一个多项式除以(2X^ 1),商式为X-1,余式为5X,求这个多项式
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(3x^2-2x+4)(2x+6)+3x-1=6x^3+14x^2-x+23这个多项式为6x^3+14x^2-x+23
(x^2-2x+3)(x+1)+2x-5=(x^2-2x+1+2)(x+1)+2x-5=[(x-1)^2+2](x+1)+2x-5=(x-1)^2(x+1)+2(x+1)+2x-5=(x-1)^2(x
2x³-4x²-1=A(2x)+x-1;2Ax=2x³-4x²-1-x+1;A=(2x³-4x²-x)/(2x);如果本题有什么不明白可以追
2x³-4x²-1=A×2x+(x-1)所以A=[(2x³-4x²-1)-(x-1)]/2x=(2x³-4x²-1-x+1)/2x=(2x&
多项式2x²-4x-1除以一个多项式A,得商式为2x,余式为x-12x²-4x-1=(2x)A+x-1(2x)A=2x²-4x-1-x+1=2x²-5x(2x)
1:2x³-4x²-1-2Ax=x-1.解得A=x²-2x-1/22:a²+3ab+b²=a²-2ab+b²+5ab=(a-b)^
(x-3)(3x-2)-5=3x²-2x-9x+6-5=3x²-11x+1
(2x^2+1)*(x-1)+5x=2x^3-2x^2+x-1+5x=2x^3-2x^2+6x-1
(2x³-4x²-1-x+1)÷2x=(2x³-4x²-x)÷2x=x²-2x-1/2除式=(被除式-余式)÷商式
(2x-1)(x²+1)+5x=2x³+2x-x²-1+5x=2x³-x²+7x-1答:这个多项式是2x³-x²+7x-1.您好,
ax^3+bx^2+cx+d-(5x+4)=ax^3+bx^2+(c-5)x+(d-4)整除(x^2-x+2)(x^2-x+2)(ax+e)=ax^3-ax^2+2ax+ex^2-ex+2e=ax^3
设多项式为ax^3+bx^2+cx+d显然ax^3+bx^2+cx+d=(x^2-x+2)[ax+(a+b)]+(c-a+b)x+d-a-b由于余式等于5x+4所以c-a+b=5,d-a-b=4ax^
根据题意,得:(2x+6)(3x2-2x+4)+(3x-1)=6x3+14x2-x+23所以,这个多项式为:6x3+14x2-x+23.
因为4x³+2x²+2除以一个多项式A,余式为2(4x³+2x²+2-2)/A=2x²A=(4x³+2x²)/2x²=2
2x(2x²+1)+3x-2=4x³+2x+3x-2=4x³+5x-2这个多项式为4x³+5x-2
因为:被除式=商式×除式+余式所以2x^3-4x^2-1=除式×(x-1)+2x所以除式×(x-1)=2x^3-4x^2-1-2x所以除式=(2x^3-4x^2-1-2x)/(x-1)因为(2x^3-
A=(6*x的平方+6x-3)-(6x+3)除以x+1=6*x的平方-6除以x+1=6(x的平方-1)除以x+1=6(x+1)(x-1)除以x+1=6x-6
解析:被除式=除式·商式+余式由题可得:(3x^2-1)(x-2)+(x-1)=3x³-6X²-x+2+x-1=3x³-6X²+1
(2x^2-1)*(X-2)+(X-1)=2X^3-4X^2+1
(2x)³-(4x)²-1/A=2x.2x-1(2x)³-(4x)²-1=2x×A+2x-18X³-16X²-1=2X(A+1)-18X&s