What must be subtracted from $6 x^{4} + 13 x^{3} + 13 x^{2} + 30 x + 20$ , so that the resulting polynomial is exactly divisible by $3 x^{2} + 2 x + 5$ ?

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Solution

$p (x) = g (x) \times q (x) + r (x)$ [Division algorithm for polynomials]
$\Rightarrow p (x) - r (x) = g (x) \times q (x)$
It is clear the RHS of the above equation is divisible by $g (x)$ . i.e., the divisor.
$∴$ LHS is also divisible by the divisor.
Therefore, if we subtract remainder $r (x)$ from dividend $p (x)$ , then it will be exactly divisible by the divisor $g (x)$ ]
Let us divide, $6 x^{4} + 13 x^{3} + 13 x^{2} + 30 x + 20$ by $3 x^{2} + 2 x + 5$
we get, quotient $q (x) = 2 x^{2} + 3 x - 1$ remainder $r (x) = 17 x + 25$
$∴$ If we subtract $17 x + 25$ from $6 x^{4} + 13 x^{3} + 13 x^{2} + 30 x + 20$ , it will be exactly divisible by $(3 x^{2} + 2 x + 5)$ .

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