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

##### 1 Answer
Jul 14, 2016

$- 1 , - 1 , \frac{1}{3} \mathmr{and} 5.$

#### Explanation:

Let $f \left(x\right) = 3 {x}^{4} - 10 {x}^{3} - 24 {x}^{2} - 6 x + 5 = 0$

$f \left(- x\right) = 3 {x}^{4} + 10 {x}^{3} - 24 {x}^{2} + 6 x + 5$

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

$f ' \left(- x\right) = 12 {x}^{3} - 30 {x}^{2} - 48 x - 6$

$f ' \left(- x\right) = - 12 {x}^{3} - 30 {x}^{2} + 48 x - 6$

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

Now $f \left(x\right) = {\left(x + 1\right)}^{2}$(a quadratic in x.

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

$f \left(x\right) = {\left(x + 1\right)}^{2} \left(3 {x}^{2} + k x + 5\right)$.

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

Solving the quadratic $3 {x}^{2} - 16 x + 5 = 0 , x = \frac{1}{3} \mathmr{and} 5$

The list is $- 1 , - 1 , \frac{1}{3} \mathmr{and} 5$