WIELOMIANY Z PARAMETREM – rozwiązanie zadań

Zadanie 1


W(x) = x3 − 6x2 + ax + b


W(1) = 0


W(2) = 0


$$\left\{ \begin{matrix} 1^{3} - 6*1^{2} + a*1 + b = 0 \\ 2^{3} - 6*2^{2} + a*2 + b = 0 \\ \end{matrix} \right.\ $$


$$\left\{ \begin{matrix} 1 - 6 + a + b = 0 \\ 8 - 24 + 2a + b = 0 \\ \end{matrix} \right.\ $$


$$\left\{ \begin{matrix} a + b = 5 \\ 2a + b = 16 \\ \end{matrix} \right.\ $$


−a = −11


a = 11


b = −6


$$\left\{ \begin{matrix} a = 11 \\ b = - 6 \\ \end{matrix} \right.\ $$


W(x) = x3 − 6x2 + 11x − 6


x3 − 6x2 + 11x − 6     :    (x−1)(x − 2)


(x−1)(x−2) = x2 − 2x − x + 2 = x2 − 3x + 2


$$\frac{x - 3}{\begin{matrix} x^{3} - 6x^{2} + 11x - 6\ \ \ :\ \ \ x^{2} - 3x + 2 \\ \frac{- x^{3} + {3x}^{2} - 2x}{\begin{matrix} = \ - 3x^{2} + 9x - 6 \\ \frac{3x^{2} - 9x + 6}{= = =} \\ \end{matrix}} \\ \end{matrix}}$$


x = 3

Zadanie 2


x3 + 4x2 + ax + b = (x − 2)2(cx + d)


x3 + 4x2 + ax + b = (x2−4x+4)(cx+d)


x3 + 4x2 + ax + b = cx3 + dx2 − 4cx2 − 4dx + 4cx + 4d


x3 + 4x2 + ax + b = cx3 + x2(d−4c) + x(−4d+4c) + 4d


c = 1


4 = d − 4c


4 = d − 4


d = 8


a = −4d + 4c


a = −32 + 4


a = −28


b = 4d


b = 32


$$\text{Odp.\ }\left\{ \begin{matrix} a = - 28 \\ b = 32 \\ \end{matrix} \right.\ $$

Zadanie 3


$$\frac{W\left( x \right)}{Q\left( x \right)} = P\left( x \right) + \frac{R\left( x \right)}{Q\left( x \right)}\ \ \ \ \ |*Q(x)$$


P(x) → czesc podzielna


$$\frac{R\left( x \right)}{Q\left( x \right)} \rightarrow czesc\ niepodzielna$$


W(x) = P(x) * Q(x) + R(x)


x1 − jest pierwiastkiem


x2 − jest pierwiastkiem


W(x1) = P(x1) * Q(x1) + R(x1)


P(x1) = 0


W(x1) = 0 * Q(x1) + R(x1)


W(x1) = R(x1)


Jezeli dane sa dwa warunki to reszta jest postaci:


R(x) = ax + b


Jezeli dane sa trzy warunki to reszta jest postaci:


R(x) = ax2 + bx + c


R(1) = 2


R(2) = 0


R(x) = ax + b


$$\left\{ \begin{matrix} a*1 + b = 2 \\ a*2 + b = 0 \\ \end{matrix} \right.\ $$


$$\left\{ \begin{matrix} a + b = 2 \\ 2a + b = 0 \\ \end{matrix} \right.\ $$


−a = 2


a = −2


−2 + b = 2


b = 4


$$\left\{ \begin{matrix} a = - 2 \\ b = 4 \\ \end{matrix} \right.\ $$

Zadanie 4


R(x) = ax + b


$$\left\{ \begin{matrix} R\left( - 3 \right) = 3 \\ R\left( 2 \right) = 2 \\ \end{matrix} \right.\ $$


$$\left\{ \begin{matrix} - 3a + b = 3 \\ 2a + b = 2 \\ \end{matrix} \right.\ $$


−5a = 1


$$a = - \frac{1}{5}$$


$$- \frac{2}{5} + b = 2$$


$$b = 2\frac{2}{5}$$


$$\left\{ \begin{matrix} a = - \frac{1}{5} \\ b = 2\frac{2}{5} \\ \end{matrix} \right.\ $$


$$Odpowiedz:\ \ \ R\left( x \right) = - \frac{1}{5}x + 2\frac{2}{5}$$

Zadanie 5


x2 = t


t > 0


t2 + (m−2)t + 4 = 0


1   a ≠ 0


1 ≠ 0


m ∈ R


2    > 0


>0  ⇔ b2 − 4ac > 0


(m−2)2 − 4 * 4 > 0


m2 − 2m + 4 − 16 > 0


m2 − 2m − 12 > 0


=16 + 48 = 64


$$\sqrt{} = \sqrt{64} = 8$$


$$m_{1} = \ \frac{4 - 8}{2} = - 2$$


$$m_{2} = \frac{4 + 8}{2} = 6$$


m ∈ (−∞, −2) ∪ (6, +∞)


