What must be added to $x^4+2x^3-2x^2+x-1$ so that the result is exactly divisible by $x^2+2x-3$?

To find what must be added so that $x^4+2x^3-2x^2+x-1$ is exactly divisible by $x^2+2x-3$, perform polynomial long division and find the remainder.

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Dividing $x^4+2x^3-2x^2+x-1$ by $x^2+2x-3$:

$x^4 \div x^2 = x^2$

$x^2(x^2+2x-3) = x^4+2x^3-3x^2$

$(x^4+2x^3-2x^2+x-1) - (x^4+2x^3-3x^2) = x^2+x-1$

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$x^2 \div x^2 = 1$

$1(x^2+2x-3) = x^2+2x-3$

$(x^2+x-1) - (x^2+2x-3) = -x+2$

Since the degree of $(-x+2)$ is less than the degree of $(x^2+2x-3)$, the division stops.

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The remainder is $-x+2$.

For exact divisibility, the remainder must be zero. To eliminate this remainder, add its negative:

What to add $= -(-x+2)$

What to add $= x-2$

Therefore, $x-2$ must be added.