Solving Polynomial Equations. Solve 64x^3+8=0.

Question

Please help.

Answer 1

It appears that Roman C made an error in

factoring the polynomial:

 

64x³ + 8 = 0

64x³ = -8

x³ = -8/64

x³ = -1/8

x = -1/2

This is the real root

 

 

Since x=-1/2 is a root then (x+1/2) is a factor

 

64x³ + 8 = (x+1/2)(___??____) = 0

 

Using synthetic division I can find the missing factor

 

-1/2 |  64   0    0   8

       |       -32  16 -8

       --------------------

          64  -32  16  0

 

The polynomial we get in factoring out (x+1/2) is

 

64x³ + 8 = (x+1/2)(64x² - 32x + 16)

 

Note that we can factor 16 out of each term in the 2nd polymomial

 

(16)(x+1/2)(4x² - 2x + 1) = 0

Divide both sides by 16

 

(x+1/2)(4x² - 2x + 1) = 0

We already know that x=-1/2 is a root.

We need to find the roots of 4x²-2x+1=0

 

Using quadratic equation

x =[ -b ±√(b²-4ac)]/2a    a=4, b=-2, c=1

 

x = [2 ±√[(-2)²-4(4)(1)]]/2(4)

x = (2±√(4-16))/8

x = (2±√-12)/8

x = (2±2i√3)/8

 

x = (1 ± i√3)/4

 

x = 1/4 ± (√3/4)i

 

Due to the dropping of the negative sign on his factoring,

Roman C. had -1/4 instead of 1/4 in the final answer

 

 

Answer 2

64x³ + 8 = 0

 

8x³ + 1 = 0

 

(2x + 1)(4x² + 2x + 1) = 0

 

Real root: x = -1/2

 

Complex roots:

 

4x² + 2x + 1 = 0

 

x = (-2 ± √(2² - 4·4·1))/(2·4)

 

= (-2 ± √(-12))/(8)

 

= (-1 ± √(-3))/(4)

 

= -1/4 ± (√3 / 4) i