Find all real and complex solutions of the polynomial equation. x^3-2x^2+4x-8=0 A. -2, 2i, -2iB. 2, 2i, -2iC. 2, -2, -2iD. 2, -2, 2i, -2i
Accepted Solution
A:
First we want to solve by factoring. So take x^3 - 2x^2 + 4x - 8 = 0, add parenthesis > (x^3 - 2x^2) + (4x - 8) Now factor out x^2 from x^3 - 2x^2 which is x^2(x - 2), factor out 4 from 4x - 8 which is 4(x - 2) > Now we have > x^2(x - 2) + 4(x - 2) [Factor out the common term of (x - 2)] > (x - 2)(x^2 + 4) = 0 Okay now we want to apply the zero factor principle. x - 2 = 0 > Add 2 to both sides > x - 2 + 2 = 0 + 2 > x = 2.
Now we have to plug into the quadratic equation where a = 1, b = 0, c = 4 We get -0 + sqrt0 - 1 * 4 * 4/1 * 2 after applying the rule of 0^a = 0 to -0 + sqrt0^2 - 4 * 1 * 4/2 * 1. Multiply the numbers now... sqrt0 - 4 * 1 * 4 > sqrt0 - 16 > subtract > sqrt-16. Apply the rad rule of.. [tex] \sqrt{-a}= \sqrt{-1} \sqrt{a} [/tex] sqrt-1, sqrt-16, Apply the imaginary rule sqrt-1 = i > sqrt16i Now we have sqrt16i/2, so we want to factor 16 which is 2^4 because 2 * 2 * 2 * 2 = 16. Assuming a > 0 we want to apply [tex] \sqrt[n]{a^m} = a^ \frac{m}{n} \ \textgreater \ 2^ \frac{4}{2} \ \textgreater \ Divide \ \textgreater \ \frac{4}{2} = 2 \ \textgreater \ 2^2[/tex] 2^2i/2 > cancel the common factor of 2 > 2i For the next part, do the exact same steps except when you get to -sqrt16i/2, apply the fraction rule > [tex] \frac{-a}{b} = - \frac{a}{b} [/tex] To get -2i. Therefore our final solutions are, (2, 2i, -2i)