Given that when (x^2+x-2) divides f(x)=x^3+ax+b and g(x)=2x^3+3x^2-1 respectively, the remainders are the same. Find the value of a-b ?

2 Answers
Jun 26, 2017

a-b=-1

Explanation:

x^2+x-2=(x+2)(x-1)

Now when f(x)=x^3+ax+b and g(x)=2x^3+3x^2-3 are divided by x^2+x-2 remainder is same,

hence x^2+x-2 completely divides

g(x)-f(x)=x^3+3x^2-ax-b-3 i.e. x+2 and x-1 are factors of h(x)=x^3+3x^2-ax-b-3

therefore using factor theorem we should have h(-2)=0 and h(1)=0

i.e. (-2)^3+3(-2)^2-a(-2)-b-3=0

or 2a-b=-1 ...........................(1)

and 1^3+3xx1^2-a-b-3=0

or a+b=1 ...........................(2)

Adding (1) and (2), we get 3a=0 i.e. a=0 and putting this in (2), b=1

and a-b=-1

Jun 26, 2017

a-b=-1

Explanation:

Here we have x^2+2-2=(x+1)(x-2) so we have

{(f(x) = Q_1(x)(x+1)(x-2)+r(x)),(g(x)=Q_2(x)(x+1)(x-2)+r(x)):}

so we have

{(f(-1)=-1-a+b=r(-1)),(g(-1)=0=r(-1)):}

then

r(-1)=0 rArr a-b=-1