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Question

Let r1 and r2 be the remainders when the polynomials

p(x)= x3+x2-5kx-7 and q(x)=x3+kx2-12x+6

are divided by x+1 and x-2 respectively.

Find the value of k if 2r1-r2=10

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Solution

Remainder Theorem : Let p(x) be an polynomial of degree greater than or equal to one.Let 'a' be any real number.If p(x) is divisible by (x−a) then the remainder is p(a).

Given

p(x)= x3+x2-5kx-7 is divisible byx+1

Then by remainder theorem the remainder is p(-1)=(-1)^3+1^2-(5k*-1)-7=-1+1+5k-7=5k-7

Given it is r1

There fore r1=5k-7

q(x)=x3+kx2-12x+6 is divisible by x-2

Then by remainder theorem the remainder is q(2)=2^3+k2^2-12*2+6=8+4k-24+6= 4k-10

Given it is r2

There fore r2=4k-10

Given 2r1-r2=10

2(5k-7)-(4k-10)=10

10k-14-4k+10=10

6k=14

K=14/6=7/3


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