0 0

# how do you find the zeros for f(x)=6x^4-13x^3+13x-6

need to know how to find zeros

Finding zeroes means that you are trying to find values along the x-axis where f(x) = 0. The number of zeroes is determined by the highest power of x in the equation. Therefore "x^4" part of the equation means there are 4 zeroes. To determine the values of the zeroes, break the x^0 coefficient (that is -6) down into its simplest multipliers (-1, 1, 2, -2, 3, -3, 6, -6) and break the x^4 coefficient (in this case 6) into its simplest multipliers which for this equation is the same. All the real possible answers are the multiples of the coefficient of the x^4 divided by the multiples of the x^0 coefficient. Possible zeroes therefore are (-1/-1, 1/-1, 2/-1, -2/-1, 3/-1, -3/-1, 6/-1, -6/-1, -1/2, 1/2, 2/2, -2/2, 3/2, -3/2, 6/2, -6/2, ..., -1/-6, 1/-6, 2/-6, -2/-6, 3/-6, -3/-6, 6/-6 and -6/-6) which simplifies to the following (-1, 1, 2, -2, 3, -3, 6, -6, -1/2, 1/2, -1/3, 1/3, -1/6, 1/6, 3/2, -3/2, 2/3, -2/3). Now plug in the values one at a time to see which values of x produce f(x)=0. Plug in x=1, f(x)=6-13+13-6=0. Therefore one of the zeroes is 1. Try x= 6, f(x) = 5,040. This is not a zero. The other zeroes are -1, 2/3, and 3/2.

You can use factor theorem.

Since f(+/-1) = 0, (x+1)(x-1) = x2 - 1 is a factor of f(x).

Now you can use synthetic division or long division or factoring to get the other factor.

By factoring,

6x^4-13x^3+13x-6

= 6x^4-6x^2-13x^3+13x + 6x^2-6

= 6x^2(x^2-1)-13x(x^2-1)+6(x^2-1)

= (x^2-1)(6x^2-13x+6)

= (x^2-1)(2x-3)(3x-2)

Solve (x^2-1)(2x-3)(3x-2) = 0

Answer: x = -1, 1, 2/3, 3/2

Without a program this is not easily solved.

It will not factor cleanly.

Using octave at home:

-2.558, ±i, .391