Roots of $x^{4} -28 x^{2}+49$ with Horner

Question

I am studying Horner's algorithm and I got a problem I can't solve.

The polynomal is $x^{4} -28 x^{2}+49$. After trying $\pm 1, \pm 7, \pm49$ with Horner I couldn't find any solution. Wolfram alpha gives solutions that include squares and I can't see how could I come to such solution. What's could be my next step?

Thanks in advance :)

Answer 1

Hint:

Set $u = x^{2}$ if you make this change you will have:

$$u^{2} - 28 u +49$$

from here you can use the quadratic formula to factorize it and solve it, when you solve it for $u$ remember to go back to $x$

Answer 2

Put $x^2=y$ now the equation becomes quadratic so solve for y. Then after getting two roots now substitute $x^2$ in place of y and again solve for x and get the roots of the given polynomial.

Answer 3

In general, a case like this is fairly easy. These are in fact known as biquadratic equations, or in other words when you have only $x^4$, $x^2$ and a constant term. In this case all you have to do is to let $t=x^2$ and solve the auxiliary equation.

However, if you interested in using Horner's algorithm, you first need to identify the zeros of the function. To do so, you need to use the rational root theorem, see here in case you don't know it. This solves our problem for any rational root. Nonetheless, we still have a problem for any other root. For that, this page will give you the method to work those out using Horner's rule.