Question

Use the given zero to find the remaining complex zeros of the polynomial function and write the polynomial as a product of linear factors.
f(x)=x^3+8x^2+6x-52; zero:2

Answer 1

Answer:

(x - 2)(x - 5 + 2i)(x - 5 - 2i)

Step-by-step explanation:

If one zero is 2, then one factor is (x - 2).

Use synthetic division here.  Taking the coefficients 1, 8, 6 and -52, divide 2 into them:

2    /  1      8      6     -52

                2     20     52

      -----------------------------

          1     10     26     0

Because the remainder is zero, we know that 2 is a root and (x - 2) is a factor, and that the coefficients of the quadratic quotient are 1, 10 and 26.

The discriminant is thus b^2 - 4ac, or 10^2 - 4(1)(26), or  100 - 104, or -4.

Because the discriminant is negative, we know that there will be two factors which themselves are complex conjugates.  One such factor is

      10 ± √-4         10 ± 2i

x = --------------- = ---------------

             2                   2

The complex zeros are thus x = 5 ± 2i.

The corresponding factors are (x - 5 + 2i) and (x - 5 - 2i)

and so the polynomial is

(x - 2)(x - 5 + 2i)(x - 5 - 2i)