如果多项式x^2-(a+5)x+5a-1能分解成两个一次因式(x+b)和(x+c)的乘积(b,c为整数),则a=?因x^2-(a+5)x+5a-1=(x+b)(x+c)= x^2+(b+c)x+bc 故有b+c=-a-5,bc=5a-1 消去a,变形得 (b+5)(c+5)=-1 因 b,c是整数,故有b=-4,c=-6 或b=-6,c=
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![如果多项式x^2-(a+5)x+5a-1能分解成两个一次因式(x+b)和(x+c)的乘积(b,c为整数),则a=?因x^2-(a+5)x+5a-1=(x+b)(x+c)= x^2+(b+c)x+bc 故有b+c=-a-5,bc=5a-1 消去a,变形得 (b+5)(c+5)=-1 因 b,c是整数,故有b=-4,c=-6 或b=-6,c=](/uploads/image/z/5628286-46-6.jpg?t=%E5%A6%82%E6%9E%9C%E5%A4%9A%E9%A1%B9%E5%BC%8Fx%5E2-%28a%2B5%29x%2B5a-1%E8%83%BD%E5%88%86%E8%A7%A3%E6%88%90%E4%B8%A4%E4%B8%AA%E4%B8%80%E6%AC%A1%E5%9B%A0%E5%BC%8F%28x%2Bb%29%E5%92%8C%28x%2Bc%29%E7%9A%84%E4%B9%98%E7%A7%AF%28b%2Cc%E4%B8%BA%E6%95%B4%E6%95%B0%29%2C%E5%88%99a%3D%3F%E5%9B%A0x%5E2-%28a%2B5%29x%2B5a-1%3D%28x%2Bb%29%28x%2Bc%29%3D+x%5E2%2B%28b%2Bc%29x%2Bbc+%E6%95%85%E6%9C%89b%2Bc%3D-a-5%2Cbc%3D5a-1+%E6%B6%88%E5%8E%BBa%2C%E5%8F%98%E5%BD%A2%E5%BE%97+%EF%BC%88b%2B5%EF%BC%89%28c%2B5%29%3D-1+%E5%9B%A0+b%2Cc%E6%98%AF%E6%95%B4%E6%95%B0%2C%E6%95%85%E6%9C%89b%3D-4%2Cc%3D-6+%E6%88%96b%3D-6%2Cc%3D)
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