S sarrudav New member Joined Mar 27, 2010 Messages 5 Apr 11, 2010 #1 9x^2+12x+16 The entire problem is: x/9x^2+12x+16 - 3x+4/27x^3-64 I need to find the LCD by factoring.
9x^2+12x+16 The entire problem is: x/9x^2+12x+16 - 3x+4/27x^3-64 I need to find the LCD by factoring.
W wjm11 Senior Member Joined Nov 13, 2004 Messages 1,417 Apr 12, 2010 #2 9x^2+12x+16 Click to expand... Not factorable. Is it possible that you have a typo? Might it be 9x^2+24x+16? That is factorable to (3x + 4)^2. Or 9x^2-24x+16? That is factorable to (3x - 4)^2.
9x^2+12x+16 Click to expand... Not factorable. Is it possible that you have a typo? Might it be 9x^2+24x+16? That is factorable to (3x + 4)^2. Or 9x^2-24x+16? That is factorable to (3x - 4)^2.
S sarrudav New member Joined Mar 27, 2010 Messages 5 Apr 12, 2010 #3 Thank you. There is no chapter 4 in the whole book, so it is entirely possible that the rest of the book is not mistake-free.
Thank you. There is no chapter 4 in the whole book, so it is entirely possible that the rest of the book is not mistake-free.
D Deleted member 4993 Guest Apr 12, 2010 #4 sarrudav said: 9x^2+12x+16 The entire problem is: x/9x^2+12x+16 - 3x+4/27x^3-64 I need to find the LCD by factoring. Click to expand... \(\displaystyle \frac{x}{9x^2+12x+16} - \frac{3x+4}{(3x)^3 - (4)^3}\) \(\displaystyle = \frac{x}{9x^2+12x+16} - \frac{3x+4}{(3x - 4)(9x^2+12x+16)}\) \(\displaystyle = \frac{x(3x -4) - 3x - 4}{(3x - 4)(9x^2+12x+16)}\)
sarrudav said: 9x^2+12x+16 The entire problem is: x/9x^2+12x+16 - 3x+4/27x^3-64 I need to find the LCD by factoring. Click to expand... \(\displaystyle \frac{x}{9x^2+12x+16} - \frac{3x+4}{(3x)^3 - (4)^3}\) \(\displaystyle = \frac{x}{9x^2+12x+16} - \frac{3x+4}{(3x - 4)(9x^2+12x+16)}\) \(\displaystyle = \frac{x(3x -4) - 3x - 4}{(3x - 4)(9x^2+12x+16)}\)