1. **State the problem:** Prove by contradiction that a polynomial equation with integer coefficients $$x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 = 0$$ cannot have a rational root unless that root is an integer. 2. **Assume the contrary:** Suppose the polynomial has a rational root $\frac{p}{q}$ in lowest terms, where $p$ and $q$ are integers with $\gcd(p,q) = 1$ and $q \neq 1$. 3. **Substitute the root:** Substitute $x = \frac{p}{q}$ into the polynomial: $$\left(\frac{p}{q}\right)^n + a_{n-1} \left(\frac{p}{q}\right)^{n-1} + \cdots + a_1 \frac{p}{q} + a_0 = 0$$ 4. **Multiply both sides by $q^n$ to clear denominators:** $$p^n + a_{n-1} p^{n-1} q + \cdots + a_1 p q^{n-1} + a_0 q^n = 0$$ 5. **Rearrange:** $$p^n = - (a_{n-1} p^{n-1} q + \cdots + a_1 p q^{n-1} + a_0 q^n)$$ 6. **Note divisibility:** The right side is divisible by $q$, so $q$ divides $p^n$. 7. **Since $p$ and $q$ are coprime, $q$ divides $p^n$ implies $q$ divides 1:** Because $\gcd(p,q) = 1$, $q$ must divide 1, so $q = \pm 1$. 8. **Conclusion:** The rational root $\frac{p}{q}$ is actually an integer. 9. **Integral root divisibility:** If $r$ is an integer root, substitute into the polynomial: $$r^n + a_{n-1} r^{n-1} + \cdots + a_1 r + a_0 = 0$$ Rearranged: $$a_0 = - (r^n + a_{n-1} r^{n-1} + \cdots + a_1 r)$$ Since the right side is divisible by $r$, $r$ divides $a_0$. 10. **Apply to (a):** $$x^4 + 6x^3 + 4x^2 + 5x + 4 = 0$$ Possible integer roots are divisors of 4: $\pm1, \pm2, \pm4$. Check each: - $x=1$: $1 + 6 + 4 + 5 + 4 = 20 \neq 0$ - $x=-1$: $1 - 6 + 4 - 5 + 4 = -2 \neq 0$ - $x=2$: $16 + 48 + 16 + 10 + 4 = 94 \neq 0$ - $x=-2$: $16 - 48 + 16 - 10 + 4 = -22 \neq 0$ - $x=4$: $256 + 384 + 64 + 20 + 4 = 728 \neq 0$ - $x=-4$: $256 - 384 + 64 - 20 + 4 = -80 \neq 0$ No integer roots, so no rational roots. 11. **Apply to (b):** $$x^4 + 5x^3 + 2x^2 - 10x + 6 = 0$$ Possible integer roots are divisors of 6: $\pm1, \pm2, \pm3, \pm6$. Check each: - $x=1$: $1 + 5 + 2 - 10 + 6 = 4 \neq 0$ - $x=-1$: $1 - 5 + 2 + 10 + 6 = 14 \neq 0$ - $x=2$: $16 + 40 + 8 - 20 + 6 = 50 \neq 0$ - $x=-2$: $16 - 40 + 8 + 20 + 6 = 10 \neq 0$ - $x=3$: $81 + 135 + 18 - 30 + 6 = 210 \neq 0$ - $x=-3$: $81 - 135 + 18 + 30 + 6 = 0$ So $x = -3$ is a root. 12. **Factor out $(x + 3)$ from (b):** Divide polynomial by $(x + 3)$: $$x^4 + 5x^3 + 2x^2 - 10x + 6 = (x + 3)(x^3 + 2x^2 - 4x + 2)$$ 13. **Find roots of cubic $x^3 + 2x^2 - 4x + 2 = 0$:** Possible integer roots: divisors of 2: $\pm1, \pm2$. Check: - $x=1$: $1 + 2 - 4 + 2 = 1 \neq 0$ - $x=-1$: $-1 + 2 + 4 + 2 = 7 \neq 0$ - $x=2$: $8 + 8 - 8 + 2 = 10 \neq 0$ - $x=-2$: $-8 + 8 + 8 + 2 = 10 \neq 0$ No integer roots, so no further rational roots. **Final answers:** - (a) No rational roots. - (b) Rational root is $x = -3$ only.