and 5$

How do you list all possible roots and find all factors and zeroes of 3x^4-10x^3-24x^2-6x+53x^4-10x^3-24x^2-6x+5?