How would I solve this question below?

Question

Find the exact solutions to f(x)=0 in the complex number system

 

x^3+x^2+2x -4=0

 

I looked in the back of the book and it says the answers are x=1, x=-1, and x= *plus or minus* ¡?3

¡: imaginary

?: squareroot

 

I'm just not completely sure how they got that answer

Answer 1

x³ + x² + 2x - 4

Descartes rule promises one positive real root and two, or zero, negative real roots.

The Rational root theorem says we should test -4, -2, -1, 1, 2, and 4 as the only possible rational roots. Since we are assured of one positive real root, we should begin with those. Testing 1 first because it is easy and it is obvious that x³ alone overwhelms the lone negative term for any other positive candidate for the root. Using synthetic division

1|  1  1  2  -4
         1  2   4
     1  2   4 |0

the remaining equation is x² + x + 4 = 0

Descartes says two or zero negative real roots. The Rational Root Theorem allows -1, -2, and -4 as candidates (we don't worry about the positive candidates because Descartes says no.)

You can verify that none will work by synthetic division, but you can also skip that step if you notice that x² + 4, which is necessarily positive for real value of x will overwhelm the x term meaning the function will have a value greater than zero for all real values x | x < 0.

 

the Quadratic formula gives

 

x = {-1 ± √[1² - 4(1)(4)]}/2(1)

x = (-1 ± √-15)/2

x = (-1 ± i√15)/2

 

The answer you had was wrong.  (The first clue is that it gave you four roots for a cubic equation.)

Answer 2

First, we need to completely factor f(x).  Just by looking at the function, it looks like we can factor by grouping.  However, we cannot.  So we use a method called synthetic division.  The process divides the coefficients of f(x) by a possible root c.

 

c = ±1, ±2, ±4

 

We try c=1.   Therefore

 

 

 

1 | 1       1         2        -4

 

              1         2         4

_________________________

     1       2         4         0

 

 

Since the last digit under the line is zero, and it represents the remainder of the quotient, we found our factors.

 

f(x) = (x - 1)(x² + 2x + 4)

 

0 = (x - 1)(x² + 2x + 4)

 

Setting  (x - 1) equal to zero give us         x=1.

 

Use the quadratic formula to find the roots from the quadratic factor (x² + 2x + 4):

 

x = (-b ± √(b² - 4ac)) / 2a

 

where:

a = 1

b = 2

c = 4

 

 

So we have

 

x = (-2 ± √(4 - 16)) / 2

 

x = (-2 ± √(-12)) / 2

 

x = (-2 - 2i√3) / 2             and              x = (-2 + 2i√3) / 2

 

x = -1 - i√3                       and              x = -1 + i√3

 

 

You have one real root and two complex roots.