SOLUTION: Divide. 6x3-7x2-9/ 3x+1 Please help thank you! that / is suppose to be in the middle of 6x3-7x2-9 and 3x+1 the 6x3 the 3 is suppose it be an exponent also the 7x2 the 2 is sup

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Divide. 6x3-7x2-9/ 3x+1 Please help thank you! that / is suppose to be in the middle of 6x3-7x2-9 and 3x+1 the 6x3 the 3 is suppose it be an exponent also the 7x2 the 2 is sup      Log On


   

Question 722429: Divide.
6x3-7x2-9/
3x+1
Please help thank you!
that / is suppose to be in the middle of 6x3-7x2-9 and 3x+1
the 6x3 the 3 is suppose it be an exponent also the 7x2 the 2 is suppose to be an exponent
**Sorry I dont know how to do exponents on this website.
Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
This is what you want:
%286x%5E3-7x%5E2-9%29%2F%283x%2B1%29, which is understood as numerator divided by denominator.

In case you are interested simply in polynomial division, long division, you would do this the same way as you would for ordinary long division. Show the dividend in its complete expanded form:

Dividend is: 6x%5E3-7x%5E2%2B0%2Ax-9.

Begin: How many times does 3x go into 6x%5E3? 2x%5E2 times. Multiply 2x%5E2 by 3x%2B1 and subtract from the dividend's 6x%5E3-7x%5E2. Bring down the next term, including its sign.
Second Partial Division: How many times does 3x go into the leading term of the current partial dividend? Continue the process...

My result is the quotient 2x%5E2-3x%2B1-10%2F%283x%2B1%29