what are the real or imaginary solutios of the polynomial equation
x^3-8=0
I came up with 1+i sgrt 3 and 1-i sgrt 3
could some see if I'm right
thank you
To find the solutions of the polynomial equation x^3 - 8 = 0, we can solve it step by step.
Step 1: Factor the polynomial.
The given polynomial can be factored using the difference of cubes formula,
x^3 - 8 = (x - 2)(x^2 + 2x + 4).
Step 2: Set each factor equal to zero and solve for x.
Setting the first factor equal to zero, we have:
x - 2 = 0
x = 2
Setting the second factor equal to zero, we need to solve a quadratic equation:
x^2 + 2x + 4 = 0
Using the quadratic formula, x = (-b ± √(b^2 - 4ac)) / (2a), where a = 1, b = 2, and c = 4, we get:
x = (-2 ± √(2^2 - 4(1)(4))) / (2(1))
x = (-2 ± √(4 - 16)) / 2
x = (-2 ± √(-12)) / 2
x = (-2 ± 2i√3) / 2
x = -1 ± i√3
Therefore, the solutions to the polynomial equation x^3 - 8 = 0 are x = 2, x = -1 + i√3, and x = -1 - i√3.
So, your solution of 1 + i√3 and 1 - i√3 are incorrect. The correct solutions are x = 2, x = -1 + i√3, and x = -1 - i√3.
To find the real or imaginary solutions of the polynomial equation x^3 - 8 = 0, we can first factor the expression using the difference of cubes formula:
a^3 - b^3 = (a - b)(a^2 + ab + b^2).
Applying this formula, we have:
x^3 - 8 = (x - 2)(x^2 + 2x + 4).
Now we can set each factor equal to zero and solve for x:
1) (x - 2) = 0.
Adding 2 to both sides:
x = 2.
2) (x^2 + 2x + 4) = 0.
This quadratic equation can be solved by applying the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a).
In this case, a = 1, b = 2, and c = 4. Substituting these values into the quadratic formula, we get:
x = (-2 ± √(2^2 - 4(1)(4))) / (2(1))
= (-2 ± √(4 - 16)) / 2
= (-2 ± √(-12)) / 2
= (-2 ± √(4 × -3)) / 2
= (-2 ± 2i√3) / 2
= -1 ± i√3.
Therefore, the solutions to the polynomial equation x^3 - 8 = 0 are:
x = 2,
x = -1 + i√3,
x = -1 - i√3.
So, your solutions of 1 + i√3 and 1 - i√3 are not correct. The correct solutions are 2, -1 + i√3, and -1 - i√3.
First byou must find real solution.
x ^ 3 - 8 = 0 Add 8 to both sides
x ^ 3 - 8 + 8 = 0 + 8
x ^ 3 = 8
x = third root of 8 = 2
In this case one rational zero is x = 2
Now you divide polynomial x ^ 3 - 8 with ( x - 2 )
( x ^ 3 - 8 ) / ( x - 2 ) = x ^ 2 + 2 x + 4
OR
( x ^ 3 - 8 ) = ( x ^ 2 + 2 x + 4 ) * ( x - 2 ) = 0
Youe equation have roots when :
x ^ 2 + 2 x + 4 = 0
AND
x - 2 = 0
Solutions of equation x ^ 2 + 2 x + 4 = 0
are
1 + i sgrt 3 and 1 - i sgrt 3
Soluton of equation x - 2 = 0
are
x = 2
Your equation have 3 solutions.
One real solution x = 2
and two imaginary solutions :
1 + i sgrt 3 and 1 - i sgrt 3