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Question
If the polynomials az3 + 4z2 + 3z – 4 and z3 – 4z + a leave the same remainder when divided by z – 3, find the value of a.
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Solution
Let p1(z) = az3 + 4z2 + 3z – 4 and p2(z) = z3 – 4z + 0
When we divide p1(z) by z – 3, then we get the remainder p,(3).
Now, p1(3) = a(3)3 + 4(3)2 + 3(3) – 4
= 27a + 36 + 9 – 4
= 27a + 41
When we divide p2(z) by z – 3 then we get the remainder p2(3).
Now, p2(3) = (3)3 – 4(3) + a
= 27 – 12 + a
= 15 + a
According to the question,
Both the remainders are same.
p1(3) = p2(3)
27a + 41 = 15 + a
27a – a = 15 – 41
26a = –26
a = `(-26)/26`
a = –1
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