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Simplify (x+1)(2x+3)(3x+4) = (6x^{2}+5)(x-3)$ to the form $0 = Ax^{2}+Bx+C$, where A, B, and C are positive integers with a greatest common divisor of 1. What is A+B+C?

 May 21, 2019
 #1
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(x + 1)(2x + 3)(3x + 4)  =  (6x2 + 5)(x - 3)

                                                                                            Multiply  (x + 1)(2x + 3)  and  (6x2 + 5)(x - 3)

(2x2 + 5x + 3 )(3x + 4)  =  6x3 - 18x2 + 5x - 15

                                                                                            Multiply  (2x2 + 5x + 3 )(3x + 4)

6x3 + 8x2 + 15x2 + 20x + 9x + 12  =  6x3 - 18x2 + 5x - 15

                                                                                            Combine like terms on the left side.

6x3 + 23x2 + 29x + 12  =  6x3 - 18x2 + 5x - 15

                                                                            Subtract  6x3  from both sides of the equation.

23x2 + 29x + 12  =  -18x2 + 5x - 15

                                                                            Add  18x2  to both sides of the equation.

41x2 + 29x + 12  =  5x - 15

                                                                            Subtract  5x  from both sides.

41x2 + 24x + 12  =  -15

                                                                            Add  15  to both sides.

41x2 + 24x + 27  =  0

                                                                            The GCF of  41, 24, and 27  is  1

0  =  41x2 + 24x + 27

 

Now it is in the form   0  =  Ax2 + Bx + C  where  A, B, and C are positive integers with a GCF of 1.

 

A + B + C   =   41 + 24 + 27   =   92

 May 21, 2019

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