x to the power of 4 minus 13x to the power of 2 plus 36

The expression given is:

\[
x^4 - 13x^2 + 36
\]

This is a quadratic form in terms of \(x^2\). We can solve it by substituting \(y = x^2\), which transforms the expression into:

\[
y^2 - 13y + 36
\]

Now, let's solve the quadratic equation \(y^2 - 13y + 36 = 0\) by factoring:

\[
y^2 - 13y + 36 = (y - 9)(y - 4) = 0
\]

So, \(y = 9\) or \(y = 4\).

Since \(y = x^2\), we substitute back:

1. \(x^2 = 9\) gives \(x = \pm 3\).
2. \(x^2 = 4\) gives \(x = \pm 2\).

Thus, the solutions for \(x\) are:

\[
x = \pm 3, \pm 2
\]

Would you like more details on the solving process or have any questions?

Here are some related questions:

1. How do you factor quadratic equations?
2. What is the quadratic formula, and when should it be used?
3. Can you explain how to solve higher-degree polynomials?
4. What are the differences between solving quadratic equations and quartic equations?
5. How does substitution simplify solving polynomial equations?

**Tip:** Always check all potential solutions in the original equation to ensure they are valid, especially in higher-degree polynomials.

Learn how to solve the polynomial equation x^4 - 13x^2 + 36 step-by-step. This problem involves factoring a quadratic polynomial derived from the original equation, suitable for advanced algebra students in Grades 9-12.

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