Posted by Matt on Monday, November 9, 2009 at 1:25pm.
I need to find the critical number of 4x^336x^2+96x64 by factoring. Basically I need to know the zeros.

calculus  Jeff, Monday, November 9, 2009 at 5:43pm
If you plug in x=1, you find that f(x)=4x^336x^2+96x64 evaluates to zero. From here, use either synthetic division or long division to get the quadratic 4x^232x+64. Reduce to x^28x+16 and solve by factoring into (x4)^2.
So 4x^336x^2+96x64=4(x1)(x4)^2. 
calculus  Thomas, Monday, November 9, 2009 at 5:45pm
The critical numbers are the values for x in which the function has a horizontal tangent line, so where the first derivative is zero.
F1(x)=12x^272x+96
0=12x^272x+96
0= x^26x+8
0=(x4)(x2)
The critical numbers are
x=4 and x=2 
calculus  Thomas Kilbride, Monday, November 9, 2009 at 5:46pm
If you need to find the zeros of a cubic by factoring then the resault is:
4(x4)^2(x1), so the zeros are at 4, and 1, and these are also the critical points.
You did not specify if 4x^336x^2+96x64 was your function, or the first derivative of the function, however if it is not then you should take the first derivative which should leave you a quadratic function, then use the quadratic formula on the remaining seconddegree polynomial. 
calculus  TeacherJim, Monday, November 9, 2009 at 5:53pm
Well, everything divides by 4, so you can take that out for a start, leaving
x^3  9x^2 + 24x  16
We need three numbers with a product of 16 (1, 2, 4, 8 and 16 are the only possibilities) that sum to 9.
Shouldn't be too hard to figure out from there.