Question

If m(z)=z^3+az^2+bz +6 leaves remainders 3 and 0 when divided by (z-3) and (z-2) respectively, then find the
values of a and b.
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Answer 1

a=-3 and b=-1

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Answer 2

$m(z) = {z}^{3} + {az}^{2} + bz + 6$

Dividing by (z-2):

${2}^{3} + a \times {2}^{2} + b \times 2 + 6 = 0 \\ or \: 8 + 4a + 2b + 6 = 0 \\ or \: 14 + 4a + 2b = 0 \\ or \: 2(7 + 2a + b) = 0 \\ or \: 2a + b = - 7 \\ or \: b = - (7 + 2a)$

Dividing by (z-3):

${3}^{3} + a \times {3}^{2} - (2a + 7)3 + 6 = 0 \\ or \: 27 + 9a - 6a - 21 + 6 = 0 \\ or \: 3a = - 12 \\ or \: a = - 4$

so, a = -4

and, b = 1