What are the real and complex solutions of the polynomial equation? x3−64=0 (1 point) Responses 4,−2+2i3,−−√ 2+2i3–√ 4,−2+2i3,−−√ 2+2i3–√ 4,−1+2i3,−−√ −1+2i3–√ 4,−1+2i3,−−√ −1+2i3–√ 4,−2+2i3,−−√ −2−2i3–√ 4,−2+2i3,−−√ −2−2i3–√ 4, 1+2i3,−−√ 1+2i3–√ 4, 1+2i3,−−√ 1+2i3–√
To find the real and complex solutions to the polynomial equation \(x^3 - 64 = 0\), we can first factor it to find the real solution, and then use complex numbers to find the complex solutions.
The equation can be factored using the difference of cubes formula:
\(A^3 - B^3 = (A - B)(A^2 + AB + B^2)\),
where for our equation, \(A = x\) and \(B = 4\) (since \(4^3 = 64\)).
This gives us:
\(x^3 - 64 = x^3 - 4^3 = (x - 4)(x^2 + 4x + 16) = 0\).
From the first factor, \(x - 4 = 0\), we get the real solution:
\(x = 4\).
Next, we need to factor and solve the quadratic equation \(x^2 + 4x + 16 = 0\).
To find the complex solutions, we can use the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\),
where \(a = 1\), \(b = 4\), and \(c = 16\) for our quadratic equation.
Plugging in the values, we get:
\(x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}\),
\(x = \frac{-4 \pm \sqrt{16 - 64}}{2}\),
\(x = \frac{-4 \pm \sqrt{-48}}{2}\).
Since we have a negative number under the square root, we will have complex solutions:
\(x = \frac{-4 \pm \sqrt{48}i}{2}\),
\(x = \frac{-4 \pm 4\sqrt{3}i}{2}\),
\(x = -2 \pm 2\sqrt{3}i\).
Thus, the solutions to the polynomial equation \(x^3 - 64 = 0\) are:
Real solution:
\(x = 4\),
Complex solutions:
\(x = -2 + 2\sqrt{3}i\),
\(x = -2 - 2\sqrt{3}i\).