For the polynomial f(x) = x^5 - 2x^4 + 8x^2 - 13x+6 answer these questions

(a) how many zeros does the function have over the set of complex numbers?

(d) what is the maximum number of local extrema (maxima or minima) the graph of the function can have?

(e) complete the following statements: as x -> ∞,f(x)-> as x-> -∞. f(x)->

(f) list the possible rational zeros of this function.

(g) factor this polynomial completely over the set of complex numbers.

  (a) 5

  (d) 4

  (e) as x → ∞, f(x) → ∞. as x → -∞, f(x) → -∞

  (f) ±1, ±2, ±3, ±6

  (g) f(x) = (x +2)(x -1)²(x -1+i√2)(x -1-i√2)

Step-by-step explanation:

(a) A 5th-degree polynomial has 5 zeros in the set of complex numbers.

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(d) The derivative of a 5th degree polynomial is 4th degree. It can have at most 4 real zeros, so the 5th degree polynomial can have at most 4 turning points (local extrema).

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(e) The positive leading coefficient of this odd-degree polynomial tells you the signs of the end behavior match. (f(x) is positive for large positive x; negative for large negative x.)

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(f) Possible rational zeros are the factors of the constant term, 6. These are ...

  ±1, ±2, ±3, ±6

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(g) I like to use a graphing calculator to show the real zeros of higher-degree polynomials. The graph has x-intercepts of -2 and +1. The graph touches, but does not cross the x-axis at x=1, so this is a root of even multiplicity. The curvature of the "touch" at x=1 tells you it has multiplicity of 2 (not higher).

The graphing calculator can show the residual quadratic after these zeros are factored out. In vertex form, it is (x -1)^2 +2. It has complex zeros of ...

  1 ±i√2.

So, the complete factorization over complex numbers is ...

  f(x) = (x +2)(x -1)²(x -1+i√2)(x -1-i√2)

Step-by-step explanation:

(a) it's a degree 5 polynomial so it will have 5 zeros.

(d) 2

(e) as  x --> ∞, f(x)  --> ∞.

   as  x --> -∞, f(x)  --> -∞.

(f) By the rational roots theorem:  they are +/- 1, +/- 2 , +/- 3, +/- 6.

(g) x^5 - 2x^4 + 8x^2 - 13x + 6 = 0

f(1) = 0 and f(-2)  = 0 so (x - 1) and (x + 2) are factors.

Dividing by (x - 1)(x - 2) gives the quotient (x - 1)(x^2 - 2x + 3) so the factors are:

= (x - 1)^2(x + 2)(x^2 - 2x + 3).

The correct answer :

a = not viable

b = 265

c = 0