Question

Find the roots of the polynomial equation.

2x3 + 2x2 – 19x + 20 = 0

Answer 1

1. a, -3,-1,1,3

2. a, 3+i/2, 3-i/2, -4

3. d, 6- sqrt 6

4. a, x3-8x2-11x+148=0

5. d, there are either 2 or 0 positive roots and one negative

Step-by-step explanation:

I did the quick check.

Answer 2

First, find any zero of the polynomial. Since you didn't ask for work, I'll assume it's okay if I use my calculator. Your given polynomial has only one real root which is x=-4.

Now we use the rule that x-a divides the polynomial where a is a zero of said polynomial.

So x+4 divides 2x^3+2x^2-19x+20.
(2x^3+2x^2-19x+20) / (x+4 equals 2x^2-6x+5).

If we factor out a two, we can use the quadratic formula.

2(x^2-3x+2.5) so we have x = (-(-3)+/-(9-4*1*2.5)^(1/2))/2*1)=(3+i)... or (3-i)/2 Where i is the square root of negative one. final answer:
2x^3+2x^2-19x+20=0

then x=-4, (3+i)/2, or (3-i)/2
we have two imaginary number.
I hope it helped you