Multiplying Binomials

[Temur]

Simplify the following:

. . . . .\(\displaystyle (x\, +\, 1)\, (2x\, +\, 3)\, (3x\,+\, 4)\, =\, (6x^2\, +\, 5)\, (x\, -\, 3)\)

to the following form:

. . . . .\(\displaystyle 0\, =\, Ax^2\, +\, Bx\, +\, C\)

where A, B, and C are positive integers with a greatest common divisor of 1.

What is the value of A + B + C ?

 

[Deleted member 4993]

Simplify the following:

. . . . .\(\displaystyle (x\, +\, 1)\, (2x\, +\, 3)\, (3x\,+\, 4)\, =\, (6x^2\, +\, 5)\, (x\, -\, 3)\)

to the following form:

. . . . .\(\displaystyle 0\, =\, Ax^2\, +\, Bx\, +\, C\)

where A, B, and C are positive integers with a greatest common divisor of 1.

What is the value of A + B + C ?

Expand the right-hand-side and the left-hand-side using distributive law of multiplication. Then collect similar terms and continue....

What are your thoughts?

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[Temur]

6x^3+5x-18x^2-15=6x^3+5x-18^2-15
I don't get what I a supposed to be doing

 

[stapel]

6x^3+5x-18x^2-15=6x^3+5x-18^2-15
I don't get what I a supposed to be doing

The instructions said to simplify to a particular form. Try moving terms around to reach that form.

 

[Steven G]

6x^3+5x-18x^2-15=6x^3+5x-18^2-15
I don't get what I a supposed to be doing

Your left hand side above and the original left hand side are not equal. Multiply out the left hand side correctly and then bring or terms to the left hand side (or to the right hand side) and then......