Question

if x^2+x-12 divides x^3+ax+bx-6 completely find the value of a and b​

Answer 1

If x^2+x-12 is a factor of x^3+ax+BC-6

then x^2+x-12=0.

x^2+4x-3x-12=0. (middle term splitting)

x(x+4) -3(x+4)=0

(x+4)(x-3)=0

x+4=0.

x=-4(neglect )

x-3=0

x=3( ans)___________(i)

x^3+ax+Bx-6

(3)^3+a×3+b×3-6. (from(i))

27+3a+3b-6

27-6=-3a-3b

-3a=27

a=27/-3

a=-9.

-3b=-6

b=2.

a=-9 and b=2