Math Questions (Roots/Polynomials)?

Question

Find the exact roots of the polynomials below:

a. 2x^3 - x² - 3x - 1

b. 2x^3-x²+8x-4

All the help is perfect. THANKS

Answer 1

Hey man, I was a math major in college so bear with me on this explanation:

a. You have to do this by grouping: (2x^3 - 3x) (-x^2 - 1). Take the first polynomial and solve for x.:

2x^2 - 3 = 0 -> 2x^2 = 3 -> x^2 = 3/2 -> x = sqrt3/2 (do not forget the OR there as you need to use 2 roots

Now the other term is (-x^2 - 1). If you dont like that negative 1 in front of the x, simply divide by -1, and you get x^2 + 1 = 0. Now, I assume you learned about imaginary numbers. So, to solve for x here you have x^2 = -1 or i.

b. Similar idea:

A little harder here but here goes: (2x^3 +8x) (-x^2 - 4) becomes: 2x(4x^2 + 4) -1(x^2 + 4), thus becoming solving for 0 in each case:

4x^2 + 4 = 0
4x^2 = -4
x^2 = -1
x = i

2x-1 = 0
2x=1
x=1/2

x^2 + 4 = 0
x^2 = -4
x=2i

Answer 2

a. This has a root at -1/2 , found by inspection.
So divide 2x^3 - x² - 3x - 1 by x+.5 and solve the resulting quadratic equation using the quadratic formula.

b. 2x^3-x²+8x-4
=x^2(2x-1) +4(2x-1)
=(x^2+4)(2x-1)
x = 1/2 , 2i, -2i

Answer 3

use Rational Root Test
possible rational roots are: 1/2,-1/2,1, -1
because you use the last coefficient over first one and just list all the different way the numbers can be matched
useing graphing calculatir may be faster and just plug in all those number

x= -1/2 the only one that works

so


then use
x= -b=+-square root b2 -4ac over 2a(quadratic formula)
then use get two irrational roots
1+-squareroot 5

Answer 4

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Answer 5

do you still need help?

Answer 6

and your question is