Question

The polynomial ax3 + 3x2 -3 and 2x3 - 5x +a when divided by x-4 leave the same remainder in each case. Find the value of a.

Answer 1

Given, f(x) = ax³+3x²-3
$\: \: \: \: \: \: g(x) = 2 {x}^{3} - 5x + a$

$If \: f(x) \: is \: divided \: by \: (x-4) \\ then \: it \:leaves \: a \: remainder \: f(4)$

$f(4) = a \times 4 ^{3} + 3(4 ^{2} ) - 3 \\ = a \times 64 + 48 - 3 \\ = 64a + 45$

$If \: g(x) \: is \: divided \: by \: (x-4) \\ then \: it \:leaves \: a \: remainder \: g(4)$

$g(4) = 2(4 ^{3} ) - 5 \times 4 + a \\ = 128 - 20 + a \\ = 108 + a \\ \\ \\ Given \: that \: remainders \: in \: both \: \\ cases \: are \: equal \: \\ \\ Now, \: \\ 108 + a = 64a + 45 \\ 63a = 108 - 45 = 63 \\ \\ a = \frac{63}{63} = 1$

Therefore a = 1 .

Answer 2

The answer is in the attachment.