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?

Guest May 21, 2019

#1**+3 **

(x + 1)(2x + 3)(3x + 4) = (6x^{2} + 5)(x - 3)

Multiply (x + 1)(2x + 3) and (6x^{2} + 5)(x - 3)

(2x^{2} + 5x + 3 )(3x + 4) = 6x^{3} - 18x^{2} + 5x - 15

Multiply (2x^{2} + 5x + 3 )(3x + 4)

6x^{3} + 8x^{2} + 15x^{2} + 20x + 9x + 12 = 6x^{3} - 18x^{2} + 5x - 15

Combine like terms on the left side.

6x^{3} + 23x^{2} + 29x + 12 = 6x^{3} - 18x^{2} + 5x - 15

Subtract 6x^{3} from both sides of the equation.

23x^{2} + 29x + 12 = -18x^{2} + 5x - 15

Add 18x^{2} to both sides of the equation.

41x^{2} + 29x + 12 = 5x - 15

Subtract 5x from both sides.

41x^{2} + 24x + 12 = -15

Add 15 to both sides.

41x^{2} + 24x + 27 = 0

The GCF of 41, 24, and 27 is 1

0 = 41x^{2} + 24x + 27

Now it is in the form 0 = Ax^{2} + Bx + C where A, B, and C are positive integers with a GCF of 1.

A + B + C = 41 + 24 + 27 = 92

hectictar May 21, 2019