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DIVIDE: (6x^2+13x-4)/(2x+5)

asked Dec 11, 2014 in ALGEBRA 1 by anonymous

1 Answer

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The expression (6x2 + 13x - 4)/(2x + 5)

Divide the first term of the dividend by the first term of the divisor (6x2)/2x = 3x.

So, the first term of the quotient is x3. Multiply (x - 3) by x3 and subtract.

2x + 5) 6x2 + 13x - 4 (3x

           6x2 + 15x

(-) ______________________

            - 2x - 4         

Divide the first term of the last row by first term of the divisor (- 2x)/(2x) = - 1.

So,the second term of the quotient is - 1. Multiply (2x + 5) by (- 1) and subtract.

2x + 5) 6x2 + 13x - 4 (3x - 1

           6x2 + 15x

(-) ______________________

            - 2x - 4 

           - 2x - 5

(-) ______________________

                 1

The result of the division is (6x2 + 13x - 4)/(2x + 5) = (3x - 1) + 1/(2x + 5).

The result of the division is (6x2 + 13x - 4)/(2x + 5) = (3x - 1)(2x + 5) + 1.

answered Dec 11, 2014 by david Expert

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