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Mathematics

Nine Standard >> Application of Remainder theorem | Part -1

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Problem:

For what value m, the remainder is 2, when \(P(x)=3x^2-mx+5\) is divided by x+3.


Solution: 


\(P(x)=3x^2-mx+5\) 
Divisor g(x)=x+3=x-(-3), a=-3
P(-3) is the remainder.
P(-3)=2....(1)

By replacing x=-3, we get

\(P(-3)=3(-3)^2-m(-3)+5\)....(2)

Form equation (1) and (2) we get

     \(3(-3)^2-m(-3)+5\)=2\)
or, \(3 \times 9-3m+5=2\)
or, 27-3m+5=2
or, 3m=27+5-2
or, 3m=30
So, m=\(\frac{30}{3}\)=10

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