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# ?

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Answer 1 · 2017-06-26T06:15:16

#a-b=-1#

#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#

Answer 2 · 2017-06-26T08:01:12

#a-b=-1#

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#