Find the Roots (Zeros) p(x)=x^3+8

Problem

p(x)=x^3+8

Solution

1. Set the polynomial to zero to find the roots of the equation.

x^3+8=0

2. Identify the expression as a sum of cubes, noting that 8=2^3

x^3+2^3=0

3. Apply the sum of cubes formula, which is a^3+b^3=(a+b)*(a^2−a*b+b^2) where a=x and b=2

(x+2)*(x^2−2*x+4)=0

4. Solve the linear factor by setting it equal to zero.

x+2=0

x=−2

5. Solve the quadratic factor using the quadratic formula x=(−b±√(,b^2−4*a*c))/(2*a) where a=1 b=−2 and c=4

x=(−(−2)±√(,(−2)^2−4*(1)*(4)))/(2*(1))

6. Simplify the discriminant and solve for the complex roots.

x=(2±√(,4−16))/2

x=(2±√(,−12))/2

x=(2±2*i√(,3))/2

x=1±i√(,3)

Final Answer

x=−2,1+i√(,3),1−i√(,3)

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