What should be added in the polynomial $x^{3} + 2 x^{2} - 9 x + 1$ so that it is completely divisible by $x + 4$ ?

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Solution

$(x + 4)) x^{3} + 2 x^{2} - 9 x + 1 (x^{2} - 2 x - 1$
$x^{3} + 4 x^{2}$
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$- 2 x^{2} - 9 x$
$- 2 x^{2} - 8 x$
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$- x + 1$
$- x - 4$
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$5$ $=$ Remainder
$∴$ For making it completely divisible from $(x + 4)$
$(- 5)$ should be added in equation.
$∴ x^{3} + 2 x^{2} + 9 x + 1$ can be expressed as $(x^{2} - 2 x - 1) (x + 4) + 5$ .

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The expression that should be added to the polynomial f(x) = x⁴ + 2x³ – 2x² + x + 1, so that it should be exactly divisible by (x² + 2x – 3) is

f (x) = 4 x^{4} + 2 x^{3} - 2 x^{2} + x - 1

g (x) = x^{2} + 2 x - 3

x^{2} + 6 x + 5

x + 3

Find teh number which should be added to $x^{2} + x + 3$ so that the resulting polynomial is completely divisible by (x+3).

8 x^{4} + 14 x^{3} - 2 x^{2} + 7 x - 8

4 x^{2} + 3 x - 2

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