1. **State the problem:** Solve the polynomial equation $$x^5 - 4x^4 + 5x^3 - 4x^2 + 4x = 0$$ for $x$. 2. **Factor out the common factor:** Notice every term contains $x$, so factor it out: $$x(x^4 - 4x^3 + 5x^2 - 4x + 4) = 0$$ 3. **Set each factor equal to zero:** - From $x = 0$, we get one root: $$x = 0$$ - Now solve the quartic: $$x^4 - 4x^3 + 5x^2 - 4x + 4 = 0$$ 4. **Try to factor the quartic:** Look for quadratic factors: Assume $$ (x^2 + ax + b)(x^2 + cx + d) = x^4 - 4x^3 + 5x^2 - 4x + 4 $$ Expanding: $$x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd = x^4 - 4x^3 + 5x^2 - 4x + 4$$ Match coefficients: - $a + c = -4$ - $ac + b + d = 5$ - $ad + bc = -4$ - $bd = 4$ 5. **Find integer factors of 4 for $b$ and $d$:** Possible pairs: (1,4), (2,2), (4,1), (-1,-4), (-2,-2), (-4,-1) Try $b=2$, $d=2$: - $bd=4$ correct - $ac + 2 + 2 = 5 ightarrow ac + 4 = 5 ightarrow ac = 1$ - $a + c = -4$ - $ad + bc = a*2 + b*c = 2a + 2c = -4$ From $a + c = -4$, multiply both sides by 2: $$2a + 2c = -8$$ But from above $2a + 2c = -4$, contradiction. Try $b=1$, $d=4$: - $bd=4$ correct - $ac + 1 + 4 = 5 ightarrow ac = 0$ - $a + c = -4$ - $ad + bc = 4a + c = -4$ From $a + c = -4$, $c = -4 - a$ Substitute into $4a + c = -4$: $$4a + (-4 - a) = -4 ightarrow 3a - 4 = -4 ightarrow 3a = 0 ightarrow a = 0$$ Then $c = -4 - 0 = -4$ Check $ac = 0 * (-4) = 0$ correct. 6. **So factors are:** $$(x^2 + 0x + 1)(x^2 - 4x + 4) = (x^2 + 1)(x^2 - 4x + 4)$$ 7. **Solve each quadratic:** - $x^2 + 1 = 0 ightarrow x^2 = -1 ightarrow x = \pm i$ - $x^2 - 4x + 4 = 0$ is a perfect square: $$ (x - 2)^2 = 0 ightarrow x = 2$$ 8. **Summary of roots:** $$x = 0, 2, 2, i, -i$$ **Final answer:** The solutions to the equation are $$x = 0, x = 2 \text{ (double root)}, x = i, x = -i$$.