2x^3+2=0
divide by 2
x^3 +1=0
now you have a sum of cubes
(x+1)(x^2-x+1)=0
x=-1 or x = (1 ± ?-3)/2 or (1 ± i?3)/2
(Factoring a sum or difference of cubes) and (solving a polynomial equation).
2x^3+2=0
2 answers
Factor the given equation into:
2x³+2 =0
=> 2(x²+1)=0
and use one of the identities:
x³+y²=(x+y)(x²-xy+y²)
x³-y²=(x-y)(x²+xy+y²)
to factor the given equation to linear and quadratic factors.
Hence solve the resulting linear and quadratic equations.
2x³+2 =0
=> 2(x²+1)=0
and use one of the identities:
x³+y²=(x+y)(x²-xy+y²)
x³-y²=(x-y)(x²+xy+y²)
to factor the given equation to linear and quadratic factors.
Hence solve the resulting linear and quadratic equations.
1 answer