Question

if x^3+ax^2+bx=6 is exactly divisible by (x-1), prove that a+b=5.​

Answer 1

Step-by-step explanation:

x-1=0

x=1

subtitute value of x

p(x)=x^3+ax^2+bx

p(1)=(1)^3+a(1)^2+b(1)

p(1)=1+a+b

as p(x) =6. (given)

1+a+b=6

a+b=6-1

a+b=5