3   t1 * t2 > 0


$$t_{1}*t_{2} > 0\ \Leftrightarrow \frac{c}{a} > 0$$


4 > 0


m ∈ R


4   t1 + t2 > 0


$$t_{1} + t_{2} > 0\ \Leftrightarrow \ - \frac{b}{a} > 0$$


$$\frac{- m + 2}{1} > 0$$


−m + 2 > 0


−m > −2


m < 2


m ∈ ( − ∞, 2) 


zestawienie zalozen


Odpowiedz :    m ∈ (−∞,−2)

Zadanie 6


( − 2x4 + 6x2 − 4)19 = (−2 + 6 − 4)19 = 019 = 0

Zadanie 7


W(x) = x3 + 4ax2 − 11x − 12


W(−1) = 0


W(x) = x3 + 4ax2 − 11x − 12


−1 + 4a + 11 − 12 = 0


4a = 2


22a = 2


2a = 1


$$a = \frac{1}{2}$$


$$x^{3} + 4^{\frac{1}{2}}x^{2} - 11x - 12 = 0$$


x3 + 2x2 − 11x − 12 = 0


1
2
−11
−12

−1
1
1
−12
0


(x+1)(x2+x−12) = 0


x = −1


x2 + x − 12 = 0


=1 + 4 * 12 = 49


$$\sqrt{} = \sqrt{49} = 7$$


$$x_{1} = \frac{- 1 - 7}{2} = - 4$$


$$x_{2} = \frac{- 1 + 7}{2} = 3$$

Zadanie 8


m2x2 + (m2+6m)x2 + (m+6)x = 0


x[m2x2+(m2+6m)x+(m+6)] = 0


I pierwiastek → x = 0


m2x2 + (m2+6m)x + (m+6) = 0


1   m2 ≠ 0


m ≠ 0


m ∈ R − {0}


2    > 0


a = m2


b = m2 + 6m


c = m + 6


>0  ⇔  b2 − 4ac > 0


(m2+6m)2 − 4m2(m+6) > 0


m4 + 12m3 + 36m2 − 4m3 − 24 > 0


m2(m+6)2 − 4(m+6) > 0


m2(m+6)(m+6−4) > 0


m = 0


m = −6


m = −2


m ∈ (−∞,−6) ∪ (−2,0) ∪ (0, +∞)


Odpowiedz :    m ∈ (−∞,−6) ∪ (−2,0) ∪ (0, +∞)

Zadanie 9


x4 + ax3 + bx2 + cx − 1 = 0


x4 + ax3 + bx2 + cx − 1 = (x+1)3(dx + e)


x4 + ax3 + bx2 + cx − 1 = (x3+3x2+3x+1)(dx+e)


x4 + ax3 + bx2 + cx − 1 = dx4 + 3x2 + 3x2 + dx + ex3 + 3ex2 + 3ex + e


x4 + ax3 + bx2 + cx − 1 = dx4 + x3(3+e) + x2(3+3e) + x(d+3e) + e


x4 = dx4


d = 1


ax3 = x3(3+e)


a = 3 + e


e = −1


a = 2


bx2 = x2(3+3e)


b = 3 + 3e


b = 3 + (−3)


b = 0


cx = x(d+3e)


c = d + 3e


c = 1 + (−3)


c = −2


Odpowiedz :    a = 2,  b = 0,  c = −2.

Zadanie 10


f(x) = (x−a)(x−b) + (x−b)(x−c) + (x−c)(x−a) = (x2 − bx − ax + ab)+(x2−cx−bx+bc) + (x2−ax−cx+ac) = x2 − bx − ax + ab + x2 − cx − bx + bc + x2 − ax − cx + ac = 3x2 − 2ax − 2bx − 2cx + ab + bc + ac = 3x2 − (2a+2b+2c)x + ab + bc + ac


$$\frac{a = 3}{\begin{matrix} b = - \left( 2a + 2b + 2c \right) \\ c = ab + bc + ac \\ \end{matrix}}$$


