Question

3x^3+x^2=48x+16?

solve polynomial equations setting all terms to zero

Answer 1

`x_1=-4`, `x_2=-1/3` and `x_3=4`

Explanation:

`3x^3+x^2=48x+16`

`3x^3-48x+x^2-16=0`

`3x*(x^2-16)+x^2-16=0`

`(3x+1)*(x^2-16)=0`

`(3x+1)(x+4)(x-4)=0`

Hence `x_1=-4`, `x_2=-1/3` and `x_3=4`