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December 7, 2016
Posted by **Matt** on Monday, November 9, 2009 at 1:25pm.

- calculus -
**Jeff**, Monday, November 9, 2009 at 5:43pmIf you plug in x=1, you find that f(x)=4x^3-36x^2+96x-64 evaluates to zero. From here, use either synthetic division or long division to get the quadratic 4x^2-32x+64. Reduce to x^2-8x+16 and solve by factoring into (x-4)^2.

So 4x^3-36x^2+96x-64=4(x-1)(x-4)^2. - calculus -
**Thomas**, Monday, November 9, 2009 at 5:45pmThe 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^2-72x+96

0=12x^2-72x+96

0= x^2-6x+8

0=(x-4)(x-2)

The critical numbers are

x=4 and x=2 - calculus -
**Thomas Kilbride**, Monday, November 9, 2009 at 5:46pmIf you need to find the zeros of a cubic by factoring then the resault is:

4(x-4)^2-(x-1), so the zeros are at 4, and 1, and these are also the critical points.

You did not specify if 4x^3-36x^2+96x-64 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 second-degree polynomial. - calculus -
**TeacherJim**, Monday, November 9, 2009 at 5:53pmWell, 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.