Find the value of p when 4 x^3 + 4^2 + px + 6 is divided by 2x-3 leaving a remainder -12.
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Remainder theorem:
When we divide f(x)f(x) by (x−c)(x−c) , the remainder f(c)f(c)
When p(x)=4x3−2x2+px+5p(x)=4x3−2x2+px+5 is divided by (x+2)(x+2), the remainder is
p(−2)=4(−2)3−2(−2)2+p(−2)+5p(−2)=4(−2)3−2(−2)2+p(−2)+5
=−2p−35=−2p−35
So,
a=−2p−35a=−2p−35
When q(x)=x3+6x2+pq(x)=x3+6x2+p is divided by (x+2)(x+2), the remainder is
q(−2)=(−2)3+6(−2)2+pq(−2)=(−2)3+6(−2)2+p
=p+16=p+16
So,
b=p+16b=p+16
Now
a+b=0a+b=0
−2p−35+p+16=0−2p−35+p+16=0
−p−19=0−p−19=0
P=-19
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Answer:
c) -27 is the correct answer
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