How would you find the zeros of this polynomial?

Question

-x^6+3x^4-48x^2+16=0

Answer 1

Angel, sometimes you just have to use a graphing calculator to find the approximate zeros. I used MATHCAD to determine the zeros, and got.

 

x1≈-0.583 and x2≈0.583

 

I don't know if you are familiar Newton's method in determining the zeros of a function. This is where you could use calculus. First of all, let

 

f(x)=-x^6+3x^4-48x^2+16.

 

Newton's method involves the following steps:

 

1) Take the derivative of the function

2) Find the equation of the line that passes through the point tangent to the curve of the original function in which its slope is the derivative at the point of the curve.

3) Determine what value of x in which the line passes through the x-axis.

 

The general formula is:

 

x_n+1=x_n- f(x_n)/f'(x_n)

 

Start with x₀=1, you'll get x₁=0.667. Since this value is less than the one guessed, then

 

x₂=0.667-f(0.667)/f'(0.667) Just keep plugging in the numbers until you get a number close to zero.

 

Hope this helps.