Qoldiqli bo`lish haqida teorema

1-misol. P(x) ko’phadni (x-3)² ga bo’linganda qoldiq (x-1) bo’lsa P²(x) ko’phadi (x-3)² ga bo’lingandagi qoldiqni toping.


Yechish. P²(x)=

X^2-2x+1 X^2-6x+9	X^2-6x+9
	1
4x-8=r(x)

2-misol. P(x)=(x²-3x+2)³+2(x²-3x+2)²+2x²+x+10 ko’phadi (x²-3x+4) ga bo’linganda qolgan qoldiqni toping .
Yechish: P(x)=(x²-3x+2)³+2(x²-3x+2)²+2x²+x+10=T³-3t²*2+3*2²*t-8+2(t²-4t+4)+2x²+x+10=

2x^2+x+10 2x^2-6x+8	X^2-3x+4
	2
7x+2

R(x)=7x+2
3-misol. P(x)=x⁶-1 ko’phadni x²+x+1 ga bo’lingandagi qoldiqni aniqlang.
Yechish: P(x)=x⁶-1=(x³-1)(x³+1)=(x-1)(x²+x+1)(x+1)(x²-x+1);
(x²+x+1) r(x)=0.
4-misol. P(x+3)=x²-x+n bo’lsa P(x-2) ko’phadni x-3 ga bo’lganimizda qoldiq 10, n=?
Yechish:
P(x-2)=(x-3)*Q(x)+10
X=3 P(1)=0*Q(x)+10=10
X=-2 P(-2+3)=4+2+n
P(1)=6+n=10 n=4.


5-misol. P(x)=ax³+bx²+7x-12 ko’phadni Q(x) ko’phadga bo’linganda bo’linma x bo’lsa qoldiqni toping.
Yechish: P(x)=
r(x)=-12
Javob: r(x)=-12
6-misol. P(x)=x¹³+4x¹⁰+x⁸+5x⁷-2x³+3x-1 ko’phadni x⁴ ga bo’lgandagi qoldiq qancha.
Yechish: P(x)=x¹³+4x¹⁰+x⁸+5x⁷-2x³+3x-1=x⁴*Q(x)+r(x)

d

P(x)=x⁴(x⁹-4x⁶+5x)
-2x³+3x-1=x⁴Q(x)+r(x)
r(x)=-2x³+3x-1


Javob: r(x)=-2x³+3x-1