Factor completely: . . .4x^3 - 11x^2 - 6x + 9 I just don't know how to get started.
L Laila New member Joined Jul 10, 2006 Messages 8 Jul 10, 2006 #1 Factor completely: . . .4x^3 - 11x^2 - 6x + 9 I just don't know how to get started.
pka Elite Member Joined Jan 29, 2005 Messages 11,995 Jul 10, 2006 #2 The roots are: −1,3,34.\displaystyle - 1,3,\frac{3}{4}.−1,3,43.
L Laila New member Joined Jul 10, 2006 Messages 8 Jul 10, 2006 #3 pka said: The roots are: −1,3,34.\displaystyle - 1,3,\frac{3}{4}.−1,3,43. Click to expand... How did you find that?
pka said: The roots are: −1,3,34.\displaystyle - 1,3,\frac{3}{4}.−1,3,43. Click to expand... How did you find that?
pka Elite Member Joined Jan 29, 2005 Messages 11,995 Jul 10, 2006 #4 One finds roots by factoring! RIGHT? So if we know the roots the we know the factors. RIGHT?
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,582 Jul 10, 2006 #5 Re: factoring 4x^3 - 11x^2 - 6x + 9 completely Laila said: i just dont know how to get started Click to expand... I would suggest starting with the Rational Roots Test, and then applying synthetic division to the potential roots that the Test gives you. If you get stuck in this process, or if you are supposed to use some other methodology, please reply showing all of your progress so far. Thank you. Eliz.
Re: factoring 4x^3 - 11x^2 - 6x + 9 completely Laila said: i just dont know how to get started Click to expand... I would suggest starting with the Rational Roots Test, and then applying synthetic division to the potential roots that the Test gives you. If you get stuck in this process, or if you are supposed to use some other methodology, please reply showing all of your progress so far. Thank you. Eliz.