Question

if the polynomials az cube + 4 z square + 3 z minus 4 and Z cube minus 4z + a leave the same remainder when divided by Z minus 3 find the value of a

Answer 1

$z - 3 = 0 \\ z = 3 \\az + 4 {z}^{3} + 4 {z}^{2} + 3z - 4 \\ a \times 3 + 4( {3}^{3} ) + 4 {}^{2} + 3 \times 3- 4 \\ 3a + 108 + 16 + 9 - 4 \\ 3a + 129.....1$
$z = 3 \\ {z}^{3} - 4z + a \\ (3) {}^{3} - 4 \times 3 + a \\ 27 - 12 + a \\ 15 + a....2$
Since, 1=2
$3a + 129 = 15 + a \\ 2a = - 114 \\ a = - 57$
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Answer 2

See attachment it can help you