Use long division to divide $\left(x^{2}+3 x-3\right) \div(x+3)$. Find the quotient and remainder. - Quotient: $\square$ help (formulas) - Remainder: $\square$ help (formulas)

Step 1 ：Set up the long division. We are dividing \(x^{2}+3x-3\) by \(x+3\).

Step 2 ：Divide the first term in the dividend (\(x^{2}\)) by the first term in the divisor (\(x\)). This gives us \(x\). Write this above the long division line.

Step 3 ：Multiply the divisor (\(x+3\)) by the result from step 2 (\(x\)). This gives us \(x^{2}+3x\). Write this under the dividend and draw a line.

Step 4 ：Subtract the result from step 3 from the dividend. This gives us \(-3 - (x^{2}+3x) = -x^{2} - 6x\).

Step 5 ：Bring down the next term from the dividend. In this case, there are no more terms to bring down.

Step 6 ：Repeat steps 2-5 until the degree of the remainder is less than the degree of the divisor. In this case, we are already done because the degree of the remainder is less than the degree of the divisor.

Step 7 ：So, the quotient is \(x\) and the remainder is \(-x^{2} - 6x\).

Step 8 ：Check our work: If we multiply our quotient by our divisor and add our remainder, we should get our original dividend: \((x)(x+3) + (-x^{2} - 6x) = x^{2}+3x - x^{2} - 6x = x^{2}+3x-3\)

Step 9 ：Final Answer: Quotient: \(\boxed{x}\), Remainder: \(\boxed{-x^{2} - 6x}\)