You can put this solution on YOUR website! i need help solving 2x^3+2x^2-19x+20=0
..
You can use the rational roots theorem to find a root, but it is a long process. So instead, I will use a graphing calculator to see if there are any real roots then proceed from there. After you enter the above function into the graphing calculator, you will note it has only one real root,x=-4. The next step is to divide x+4 into 2x^3+2x^2-19x+20 either by long division or synthetic division. The quotient will come out to 2x^2-6x+5. So now you have:
(x+4)(2x^2-6x+5)=0
2x^2-6x+5=0
using following quadratic formula to solve
a=2, b=-6, c=5
x=[-(-6)+-sqrt((-6)^2-4*2*5)]/2*2
x=[6+-sqrt(36-40)]/4
x=(6+-sqrt(-4))/4
x=(6+-√2 i)/4
Ans:Function has 1 real root and two imaginary roots:
x=-4
x=(6+√2 i)/4
x=(6-√2 i)/4
(