如果多项式(a-2)x4-1 2xb x2-3
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![如果多项式(a-2)x4-1 2xb x2-3](/uploads/image/f/3695813-53-3.jpg?t=%E5%A6%82%E6%9E%9C%E5%A4%9A%E9%A1%B9%E5%BC%8F%28a-2%29x4-1+2xb+x2-3)
4a-2b+4a=8a-2b多项式是8a-2
x^4+x^3+2x^2-3x+2-ax^3-3x^2-bx-1=x^4+(1-a)x^3-x^2+(-3-b)x+1因为该多项式不含x^3项和x项,所以1-a=0a=1-3-b=0b=-3,把a=1
因为X4-(a-1)X3+5X2+(b+3)x-1不含X3项和x项所以a-1=0,b+3=0所以a=1,b=-3所以3a+b=1×3+(-3)=0
∵2x2-x=1,∴4x4-4x3+3x2-x-1,=2x2(2x2-x)-x(2x2-x)+(2x2-x)-1,=2x2-x,=1.故答案为:1.
∵多项式2x4-(a+1)x3+(b-2)x2-3x-1,不含x2、x3项,∴a+1=0,b-2=0,解得a=-1,b=2.∴ab=-2.故答案为:-2.
答案:-72a^2-3a+4=3则,2a^2-3a=-1则,4a^2-6a=-2则,4a^2-6a-5=-2-5=-7
解题思路:系数为0解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq.ph
设另一个因式是:2x2+mx+n,则(x2+x-6)(2x2+mx+n)=2x4+(m+2)x3+(m+n-12)x2+(n-6m)x-6n则:m+2=1m+n−12=−an−6m=ba+b−1=−6
(-1+x)2除多项式x4+ax3-3x2+bx+3所得的余式是x+1不妨设商式为x^2+cx+d.由已知有 x^4+ax^3-3x^2+bx+3 =(-1+x)^2(x^2+cx+d)+(
x4+2x2-3=(x2+3)(x2-1)=(x2+3)(x-1)(x+1).故选C.
|a|-3=0(x^3系数为0)a=-3(x^2系数不为0)a^2-2a-3=9+6-3=12
9,2*《2a²-3a》+7
据题意得(2x4-4x2-1)-(x4-2x2-4)=2x4-4x2-1-x4+2x2+4=x4-2x2+3=x4-2x2+1-1+3=(x2-1)2+2≥2>0∴多项式2x4-4x2-1的值总大于x
4a²-2a+5=7∴4a²-2a=2∴2a²-a=1∴2a²-a+1=1+1=2
因为(a+b)x^4+(b-2)x^3-2(a-1)x^2+ax-3不含x3次方项和x2,所以可得b-2=a-1=0,即a=1,b=2将a=1,b=2代入(a+b)x^4+(b-2)x^3-2(a-1
如果多项式x4的平方-(a-1)x³+5x²-(b+3)x+6中,不含x³和x的项,则a=【1】,b=【-3】.根据题意得:a-1=0,a=1b+3=0,b=-3所以,a
设这个多项式是A,∵A+(2x2-x3-5-3x4)=3x4-5x3-3,∴A=3x4-5x3-3-(2x2-x3-5-3x4)=3x4-5x3-3-2x2+x3+5+3x4=6x4-4x3-2x2+
conv([1,8,0,0,-10],[2,-1,3])结果ans=215-524-2010-30即2*x6+15*x5-5*x4+24*x3-20*x2+10*x-30再问:如何输让215-524-
a+2=0,a=﹣2,b=3∴3a+ab+7=﹣6-6+7=﹣5.