1. **State the problem:** Factorize the cubic polynomial $$3x^3 + 26x^2 + 61x + 30$$. 2. **Recall the method:** To factor a cubic polynomial, we try to find rational roots using the Rational Root Theorem, then perform polynomial division or synthetic division to factor out the root. 3. **Possible rational roots:** Factors of the constant term 30 divided by factors of the leading coefficient 3 give possible roots: $$\pm1, \pm2, \pm3, \pm5, \pm6, \pm10, \pm15, \pm30, \pm\frac{1}{3}, \pm\frac{2}{3}, \pm\frac{5}{3}, \pm\frac{10}{3}$$. 4. **Test roots:** Substitute values into the polynomial to find a root. - For $$x = -2$$: $$3(-2)^3 + 26(-2)^2 + 61(-2) + 30 = 3(-8) + 26(4) - 122 + 30 = -24 + 104 - 122 + 30 = -12$$ (not zero). - For $$x = -3$$: $$3(-3)^3 + 26(-3)^2 + 61(-3) + 30 = 3(-27) + 26(9) - 183 + 30 = -81 + 234 - 183 + 30 = 0$$ (root found). 5. **Divide polynomial by $$x + 3$$:** Using synthetic division: Coefficients: 3 | 26 | 61 | 30 Bring down 3. Multiply by -3: 3 * -3 = -9; add to 26: 26 + (-9) = 17. Multiply 17 * -3 = -51; add to 61: 61 + (-51) = 10. Multiply 10 * -3 = -30; add to 30: 30 + (-30) = 0 remainder. Quotient polynomial: $$3x^2 + 17x + 10$$. 6. **Factor the quadratic:** $$3x^2 + 17x + 10$$. Find two numbers that multiply to $$3 \times 10 = 30$$ and add to 17: 15 and 2. Rewrite: $$3x^2 + 15x + 2x + 10 = 3x(x + 5) + 2(x + 5) = (3x + 2)(x + 5)$$. 7. **Final factorization:** $$3x^3 + 26x^2 + 61x + 30 = (x + 3)(3x + 2)(x + 5)$$. This is the fully factorized form.