Question

P(x) = x³+8 A polynomial P is given. (a) Find all zeros of P , real and complex.

Answer 1

To find all the zeros of the polynomial P(x) = x³+8, set P(x) equal to zero and solve for x. The zeros are x = -2 and x = 1±√3i.

Step-by-step explanation:

To find all the zeros of the polynomial P(x) = x³+8, we set P(x) equal to zero and solve for x.

So, x³+8 = 0. To solve this equation, we can use the fact that a³+b³ = (a+b)(a²-ab+b²). In this case, a = x and b = 2.

Using this formula, we have (x+2)(x²-2x+4) = 0. Equating each factor to zero gives us x+2 = 0 and x²-2x+4 = 0. Solving these equations, we find that x = -2 and x = 1±√3i.