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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