Yuqori darajali tenglamalar

Misol-3. (x²–4x+6)²–4(x²–4x+6)+6=x.


Mashqlar
Bikvadrat tenglamalarni yeching.

6.65. t^4 – 2t^2– 3=0

6.66.2x^4 – 9x^2+4=0

6.67.5 y^4 – 5y^2+2=0

6.68.x^4 – 4x^2+ 4=0

Qaytma tenglamalarni yeching.

a)x^4+5x^3+2x^2+5x+1=0	b)2x^4+3x^3–3x^2+3x+2=0
d)4x^4+2x^3+3x^2+2x+4=0	e)3x^4–2x^3+x^2–2x+3=0

6.94. f(f(x)) = xko‘rinishidagitenglamaniyeching.

a) (x^2+2x–5)^2+2(x^2+2x–5)–5=x	b) (x^2–8x+18)^2–8(x^2–8x+18)+18=x
d) (x^2–3x+3)^2–3(x^2–3x+3)+3=x	e) (x^2–x–3)^2–(x^2–x–3)–3=x

12-mavzu.Bikvadrat tenglama. Qaytma tenglama. f(f(x)) = x ko‘rinishidagi tenglama. Mashqlar
Bikvadrat tenglamalarni yeching.

6.65. t^4 – 2t^2– 3=0

6.66. 2x^4 – 9x^2+4=0

6.67. 5 y^4 – 5y^2+2=0

6.68. x^4 – 4x^2+ 4=0

Qaytma tenglamalarni yeching.

a)x^4+5x^3+2x^2+5x+1=0

b) 2x^4+3x^3–3x^2+3x+2=0

d) 4x^4+2x^3+3x^2+2x+4=0

e) 3x^4–2x^3+x^2–2x+3=0

Bezu¹ teoremasi
P(x) ko‘phadnix–aikkihadgabo‘lgandabo‘linmaQ(x) va qoldiqR(x) bo‘lsin:
P(x) = (x–a) Q(x) + R(x).
Agarbumunosabatgax=aqo‘yilsa,P(a) = 0 ∙ Q(a) +R(a) = R(a) = rhosilbo‘ladi.Shutariqaushbuteoremaisbotlanadi:
Teorema (Bezu). P(x) = a₀xⁿ + a₁x^n–1 + ... + a_n–1x + a_nko‘phadnix–aga(a  0)bo‘lishdanchiqadiganrqoldiqshuko‘phadningx = adagiqiymatigateng,r = P(a).
Misol-1. 1)x⁵+x+20nix+2gabo‘lishdanchiqadiganqoldiqr=(–2)⁵+(–2)+20= –14.
2)x⁵+x+34nix+2gabo‘lishdanchiqadiganqoldiq r=(–2)⁵+(–2)+34=0.Demak, x=–2soni shu ko‘phadning ildizi.
Natijalar.nNbo‘lganda:
1) xⁿ–aⁿikkihad x–a ga bo‘linadi. Haqiqatan, P(a)=aⁿ–aⁿ=0;
2)xⁿ+aⁿikkihadx–agabo‘linmaydi.Haqiqatan,P(a)=aⁿ+aⁿ=2aⁿ0;
3)x²ⁿ–a²ⁿikkihadx+agabo‘linadi.Haqiqatan,P(–a)=(–a)²ⁿ–a²ⁿ=0;
4)x²ⁿ⁺¹–a²ⁿ⁺¹ikkihadx+agabo‘linmaydi.Haqiqatan,P(–a)=(–a)²ⁿ⁺¹–a²ⁿ⁺¹=–2a²ⁿ⁺¹0;
5)x²ⁿ⁺¹+a²ⁿ⁺¹ikkihadx+agabo‘linadi.Haqiqatan,P(–a)=(–a)²ⁿ⁺¹+a²ⁿ⁺¹=0;
6)x²ⁿ+a²ⁿikkihadx+agabo‘linmaydi.Haqiqatan,P(–a)=a²ⁿ+a²ⁿ=2a²ⁿ0.


