What must be substracted from $4 x^{3} + 16 x^{2} - x + 5$ to obtain a polynomial which is exactly divisible by $x + 5 ?$

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Solution

$4 x^{2} - 4 x + 19$
$x + 5) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 4 x^{3} + 16 x^{2} - x + 5 ($
$4 x^{3} + 20 x^{2} - ---------- -$
$- 4 x^{2} - x$
$- 4 x^{2} - 20 x - ----------- -$
$19 x + 5$
$19 x + 95 - -------- -$
$- 90 - --- -$
Hence, $- 90$ must be subtracted from $4 x^{3} + 16 x^{2} - x + 5$ so that it in exactly divisible by $x + 5$

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