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December 20, 2014

December 20, 2014

Posted by **Tammy** on Friday, August 27, 2010 at 12:29pm.

(25b^3+5b^2+34b+29)/(5b+3)

- Algebra -
**MathMate**, Friday, August 27, 2010 at 2:04pmYou would use long division for polynomials, very similar to the long division for numbers.

Each digit in the long division will be replaced by the coefficient of the powers of b.

(25b^3+5b^2+34b+29)

Dividing by 5b+3,

gives the first "digit" of the quotient as 5b²

and 5b²*3=15b²

Subtract 25b³+15b² from the original expression gives an expression one degree lower:

(25b^3+5b^2+34b+29) - 25b³+15b²

=-10b²+34b+29

Continuing this way will give the complete quotient.

See for reference:

http://www.sosmath.com/algebra/factor/fac01/fac01.html

- Algebra -
**Cody**, Friday, August 27, 2010 at 2:10pm25b^3+5b^2+34b+29

go to webmathcom, click algebra and under simplifying expressions click anything else and enter your promblem and hit try to simplify it and it will explain everything

- Cody - Algebra -
**Reiny**, Friday, August 27, 2010 at 2:48pmYour webpage did a lousy job of the problem,

your answer is not correct, it is simply the original numerator.

(How can the exponent of the first term stay the same ?????)

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