(4a³b + 5a²b² + a⁴ + 2ab³)÷ (a² + 2b² + 3ab)
Please use the long division method for this problem.
Awfully hard to show algebraic long division in this text forum.
Google "algebraic long division" and follow the method shown by numerous excellent pages.
Make sure you write it down as
(a^4 + 4a^3b + 5a^2b^2 + 2ab^3) ÷ (a^2 + 3ab + 2b^2)
I had a+ab
Oops. That'd be a2+ab
To divide the expression (4a³b + 5a²b² + a⁴ + 2ab³) by (a² + 2b² + 3ab), you can use long division. Here's how to do it step by step:
Step 1: Divide the first term of the numerator by the first term of the denominator. In this case, divide 4a³b by a². The result is 4ab.
Step 2: Multiply the quotient (4ab) by the divisor (a² + 2b² + 3ab). The result is 4a³b + 8ab³ + 12a²b.
Step 3: Subtract the result obtained in Step 2 from the numerator. Subtract (4a³b + 8ab³ + 12a²b) from (4a³b + 5a²b² + a⁴ + 2ab³). The result is a⁴ + 5a²b² - 8ab³ - 12a²b.
Step 4: Bring down the next term from the numerator, which is 0.
Step 5: Repeat Steps 1-4 until you have divided all the terms of the numerator.
Step 6: Once you've gone through all the terms, you'll have the complete quotient.
In this case, the quotient is 4ab + 2b - 8ab² - 12ab.