If the polynomials az3+4z2+3z−4 and z3−4z+a leave the same remainder when divided by z−3, then find the value of a.
−1
0
3
1
A
1
B
−1
C
0
D
3
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Solution
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Here, p(z)=a(z)3+4(z)2+3z−4, q(z)=(z)3−4z+a, and the zero of z−3 is 3. So, by the given condition p(3)=q(3) a(3)3+4(3)2+3(3)−4=(3)3−4(3)+a 27a+4×9+9−4=27−12+a 27a+36+9−4=27−12+a 27a+45−4=15+a 27a+41=15+a 27a−a=15−41 26a=−26 a=−2626 a=−1
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