1. **State the problem:** Factor the polynomial equation $$x^4 - 26x + 144 = 0$$. 2. **Understand the problem:** We want to express the quartic polynomial as a product of polynomials of lower degree, ideally quadratics or linear factors. 3. **Try to find rational roots using the Rational Root Theorem:** Possible roots are factors of 144 (constant term) over factors of 1 (leading coefficient), i.e., $$\pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm9, \pm12, \pm16, \pm18, \pm24, \pm36, \pm48, \pm72, \pm144$$. 4. **Test roots:** Substitute values to check if they satisfy the equation. - For $$x=2$$: $$2^4 - 26(2) + 144 = 16 - 52 + 144 = 108 \neq 0$$ - For $$x=3$$: $$81 - 78 + 144 = 147 \neq 0$$ - For $$x=4$$: $$256 - 104 + 144 = 296 \neq 0$$ - For $$x=6$$: $$1296 - 156 + 144 = 1284 \neq 0$$ - For $$x=8$$: $$4096 - 208 + 144 = 4032 \neq 0$$ - For $$x=9$$: $$6561 - 234 + 144 = 6471 \neq 0$$ - For $$x=12$$: $$20736 - 312 + 144 = 20568 \neq 0$$ Try negative values: - For $$x=-2$$: $$16 + 52 + 144 = 212 \neq 0$$ - For $$x=-3$$: $$81 + 78 + 144 = 303 \neq 0$$ - For $$x=-4$$: $$256 + 104 + 144 = 504 \neq 0$$ - For $$x=-6$$: $$1296 + 156 + 144 = 1596 \neq 0$$ No simple rational roots found. 5. **Try factoring as a product of quadratics:** Assume $$x^4 - 26x + 144 = (x^2 + ax + b)(x^2 + cx + d)$$ Expanding: $$x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd = x^4 - 26x + 144$$ Match coefficients: - Coefficient of $$x^3$$: $$a + c = 0$$ - Coefficient of $$x^2$$: $$ac + b + d = 0$$ - Coefficient of $$x$$: $$ad + bc = -26$$ - Constant term: $$bd = 144$$ 6. **From $$a + c = 0$$, let $$c = -a$$. Then:** - $$ac + b + d = -a^2 + b + d = 0 \Rightarrow b + d = a^2$$ - $$ad + bc = a d + b(-a) = a d - a b = a(d - b) = -26$$ - $$bd = 144$$ 7. **Try integer factor pairs of 144 for $$b$$ and $$d$$:** Possible pairs: (1,144), (2,72), (3,48), (4,36), (6,24), (8,18), (9,16), (12,12) and their negatives. 8. **Check pairs where $$a(d - b) = -26$$ is an integer:** Try $$b=9$$ and $$d=16$$: - $$b + d = 25 = a^2 \Rightarrow a = \pm 5$$ - $$a(d - b) = a(16 - 9) = 7a = -26 \Rightarrow a = -\frac{26}{7}$$ (not integer) Try $$b=12$$ and $$d=12$$: - $$b + d = 24 = a^2 \Rightarrow a = \pm \sqrt{24}$$ (not integer) Try $$b=18$$ and $$d=8$$: - $$b + d = 26 = a^2 \Rightarrow a = \pm \sqrt{26}$$ (not integer) Try $$b=24$$ and $$d=6$$: - $$b + d = 30 = a^2 \Rightarrow a = \pm \sqrt{30}$$ (not integer) Try $$b=16$$ and $$d=9$$: - $$b + d = 25 = a^2 \Rightarrow a = \pm 5$$ - $$a(d - b) = a(9 - 16) = a(-7) = -26 \Rightarrow a = \frac{26}{7}$$ (not integer) Try $$b=36$$ and $$d=4$$: - $$b + d = 40 = a^2 \Rightarrow a = \pm \sqrt{40}$$ (not integer) Try $$b=48$$ and $$d=3$$: - $$b + d = 51 = a^2 \Rightarrow a = \pm \sqrt{51}$$ (not integer) Try $$b=72$$ and $$d=2$$: - $$b + d = 74 = a^2 \Rightarrow a = \pm \sqrt{74}$$ (not integer) Try $$b=144$$ and $$d=1$$: - $$b + d = 145 = a^2 \Rightarrow a = \pm \sqrt{145}$$ (not integer) 9. **Try negative pairs:** Try $$b=-9$$ and $$d=-16$$: - $$b + d = -25 = a^2 \Rightarrow a = \pm 5i$$ (imaginary) Try $$b=-12$$ and $$d=-12$$: - $$b + d = -24 = a^2 \Rightarrow a = \pm \sqrt{-24}$$ (imaginary) Try $$b=-18$$ and $$d=-8$$: - $$b + d = -26 = a^2 \Rightarrow a = \pm \sqrt{-26}$$ (imaginary) Try $$b=-16$$ and $$d=-9$$: - $$b + d = -25 = a^2 \Rightarrow a = \pm 5i$$ - $$a(d - b) = a(-9 + 16) = a(7) = -26 \Rightarrow a = -\frac{26}{7}$$ (not imaginary) 10. **Since no integer or rational factorization is found, use substitution or numerical methods:** Try substitution $$x^2 = y$$, then equation becomes: $$y^2 - 26x + 144 = 0$$ which is not simpler. 11. **Use the quartic formula or numerical methods to find roots.** 12. **Alternatively, factor by grouping or synthetic division if a root is found.** 13. **Check for possible quadratic factors:** Try to factor as $$ (x^2 + px + q)^2 = x^4 + 2px^3 + (p^2 + 2q)x^2 + 2pqx + q^2$$ Match with $$x^4 - 26x + 144$$: - $$2p = 0 \Rightarrow p = 0$$ - $$p^2 + 2q = 0 \Rightarrow 2q = 0 \Rightarrow q = 0$$ - $$2pq = -26 \Rightarrow 0 = -26$$ (contradiction) So no perfect square factorization. 14. **Conclusion:** The polynomial does not factor nicely over the rationals. **Final answer:** The polynomial $$x^4 - 26x + 144 = 0$$ has no simple rational factorization. It can be solved numerically or using the quartic formula.