设多项式f(x)除以(x-1)的余式为2,除以(x^2-2x+3)的余式为(4x+6),则f(x)除以(x-1)(x^2-2x+3)的余式是多少?我在解析里面看到 设 f(x)=(x-1)(x^2-2x+3)q(x)+a(x^2-2x+3)+4x+6,再由f(1)=2a+10=2推出a=-4,得出余式.f(x)=(x-1)(x^2
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![设多项式f(x)除以(x-1)的余式为2,除以(x^2-2x+3)的余式为(4x+6),则f(x)除以(x-1)(x^2-2x+3)的余式是多少?我在解析里面看到 设 f(x)=(x-1)(x^2-2x+3)q(x)+a(x^2-2x+3)+4x+6,再由f(1)=2a+10=2推出a=-4,得出余式.f(x)=(x-1)(x^2](/uploads/image/z/2831373-45-3.jpg?t=%E8%AE%BE%E5%A4%9A%E9%A1%B9%E5%BC%8Ff%28x%29%E9%99%A4%E4%BB%A5%28x-1%29%E7%9A%84%E4%BD%99%E5%BC%8F%E4%B8%BA2%2C%E9%99%A4%E4%BB%A5%28x%5E2-2x%2B3%29%E7%9A%84%E4%BD%99%E5%BC%8F%E4%B8%BA%284x%2B6%29%2C%E5%88%99f%28x%29%E9%99%A4%E4%BB%A5%28x-1%29%28x%5E2-2x%2B3%29%E7%9A%84%E4%BD%99%E5%BC%8F%E6%98%AF%E5%A4%9A%E5%B0%91%3F%E6%88%91%E5%9C%A8%E8%A7%A3%E6%9E%90%E9%87%8C%E9%9D%A2%E7%9C%8B%E5%88%B0+%E8%AE%BE+f%28x%29%3D%28x-1%29%28x%5E2-2x%2B3%29q%28x%29%2Ba%28x%5E2-2x%2B3%29%2B4x%2B6%2C%E5%86%8D%E7%94%B1f%281%29%3D2a%2B10%3D2%E6%8E%A8%E5%87%BAa%3D-4%2C%E5%BE%97%E5%87%BA%E4%BD%99%E5%BC%8F.f%28x%29%3D%28x-1%29%28x%5E2)
设多项式f(x)除以(x-1)的余式为2,除以(x^2-2x+3)的余式为(4x+6),则f(x)除以(x-1)(x^2-2x+3)的余式是多少?我在解析里面看到 设 f(x)=(x-1)(x^2-2x+3)q(x)+a(x^2-2x+3)+4x+6,再由f(1)=2a+10=2推出a=-4,得出余式.f(x)=(x-1)(x^2
设多项式f(x)除以(x-1)的余式为2,除以(x^2-2x+3)的余式为(4x+6),则f(x)除以(x-1)(x^2-2x+3)的余式是多少?
我在解析里面看到 设 f(x)=(x-1)(x^2-2x+3)q(x)+a(x^2-2x+3)+4x+6,再由f(1)=2a+10=2推出a=-4,得出余式.f(x)=(x-1)(x^2-2x+3)q(x)+a(x^2-2x+3)+4x+6 这一步 +a(x^2-2x+3)+4x+6 我没有看懂诶 ,
设多项式f(x)除以(x-1)的余式为2,除以(x^2-2x+3)的余式为(4x+6),则f(x)除以(x-1)(x^2-2x+3)的余式是多少?我在解析里面看到 设 f(x)=(x-1)(x^2-2x+3)q(x)+a(x^2-2x+3)+4x+6,再由f(1)=2a+10=2推出a=-4,得出余式.f(x)=(x-1)(x^2
过程确实简单了些,细化一下:
设多项式f(x)除以(x-1)(x^2-2x+3)的商式是q(x),余式是r(x),则f(x)=(x-1)(x^2-2x+3)q(x)+r(x).
因为除式(x-1)(x^2-2x+3)是三次多项式,所以余式r(x)是二次多项式.
让r(x)除以x^2-2x+3,则商式一定是个常数a,余式是个一次多项式,记为bx+c,则r(x)=a(x^2-2x+3)+bx+c.
所以,f(x)=(x-1)(x^2-2x+3)q(x)+r(x)=(x-1)(x^2-2x+3)q(x)+a(x^2-2x+3)+bx+c.
这个式子还可以看作是f(x)=((x-1)q(x)+a)(x^2-2x+3)+bx+c,所以bx+c是f(x)除以x^2-2x+3的余式,bx+c=4x+6.
所以,f(x)=(x-1)(x^2-2x+3)q(x)+a(x^2-2x+3)+4x+6.