Question

若多项式x3+5x2+ax+b可被x2-x-6整除,求a,b,并将原式分解因式.

Answer 1

设x^3+5x^2+ax+b=(x^2-x-6)(x+c)

因为x^2-x-6=(x-3)(x+2)

所以 x=3和x=-2时,右边等于0

则左边也等于0

x=3,x^3+5x^2+ax+b=27+45+3a+b=0

3a+b=-72 (1)

x=-2,x^3+5x^2+ax+b=-8+20-2a+b=0

2a-b=12 (2)

(1)+(2)

5a=-60

a=-12,b=2a-12=-36

x^3+5x^2+ax+b

=x^3+5x^2-12x-36

=(x^2-x-6)(x+c)

所以c=(-36)/(-6)=6

所以x^3+5x^2-12x-36

=(x-3)(x+2)(x+6)