# Thread: Find the roots of the cubic equation

1. ## Find the roots of the cubic equation

What are the roots of the cubic equcation x3-8x-3=0?

Ans: D. x=-2.62, -0.38, 3

1. step
x3-8x-3 = x2+3x-1 By inspection +3 is a root, and x-3 is a factor. Factor out (x-3)
x-3

Please do whole the problem. How did they get x2+3x-1=0

2. ## Re: Find the roots of the cubic equation

It supposed to be $\displaystyle x^2+3x+1$, the division is incorrect...

3. ## Re: Find the roots of the cubic equation

We know that (x-3)( )=x^3-8x+3 We want to find the empty bracket. It must start with x^2 and end in -1
So (x-3)(x^2 -1)=X^3-8x+3. We notice that there is no x^2 term in the answer. With the brackets as we have them so far we would have -3x^2 in the answer. So we still need +3x^2. This we will have if we add +3x to the second bracket.
If you know how to do long division you could get this from (x^3-8x-3) divided by (x-3)
So x-3=0 or x^2+3x+1= 0 which you solve using the quadratic equation formula.

4. ## Re: Find the roots of the cubic equation

$\displaystyle x^3-8x-3=x^3-8x-27+24=(x^3-3^3)-8(x-3)=$

$\displaystyle =(x-3)(x^2+3x+9)-8(x-3)=(x-3)(x^2+3x+1)$

Hence :

$\displaystyle x_1=3$

$\displaystyle x_{2,3}=\frac{-3 \pm \sqrt 5}{2}$

5. ## Re: Find the roots of the cubic equation

Is long division ideal to use on ployimonal that has a missing x^2 or is skips a power such it goes from x^3 to x and misses the x^2