Gorner sxemasi
P(x) = a₀xⁿ + a₁x^n–1 + ... + a_n–1x + a_nko‘phadnix–aga(a  0)bo‘lishdanchiqadiganrqoldiqni aniqlashniGorner²sxemasi debataluvchiusulniko‘rib chiqamiz.
P(x) = (x–a) Q(x) + r (1)
bo‘lsin. Bunda Q(x) = b₀x^n–1 + b₁^n–2x+... + b_n–1bo‘lsin.
(1)daxningbirxildarajalarioldidagikoeffitsiyentlarnitenglashtiribquyidagigaegabo‘lamiz:
a₀=b₀; a₁=b₁–ab₀; a₂=b₂–ab₁; … a_n–1=b_n–1–ab_n–2; a_n=r–ab_n–1.
Bundanko‘rinadiki,b₀=a₀,b_k=ab_k–1+a_k,k = 1,2,...,n–1,r = a_n+ab_n–1.
Bo‘linmavaqoldiq quyidagi jadvalyordamidatopiladi:

	a0	a1	a2	…	an–1	an
a		ab0+a1	ab1+a2	…	abn-2+a1	abn-1+an
	a0=b0	b1	b2	…	bn–1	r

Misol-2. x³+4x²–3x+5ko‘phadniGornersxemasidanfoydalanibx–1 gabo‘lishnibajaramiz:

	a0	a1	a2	a3
	1	4	–3	5
1		ab0+a1	ab1+a2	ab2+a3
	1	5	2	7
	a0=b0	b1	b2	r

Demak,x³+4x²–3x+5=(x–1)(x²+5x+2)+7.
Misol-3. P₃(x)=x³–3x²+5x+7 ni2x+1 ga bo‘lishdanhosil bo‘lgan qoldiqni toping.
Yechish. .


Teorema-2. Agar a soni P(x) ko‘phadning ildizi bo‘lsa, P(x) ko‘phad x–a ikkihadga qoldiqsiz bo‘linadi.
Isbot.Bezuteoremasigako‘ra,P(x)nix–agabo‘lishdanchiqadiganqoldiqP(a)gateng,shartbo‘yichaesaP(a)=0.Isbotlandi.


Ko‘phadni ildizi
Teorema-2 P(x)=0tenglamaniyechishmasalasiniP(x)ko‘phadnichiziqliko‘paytuvchilargaajratishmasalasigakeltirishimkoniniberadi.
Natija-1. AgarP(x) ko‘phadharxila₁,...,a_nildizlargaegabo‘lsa,u (x–a₁)∙...∙(x–a_n)ko‘paytmaga qoldiqsizbo‘linadi.
Natija-2. n-darajaliko‘phadntadanortiqharxilildizgaega bo‘la olmaydi.


Misol-3. (x²–4x+6)²–4(x²–4x+6)+6=x.


Mashqlar
6.152.P(x)ko‘phad D(x)ko‘phadgabo‘linadimi:

a)P(x)=x^100–3x+2,D(x)=x–1	b)P(x)=x^100–3x+2,D(x)=x+1
d)P(x)=x^100–3x^2+2,D(x)=x^2–1	e)P(x)=x^100–3x+2,D(x)=2x^2–1

6.155. Bo‘lishdagi qoldiqni toping:

a) x^4–3x^2+1 ni x–2 ga	b)x^5–4x^3+x^2nix–3ga
d) x^5–4x^3–x^2+1ni2x–3ga	e) x^4–3x^3+x^2–1 ni3x–4 ga

6.156. m ning qanday qiymatlarida3x⁴–2x³–m²x–2ko‘phadx–2gaqoldiqsizbo‘linadi?
6.157.mningqandayqiymatlarida3x³–4x²–mx–1ko‘phadx+1gabo‘linmaydi?
6.158.avabningqandayqiymatlarida2x⁴+ax³+bx–2ko‘phadx²–x–2uchhadgaqoldiqsizbo‘linadi?
6.159.mvanningqandayqiymatlaridax³+mx+nko‘phadx²+3x+10uchhadgaqoldiqsizbo‘linadi?
6.162.GornersxemasiyordamidaP(x)ko‘phadniD(x)ikkihadgaqoldiqlibo‘ling:

a)P(x)=x^2–5x–7,D(x)=x–1	b) P(x)=x^3–3x^2+5x–6, D(x)=x–2
d) P(x)=2x^4–3x^2–5x+2, D(x)=x+1	e) P(x)=3x^5–4x^3–x+1, D(x)=x+3
f) P(x)=3x^6–4x^5–x^4+x^3–x^2–1, D(x)=x–3	g) P(x)=x^5–x^2–5x–6, D(x)=x–2
h) P(x)=x^4–x^3+2x^2–5x–42, D(x)=x+2	i) P(x)=x^5–4x^2+5x–3, D(x)=x–3

1^EtyenBezu(1730 – 1783) – fransuzmatematigi

2^HornerUilyam(1786–1837) – inglizmatematigi