1. **State the problem:** Factor the polynomial $$x^4 + 6x^3 + 16x^2 + 18x + 7$$. 2. **Recall the formula and rules:** To factor a quartic polynomial, try to express it as a product of two quadratic polynomials: $$ (x^2 + ax + b)(x^2 + cx + d) $$. 3. **Set up the system:** Expanding the product gives: $$ x^4 + (a+c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd $$ Match coefficients with the original polynomial: - Coefficient of $x^3$: $a + c = 6$ - Coefficient of $x^2$: $ac + b + d = 16$ - Coefficient of $x$: $ad + bc = 18$ - Constant term: $bd = 7$ 4. **Find factors of 7:** Since 7 is prime, possible pairs for $(b,d)$ are $(1,7)$ or $(7,1)$ or $(-1,-7)$ or $(-7,-1)$. 5. **Try $(b,d) = (1,7)$:** - $bd = 7$ correct. - From $a + c = 6$ and $ac + 1 + 7 = 16$ so $ac = 8$. - From $ad + bc = 18$ we have $7a + 1c = 18$ or $7a + c = 18$. 6. **Solve system:** From $a + c = 6$, $c = 6 - a$. Substitute into $7a + c = 18$: $$7a + (6 - a) = 18 \Rightarrow 6a + 6 = 18 \Rightarrow 6a = 12 \Rightarrow a = 2$$ Then $c = 6 - 2 = 4$. Check $ac = 2 \times 4 = 8$ correct. 7. **Write factorization:** $$ (x^2 + 2x + 1)(x^2 + 4x + 7) $$ 8. **Verify:** Expand: $$ (x^2 + 2x + 1)(x^2 + 4x + 7) = x^4 + 4x^3 + 7x^2 + 2x^3 + 8x^2 + 14x + x^2 + 2x + 7 $$ Combine like terms: $$ x^4 + (4x^3 + 2x^3) + (7x^2 + 8x^2 + x^2) + (14x + 2x) + 7 = x^4 + 6x^3 + 16x^2 + 18x + 7 $$ Matches original polynomial. **Final answer:** $$\boxed{(x^2 + 2x + 1)(x^2 + 4x + 7)}$$