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Let f(z) = az³+4z²+3z-4
zero of the polynomial (z-3) = 3
So,replacing z by 3,
f(3) = a(3)³+4(3)²+3(3)-4
⇒f(3) = 27a+36+9-4
⇒f(3) = 27a+41
Let g(z) = z³-4z+a
zero of the polynomial (z-3) = 3
So,replacing z by 3,
g(3) = (3)³-4(3)+a
⇒g(3) = 27-12+a
⇒g(3) = 15+a
Given that the two polynomials leaves same remainder when divided by z-3
⇒15+a = 27a+41
⇒15-41 = 27a - a
⇒-26 = 26a
⇒a = -1.

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i think i did a calculation mistake i am sorry let me try again
oh man is that cube ??
very very thank you lohithanaidu
U r welcm Uganesh2005 :)
A= -13/4 eh?? i used factor theorem where i took Z=3  and substituted in both and got the answer as 9a+29 and 3+a now i equated both and got a = -26/8 and atlast i cancelled i mean divided and got -3/4
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