≥0


b2 − 4ac ≥ 0


(2a + 2b + 2c)2 − 4 * 3 * (ab + bc + ac)≥0


(2a+2b+2c)(2a+2b+2c) − 12(ab+bc+ac) ≥ 0


4a2 + 4ab + 4ac + 4b2 + 4ab + 4bc + 4ac + 4bc + 4c2 − 12(ab+bc+ac) ≥ 0  |:4


a2 + b2 + c2 + ab + ac + ab + bc + ac + bc − 3(ab+bc+ac) ≥ 0


a2 + b2 + c2 + 2ab + 2ac + 2bc − 3ab − 3bc − 3ac ≥ 0


a2 + b2 + c2 − ab − bc − ac ≥ 0


a2 − ab + b2 − bc + c2 − ac ≥ 0  |*2


2a2 − 2ab + 2b2 − 2bc + 2c2 − 2ac ≥ 0


a2 + a2 − 2ab + b2 + b2 − 2bc + c2 + c2 − 2ac ≥ 0


a2 − 2ab + b2 + b2 − 2bc + c2 + a2 − 2ac + c2 ≥ 0


(a−b)2 + (b−c)2 + (c−a)2 ≥ 0


Lewa strona jest zawsze wieksza lub rowna zero !

RÓWNANIA, NIERÓWNOŚCI WIELOMIANOWE – rozwiązanie zadań

Zadanie 1

  1. x3 + 2x2 − 4x − 8 = 0


x2(x+2) − 4(x+2) = 0


(x+2)(x2−4) = 0


x = −2


x2 = 4  ⇒ x = −2  ∨ x = 2

  1. x3 − 13x + 12 = 0


x3 − x − 12x + 12 = 0


x(x2−1) − 12(x−1) = 0


x(x+1)(x−1) − 12(x−1) = 0


(x−1)[x(x+1)−12] = 0


x = 1


x2 + x − 12 = 0


=49


$$\sqrt{} = 7$$


x1 = −4


x2 = 3

  2. x3 + 4x2 + 9x + 6 = 0


1
4
9
6

−1
1
3
6
0


(x+1)(x2+3x+6) = 0


x = −1


x2 + 3x + 6 = 0


=9 − 4 * 6 = −15


brak pierwiastkow

  2. x4 + 4x2 + 3 = 0


x2 = t


t > 0


t2 + 4t + 3 = 0


$$= 16 - 12 = 4 \Rightarrow \sqrt{} = 2$$


$$t_{1} = \frac{- 4 - 2}{2} = - 3\ \notin zalozenia$$


$$t_{2} = \frac{- 4 + 2}{2} = - 1 \notin zalozenia$$

Zadanie 2

  1. (x+1)3(2−x)(x2−4) ≤ 0


x = −1


x = 2


x2 = 4


x = 2  ∨ x = −2


x ∈ ( − ∞,  −2⟩ ∪ ⟨1,  + ∞)


Jezeli krotnosc pierwiastka jest parzysta to w okolicach tego pierwiastka nie zmienia sie znak !


np. x3(−x)x2 = −x6

  1. (3−x)3(x2−4)2(2−x) ≥ 0


x = 3


x2 = 4  ⇔ x = 2  ∨ x = −2


x = 2


x ∈ ( − ∞,  2⟩ ∪ ⟨3 , +∞)

  1. 3x4 − 10x3 + 10x − 3 < 0


3x4 − 3 − 10x3 + 10 < 0


3(x4−1) − 10x(x2−1) < 0


3(x2−1)(x2+1) − 10x(x2−1) < 0


(x2 − 1)2[3(x2+1)−10x] < 0


x2 − 1 = 0


x2 = 1  ⇔ x = 1  ∨ x = −1


3x2 − 10x + 3 = 0


=100 − 12 * 3 = 64


$$\sqrt{} = \sqrt{64} = 8$$


$$x_{1} = \ \frac{1}{3}$$


x2 = 3


$$Odpowiedz:\ \ \ \ x \in \left( - 1,\frac{1}{3} \right) \cup (1,3)$$

Zadanie 3

  1. x3 − 9x2 + 26x − 24 = 0


1
−9
26
−24

2
1
−7
12
0


(x−2)(x2−7x+12) = 0


x = 2


x2 − 7x + 12 = 0


=49 − 4 * 12 = 1


$$\sqrt{} = \sqrt{1} = 1$$


$$x_{1} = \frac{7 - 1}{2} = 3$$


$$x_{2} = \frac{7 + 1}{2} = 4$$

  2. x3 − 3x − 2 = 0


1
0
−3
−2

2
1
2
1
0


(x−2)(x2+2x+1) = 0


x = 2


x2 + 2x + 1 = 0


=4 − 4 = 0


$$x = - \frac{b}{2a}$$


$$x = - \frac{2}{2} = - 1$$

  2. x(x2 − 4x + 3)2(x + 3)3 ≤ 0


x = 0


x2 − 4x + 3 = 0


=16 − 12 = 4


$$\sqrt{} = 2$$


x1 = 3,  x2 = 1


x + 3 = 0


x = −3


Odpowiedz :    x ∈ ⟨−3, 0⟩

NIERÓWNOŚCI WIELOMIANOWE Z WARTOŚCIĄ BEZWZGLĘDNĄ – rozwiązanie zadań

Zadanie 1

  1. x3 − x2 + |x−1| > 0


1   x − 1 ≥ 0


x ≥ 1


2   x − 1 < 0


x < 1


Ad.1


x3 − x2 + x − 1 > 0


(x−1)(x2+1) > 0


x = 1


x2 = −1


brak rozwiazan


x ∈ (1,+∞)


zestawienie zalozen


x ∈ ⟨1 ,   + ∞)


