How do you divide #(2x^3-5x^2+7x+15)/(x-7) #?

To divide (2x^3-5x^2+7x+15) by (x-7), you can use long division. Here are the steps:

1. Divide the first term of the dividend (2x^3) by the first term of the divisor (x). The result is 2x^2.
2. Multiply the divisor (x-7) by the quotient obtained in step 1 (2x^2). The result is 2x^3-14x^2.
3. Subtract the product obtained in step 2 from the dividend (2x^3-5x^2+7x+15). This gives you -9x^2+7x+15.
4. Bring down the next term from the dividend, which is 7x.
5. Divide the first term of the new dividend (-9x^2) by the first term of the divisor (x). The result is -9x.
6. Multiply the divisor (x-7) by the quotient obtained in step 5 (-9x). The result is -9x^2+63x.
7. Subtract the product obtained in step 6 from the new dividend (-9x^2+7x+15). This gives you -56x+15.
8. Bring down the next term from the dividend, which is 15.
9. Divide the first term of the new dividend (-56x) by the first term of the divisor (x). The result is -56.
10. Multiply the divisor (x-7) by the quotient obtained in step 9 (-56). The result is -56x+392.
11. Subtract the product obtained in step 10 from the new dividend (-56x+15). This gives you -377.
12. Since there are no more terms to bring down, the division is complete.

The quotient is 2x^2 - 9x - 56, and the remainder is -377.