How do you find the quotient of #(14y^5+21y^4-6y^3-9y^2+32y+48)div(2y+3)# using long division?

Question

Ethan Adkins · Accepted Answer

Quotient is #7y^4-3y+2+16# #14y^5+21y^4-6y^3-9y^2+32y+48# =#7y^4*(2y+3)-3y+2*(2y+3)+16*(2y+3)# =#(2y+3)*(7y^4-3y+2+16)# Hence quotient is #7y^4-3y+2+16# and remainder is #0#.

Genesis Allen · Answer

undefined To find the quotient of (14y^5+21y^4-6y^3-9y^2+32y+48) divided by (2y+3) using long division, follow these steps:

1. Divide the first term of the dividend (14y^5) by the first term of the divisor (2y). The result is 7y^4.

2. Multiply the divisor (2y+3) by the quotient obtained in step 1 (7y^4). The result is 14y^5+21y^4.

3. Subtract the product obtained in step 2 from the dividend (14y^5+21y^4-6y^3-9y^2+32y+48) to get the new dividend: -6y^3-9y^2+32y+48.

4. Bring down the next term from the original dividend, which is -6y^3.

5. Divide the first term of the new dividend (-6y^3) by the first term of the divisor (2y). The result is -3y^2.

6. Multiply the divisor (2y+3) by the quotient obtained in step 5 (-3y^2). The result is -6y^3-9y^2.

7. Subtract the product obtained in step 6 from the new dividend (-6y^3-9y^2+32y+48) to get the new dividend: 32y+48.

8. Bring down the next term from the new dividend, which is 32y.

9. Divide the first term of the new dividend (32y) by the first term of the divisor (2y). The result is 16.

10. Multiply the divisor (2y+3) by the quotient obtained in step 9 (16). The result is 32y+48.

11. Subtract the product obtained in step 10 from the new dividend (32y+48) to get the new dividend: 0.

12. Since the new dividend is 0, the long division process is complete.

The quotient of (14y^5+21y^4-6y^3-9y^2+32y+48) divided by (2y+3) is 7y^4-3y^2+16.