What must be added to $x^{4} + 2 x^{3} - 2 x^{2} + x - 1$ . So that result is exactly divisible by $x^{2} + x - 2$ ?

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Solution

$x^{4} + 2 x^{3} - 2 x^{2} + x - 1 = x^{4} + x^{3} + x^{3} - 2 x^{2} + x^{2} - x^{2} + x + 1 = (x^{4} + x^{3} - 2 x^{2}) + (x^{3} + x^{2} - 2 x) - (x^{2} + x - 2) + (4 x - 3) = x^{2} (x^{2} + x - 2) + x (x^{2} + x - 2) - (x^{2} + x - 2) + (4 x - 3)$
Here we have to remove the last bracket terms $4 x - 3$ so that the above expression is divisible by the expression given in question
So we have to add $3 - 4 x$ to make the expression divisible

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Similar questions

4 x^{4} - 2 x^{3} - 6 x^{2} + x - 5

2 x^{2} + x - 1

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

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