Find the zeros of the polynomial. ?

Question

round to 3 decimal places

3x^3+19x^2+45x+13

there's 1 real & 2 complex 0's

Answer 1

Our equation factors to:

(3*x+1)*(x^2+6*x+13) = 0

Thus we immediately see one real solution is given by:

3*x + 1 = 0

x = -1/3

Now we find solutions for:

x^2+6*x+13 = 0

This seems to have no nice factorization so we use the quadratic formula.

x =  [-b±sqrt((b^2)-4ac)]/2a

x =  [-6±sqrt((36)-4(1)(13))]/2(1)

x = [-6±sqrt(-16)]/2

sqrt(-16) = sqrt(16)*sqrt(-1) = sqrt(16)*i = 4i

x = -3±2i

Thus we have 1 real solution x = -1/3 and two complex x = -3+2i and x = -3-2i