find all real and complex roots of the following polynomials: Hint - use Rational Root Theorum to identify potential roots.

1) 

Solve for x:
x^3-1 = 0

 

Add 1 to both sides:
x^3 = 1

 

Taking cube roots gives 1 times the third roots of unity:
Answer: |  x = 1    or     x = -(-1)^(1/3)    or    x = (-1)^(2/3)

 

2)

Solve for x:
x^4-1 = 0

 

Add 1 to both sides:
x^4 = 1

 

Taking 4^th roots gives 1 times the 4^th roots of unity:
Answer: |  x = -1    or    x = -i    or    x = i    or    x = 1

 

3)

Solve for x:
x^3-2 x^2-5 x+6 = 0

 

The left hand side factors into a product with three terms:
(x-3) (x-1) (x+2) = 0

 

Split into three equations:
x-3 = 0 or x-1 = 0 or x+2 = 0

 

Add 3 to both sides:
x = 3 or x-1 = 0 or x+2 = 0

 

Add 1 to both sides:
x = 3 or x = 1 or x+2 = 0

 

Subtract 2 from both sides:
Answer: |  x = 3    or    x = 1    or    x = -2

 

4)

Solve for x:
x^4+2 x^3-16 x^2-2 x+15 = 0

 

The left hand side factors into a product with four terms:
(x-3) (x-1) (x+1) (x+5) = 0

 

Split into four equations:
x-3 = 0 or x-1 = 0 or x+1 = 0 or x+5 = 0

 

Add 3 to both sides:
x = 3 or x-1 = 0 or x+1 = 0 or x+5 = 0

 

Add 1 to both sides:
x = 3 or x = 1 or x+1 = 0 or x+5 = 0

 

Subtract 1 from both sides:
x = 3 or x = 1 or x = -1 or x+5 = 0

 

Subtract 5 from both sides:
Answer: |  x = 3    or    x = 1    or    x = -1    or    x = -5

1) f(x) = x^3 - 1 

 

\(x^3-1=0\\x^3=1\\ x=1\\ 3 \;roots\\ 2\pi/3\\ x=e^{(2\pi/3)i}, \qquad x=e^{(4\pi/3)i} \qquad and \qquad x=1 \)

2)  g(x) = x^4 - 1

 

x^4=1

roots    1, i, -1, -i

 

because 1 is 'primary' one and there will be 4 altogether,  each will be 90 degrees apart on the complex plane so the answer was easy.