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Find the value of a Given: Let A and B be the remainders when polynomials x3+2x2-5ax-7 and x3+ax2-12x+6 are divided by x+1 and x-2 respectively. If 2A+B=6 , find the value of a.



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

Step 1:

p(x)=x3+2x2−5ax−7p(x)=x3+2x2−5ax−7

Since p(x)p(x) is divisible by (x+1)(x+1) the remainder is p(−1)p(−1)

p(−1)=−1+2+5a−7p(−1)=−1+2+5a−7

=5a−6=5a−6

=A=A

Step 2:

t(x)=x3+ax2−12x+6t(x)=x3+ax2−12x+6

Since t(x)t(x) is divisible by (x−2)(x−2) the remainder is t(2)t(2)

t(2)=(2)3+4a−24+6t(2)=(2)3+4a−24+6

=4a−10=4a−10

=B=B

Step 3:

Since 2A+B=62A+B=6

we know that A=(5a−6),B=4a−10A=(5a−6),B=4a−10

Therefore 2A+B=6⇒2(5a−6)+4a−10=62A+B=6⇒2(5a−6)+4a−10=6

Solving for ′a′′a′

10a−12+4a−10=610a−12+4a−10=6

14a−22=614a−22=6

14a=2814a=28

We get a=2