Ad.2


x2(x−1) − (x−1) > 0


(x−1)(x2−1) > 0


x = 1


x2 = 1


x = 1  ∨  x = −1


x ∈ (−1, 1) ∪ (1, +∞)


zestawienie zalozen


x ∈ (−∞,−1)


suma rozwiazan


x ∈ (−∞,−1) ∪ ⟨1 ,   + ∞)


Odpowiedz :    x ∈ (1,+∞)

  1. x3 + |3x2−4x| − 12 ≥ 0


1   3x2 − 4x ≥ 0


x(3x−4) ≥ 0


x = 0


3x = 4


$$x = \frac{4}{3}$$


$$x \in ( - \infty,\ \left. \ 0 \right\rangle \cup \left\langle \frac{4}{3}, + \infty) \right.\ $$


2   3x2 − 4x < 0


x(3x−4) < 0


x = 0


3x = 4


$$x = \frac{4}{3}$$


$$x \in (0,\frac{4}{3})$$


1 x3 + 3x2 − 4x − 12 ≥ 0


x2(x+3) − 4(x+3) ≥ 0


(x+3)(x2−4) ≥ 0


x = −3


x2 = 4


x = 2  ∨ x = −2


x ∈ ⟨−3, −2⟩  ∪  ⟨2 ,   + ∞)


2 x3 − 3x2 + 4x − 12 ≥ 0


x2(x−3) + 4(x−3) ≥ 0


(x−3)(x2+4) ≥ 0


x = 3


x2 = −4


brak rozwiazan


1   zestawienie zalozen


x ∈ ⟨−3, −2⟩  ∪ ⟨2 ,   + ∞)


2   zestawienie zalozen


x ∈ ⌀


SUMA ROZWIAZAN


Odpowiedz :     x ∈ ⟨−3, −2⟩ ∪ ⟨2,+∞ )

Zadanie 2


Jezeli wielomian jest stopnia nieparzystego to ma co najmniej jeden pierwiastek.


Jezeli wielomian jest stopnia parzystego to moze nie miec zadnego pierwiastka.


Maksymalna liczba pierwiastkow jest taka jaki jest stopien wielomianu.


2x3 − 3x2 − 12x − 5m = 0


2x3 − 3x2 − 12x = 5m


f(x) =  2x3 − 3x2 − 12x


2x3 − 3x2 − 12x = 0


x(2x2−3x−12) = 0


x = 0


2x2 − 3x − 12 = 0


=9 + 8 * 12 = 9 + 96 = 105


$$\sqrt{} = \sqrt{105}$$


$$x_{1} = \frac{3 - \sqrt{105}}{4}\ \approx - 1,8$$


$$x_{2} = \frac{3 + \sqrt{105}}{4} \approx 3,3$$


W(p,q)


$$p = - \frac{b}{2a}$$


$$q = - \frac{}{4a}$$


$$q = - \frac{\sqrt{105}}{2*4} = - \frac{\sqrt{105}}{8}$$

Zadanie 3


|x3−6x2| > 8x − 48


1 x3 − 6x2 ≥ 0


x2(x−6) ≥ 0


x = 0


x = 6


x ∈ ⟨6,+∞ )


x3 − 6x2 − 8x + 48 > 0


x2(x−6) − 8(x−6) > 0


(x−6)(x2−8) > 0


x = 6


x2 = 8


$$x = \sqrt{8} = 2\sqrt{2}\ \vee x = - \sqrt{8} = - 2\sqrt{2}$$


$$x \in \left( - 2\sqrt{2},\ 2\sqrt{2} \right) \cup \left( 6, + \infty \right)$$


zestawienie z zalozen


x ∈  ⟨6 ,   + ∞)


2 x3 − 6x2 < 0


x2(x−6) < 0


x = 0


x = 6


x ∈ (−∞, 0)


−x3 + 6x2 − 8x + 48 > 0


x3 − 6x2 + 8x − 48 < 0


x2(x−6) + 8(x−6) < 0


(x−6)(x2+8) < 0


x = 6


x2 = −8


brak rozwiazan


x ∈ (−∞,6)


zestawienie z zalozen


x ∈ (−∞,0)


SUMA ROZWIAZAN


Odpowiedz :    x ∈ (−∞,0) ∪ ⟨6 , +∞)