(Factoring a sum or difference of cubes) and (solving a polynomial equation).
2x^3+2=0
To factor a sum or difference of cubes, we follow a specific pattern. In the case of the equation 2x^3 + 2 = 0, we have a sum of cubes because we're adding a constant term (2).
To factor a sum of cubes, we use the formula:
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
Given 2x^3 + 2 = 0, we can rewrite it as:
(2x^3 + 2) = 0
Now we can identify the values of a and b from our equation:
a = ∛(2x^3) = x
b = ∛2 = ∛(2 * 1^3) = 1
Next, we use the sum of cubes formula:
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
Replacing a and b with x and 1 respectively, we have:
(2x^3 + 2) = (x + 1)(x^2 - x + 1)
Therefore, the factored form of the equation 2x^3 + 2 = 0 is:
(x + 1)(x^2 - x + 1) = 0
Now, to solve the polynomial equation (x + 1)(x^2 - x + 1) = 0, we set each factor equal to zero:
(x + 1) = 0 or (x^2 - x + 1) = 0
Solving the first equation, we get:
x + 1 = 0
x = -1
To solve the second equation, we can either apply the quadratic formula or factor the expression:
x^2 - x + 1 = 0
Using the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
For this equation, a = 1, b = -1, and c = 1. Substituting these values into the formula and simplifying further, we have:
x = (-(-1) ± √((-1)^2 - 4(1)(1)))/(2(1))
x = (1 ± √(1 - 4))/(2)
x = (1 ± √(-3))/(2)
Since the discriminant (b^2 - 4ac) is negative, the equation has no real solutions. Therefore, the only solution to the equation 2x^3 + 2 = 0 is x = -1.
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
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.