Respuesta :
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
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
Answer:
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.