How do you divide #(2x^3+4x^2-10x-9)÷(x-3)#?

Question

Eli Assouline · Accepted Answer

I found:
#(2x^3+4x^2-10x-9)-:(x-3)=2x^2+10x+20# with a remainder of #51#.

Have a look:

Nathan Axel · Answer

undefined To divide (2x^3+4x^2-10x-9) by (x-3), 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-3) by the quotient obtained in step 1 (2x^2). The result is 2x^3-6x^2.
3. Subtract the result obtained in step 2 from the dividend (2x^3+4x^2-10x-9) to get the new dividend: (4x^2-10x-9).
4. Repeat steps 1-3 with the new dividend (4x^2-10x-9).
5. Divide the first term of the new dividend (4x^2) by the first term of the divisor (x). The result is 4x.
6. Multiply the divisor (x-3) by the quotient obtained in step 5 (4x). The result is 4x^2-12x.
7. Subtract the result obtained in step 6 from the new dividend (4x^2-10x-9) to get the new dividend: (2x-9).
8. Repeat steps 1-3 with the new dividend (2x-9).
9. Divide the first term of the new dividend (2x) by the first term of the divisor (x). The result is 2.
10. Multiply the divisor (x-3) by the quotient obtained in step 9 (2). The result is 2x-6.
11. Subtract the result obtained in step 10 from the new dividend (2x-9) to get the remainder: (-3).

Therefore, the quotient is 2x^2+4x+2 and the remainder is -3.