1. **State the problem:** Factor each of the six given trinomials. 2. **Recall the factoring formula for perfect square trinomials:** - A trinomial of the form $a^2 - 2ab + b^2$ factors as $(a - b)^2$. - A trinomial of the form $a^2 + 2ab + b^2$ factors as $(a + b)^2$. 3. **Factor each trinomial step-by-step:** - For $a^2 - 20a + 100$: - Recognize $100 = 10^2$ and $-20a = -2 \times a \times 10$. - So, it factors as $(a - 10)^2$. - For $x^2 - 6x + 36$: - Note $36 = 6^2$ but $-6x \neq -2 \times x \times 6$ (which would be $-12x$), so this is not a perfect square. - Check discriminant: $\Delta = (-6)^2 - 4 \times 1 \times 36 = 36 - 144 = -108 < 0$, no real factors. - So, it cannot be factored over the reals. - For $4b^2 - 20b + 25$: - Recognize $4b^2 = (2b)^2$, $25 = 5^2$, and $-20b = -2 \times 2b \times 5$. - So, it factors as $(2b - 5)^2$. - For $16 + 12y + 9y^2$: - Rewrite as $9y^2 + 12y + 16$. - Check if perfect square: $9y^2 = (3y)^2$, $16 = 4^2$, but $12y \neq 2 \times 3y \times 4 = 24y$. - So, not a perfect square. - Check discriminant: $\Delta = 12^2 - 4 \times 9 \times 16 = 144 - 576 = -432 < 0$, no real factors. - For $c^2 - 2cd + d^2$: - Recognize $c^2 = c^2$, $d^2 = d^2$, and $-2cd = -2 \times c \times d$. - So, it factors as $(c - d)^2$. - For $4m + 2m^2 + 4$: - Rewrite as $2m^2 + 4m + 4$. - Factor out 2: $2(m^2 + 2m + 2)$. - Check discriminant of inner trinomial: $\Delta = 2^2 - 4 \times 1 \times 2 = 4 - 8 = -4 < 0$, no real factors. 4. **Final factorizations:** - $a^2 - 20a + 100 = (a - 10)^2$ - $x^2 - 6x + 36$ is prime over the reals. - $4b^2 - 20b + 25 = (2b - 5)^2$ - $16 + 12y + 9y^2$ is prime over the reals. - $c^2 - 2cd + d^2 = (c - d)^2$ - $4m + 2m^2 + 4 = 2(m^2 + 2m + 2)$, inner trinomial prime over the reals. This completes the factoring process with explanations.