Solve the equation 3x(x^2 + 6) = 8 – 17x^2?

x = -2
x = -4
x = 1/3

3x^3 + 18x = 8 - 17x^2
3x³ + 17x^2 + 18x - 8 = 0
x = -2 is ONE solution
Find other solutions by synthetic division.

-2 | 3_____ 17_____18_____-8
__ |______ - 6_____-22_____ 8
__ | 3_____ 11_____ -4_____0

(x + 2)(3x² + 11x - 4) = 0
(x + 2)(3x - 1)(x + 4) = 0
x = - 2 , x = 1/3 , x = - 4

First, use the distributive property.
3x^3+18x=8-17x^2
Move all onto one side
3x^3+17x^2+18x-8=0
Use the factor theorem, which states that
(x-b) is a factor of P(x) if P(b)=0
(which means, when you put b into your equation, it makes the equation equal 0)
So you try numbers (which are factors of the last number and the first number) in your equation, and find -2 is a factor.
Here are some possible factors for you equation
1, -1, 2, -2, 4,-4, 8, -8
So you divide into the polynomial, with x=-2 (which is x+2=0)
(x+2) into 3x^3+17x^2+18x-8
And get 3x^2+11x-4
That means (3x^2 +11x-4)(x+2)=3x^3+17x^2+18x-8
Factor 3x^2+11x-4 with quadratic formula or other method.
you get x=1/3
x=-4
and from before, x=-2

so, (x+4)(x-1/3)(x+2)=3x^3+17x^2+18x-8

Hope this helps!

3x^3+18x=8-17x^2
3x^3+17x^2+18x-8=0

Use the trial and error method, when x=-2, you should get 0.
So x+2 is a root

Now apply the long division method: 3x^3+17x^2+18x-8 divide x+2
(x+2)(3x^2+11x-4)
=(x+2)(3x-1)(x+4)

Therefore x=-2, x=1/3, x= -4
Hope it helps and good luck.