How do you find all the real and complex roots of x^6-64=0x6−64=0?
1 Answer
Explanation:
Knowing the following factoring techniques is imperative:
- Difference of squares:
a^2-b^2=(a+b)(a-b)a2−b2=(a+b)(a−b) - Sum of cubes:
a^3+b^3=(a+b)(a^2-ab+b^2)a3+b3=(a+b)(a2−ab+b2) - Difference of cubes:
a^3-b^3=(a-b)(a^2+ab+b^2)a3−b3=(a−b)(a2+ab+b2)
x^6-64=0x6−64=0
Apply difference of squares:
color(red)((x^3+8))color(green)((x^3-8))=0(x3+8)(x3−8)=0
Use both sum & difference of cubes:
color(red)((x+2)(x^2-2x+4))color(green)((x-2)(x^2+2x+4))=0(x+2)(x2−2x+4)(x−2)(x2+2x+4)=0
From here, set each portion of the product equal to
x+2=0=>color(blue)(x=-2x+2=0⇒x=−2
x-2=0=>color(blue)(x=2x−2=0⇒x=2
The following two quadratic factors can be solved via completing the square or using the quadratic formula.
Solving
color(blue)(x)=(2+-sqrt(4-16))/2=(2+-2sqrt3i)/2color(blue)(=1+-sqrt3ix=2±√4−162=2±2√3i2=1±√3i
Solving
color(blue)(x)=(-2+-sqrt(4-16))/2=(-2+-2sqrt3i)/2color(blue)(=-1+-sqrt3ix=−2±√4−162=−2±2√3i2=−1±√3i