SOLUTION: 1. A polynomial function is shown below. f(x) = x^6 + 12x^5 + 43x^4 + 22x^3 a. Find the real zeros of f (x). b. Express f (x) in terms of linear factors. c. Find all of t

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: 1. A polynomial function is shown below. f(x) = x^6 + 12x^5 + 43x^4 + 22x^3 a. Find the real zeros of f (x). b. Express f (x) in terms of linear factors. c. Find all of t      Log On


   



Question 1034208: 1. A polynomial function is shown below.
f(x) = x^6 + 12x^5 + 43x^4 + 22x^3
a. Find the real zeros of f (x).
b. Express f (x) in terms of linear factors.
c. Find all of the zeros of f (x).
(3 points)

Answer by ikleyn(52781) About Me  (Show Source):
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1. A polynomial function is shown below.
f(x) = x^6 + 12x^5 + 43x^4 + 22x^3
a. Find the real zeros of f (x).
b. Express f (x) in terms of linear factors.
c. Find all of the zeros of f (x).
(3 points)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

First, factor by taking x^3 out:

f(x) = x%5E6+%2B+12x%5E5+%2B+43x%5E4+%2B+22x%5E3  = x%5E3%2A%28x%5E3+%2B+12x%5E2+%2B+43x+%2B+22%29.

Next, check that x = -1 is the root of the polynomial in parentheses  %28x%5E3+%2B+12x%5E2+%2B+43x+%2B+22%29.

So, the polynomial in parentheses has the factor (x+1).

Make long division and factor this polynomial:

%28x%5E3+%2B+12x%5E2+%2B+43x+%2B+22%29 = %28x%2B1%29%2A%28x%5E2+%2B+11x+%2B+32%29.

The quadratic polynomial  x^2 + 11x + 32  has no real roots (since its discriminant 11%5E2+-+4%2A32 = -7 is negative).

Thus you can not factor this quadratic polynomial further over real domain.

Therefore, your final factoring over the real domain is 

f(x) = x%5E6+%2B+12x%5E5+%2B+43x%5E4+%2B+22x%5E3 = x%5E3%2A%28x%2B1%29%2A%28x%5E2+%2B+11x+%2B+32%29.

The two real roots are x = 0  and x = -1.

The complex zeroes of the polynomial x^2 + 11x + 32 are 

x%5B1%5D = %28-11+%2B+i%2Asqrt%287%29%29%2F2  and  x%5B1%5D = %28-11+-+i%2Asqrt%287%29%29%2F2.