1. **Problem:** Find the zeroes of the polynomial equation $$x^8 - 26x^4 + 25 = 0$$ and factor it. 2. **Step 1: Recognize the substitution.** Let $$y = x^4$$. Then the equation becomes a quadratic in $$y$$: $$y^2 - 26y + 25 = 0$$ 3. **Step 2: Factor the quadratic.** We look for two numbers that multiply to 25 and add to -26. These are -25 and -1. $$y^2 - 26y + 25 = (y - 25)(y - 1) = 0$$ 4. **Step 3: Solve for $$y$$.** Set each factor equal to zero: $$y - 25 = 0 \Rightarrow y = 25$$ $$y - 1 = 0 \Rightarrow y = 1$$ 5. **Step 4: Substitute back $$x^4$$ for $$y$$.** $$x^4 = 25$$ $$x^4 = 1$$ 6. **Step 5: Solve each equation for $$x$$.** For $$x^4 = 25$$: $$x = \pm \sqrt{5}$$ because $$x^4 = (x^2)^2 = 25$$ implies $$x^2 = \pm 5$$ but $$x^2$$ must be nonnegative, so $$x^2 = 5$$. Therefore, $$x = \pm \sqrt{5}$$ For $$x^4 = 1$$: Similarly, $$x^2 = \pm 1$$ but $$x^2$$ must be nonnegative, so $$x^2 = 1$$. Therefore, $$x = \pm 1$$ 7. **Step 6: Write the complete factorization.** Since $$x^4 - 25 = (x^2 - 5)(x^2 + 5)$$ and $$x^4 - 1 = (x^2 - 1)(x^2 + 1)$$, the original polynomial factors as: $$x^8 - 26x^4 + 25 = (x^4 - 25)(x^4 - 1) = (x^2 - 5)(x^2 + 5)(x^2 - 1)(x^2 + 1)$$ 8. **Step 7: Further factor where possible.** $$x^2 - 1 = (x - 1)(x + 1)$$ $$x^2 + 1$$ and $$x^2 + 5$$ do not factor over the reals. 9. **Final answer:** $$x^8 - 26x^4 + 25 = (x - \sqrt{5})(x + \sqrt{5})(x^2 + 5)(x - 1)(x + 1)(x^2 + 1)$$ The zeroes are $$x = \pm \sqrt{5}, \pm 1$$ (real zeroes).