Answer:
(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)
Answer:
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).
