Factoring Quadratic Trinomials (when a = -1) - Algebra-1

1. Fundamental Concepts

Quadratic Trinomial Definition: A polynomial of the form $-x^2 + bx + c$ (where $a = -1$，the coefficient of $x^2$ is $-1$；b is the coefficient of the linear term x；c is the constant term，and x is the variable).
Factoring Goal: Rewrite the trinomial by first factoring out $-1$ to convert it to the form $-(x^2 - bx - c)$，then further factor the resulting quadratic trinomial (with $a = 1$) into a product of two linear binomials.

Core Step: Factor out $-1$ first: Since $a = -1$，factoring out $-1$ from the entire trinomial flips the signs of all terms，turning it into $-(x^2 - bx - c)$，which is a standard quadratic trinomial with $a = 1$ (inside the parentheses).
Reuse $a = 1$ factoring rules: After factoring out $-1$，the inner trinomial $x^2 - bx - c$ follows the rules for $a = 1$：find integers m and n such that $m + n = -b$ and $mn = -c$，so $x^2 - bx - c=(x + m)(x + n)$，and the final factored form is $-(x + m)(x + n)$.

Problem: Factor $-x^2 + 5x + 6$
Solution:
1. Factor out $-1$：$-(x^2 - 5x - 6)$；
2. Factor $x^2 - 5x - 6$：find $m, n$ where $m + n = -5$，$mn = -6$ (pairs: $-6$ and 1，since $-6 + 1 = -5$，$-6×1 = -6$)；
3. Final result：$-(x - 6)(x + 1)$.

Problem: Factor $-x^2 - 2x + 8$
Solution:
1. Factor out $-1$：$-(x^2 + 2x - 8)$；
2. Factor $x^2 + 2x - 8$：find $m, n$ where $m + n = 2$，$mn = -8$ (pairs: 4 and $-2$，since $4 + (-2) = 2$，$4×(-2) = -8$)；
3. Final result：$-(x + 4)(x - 2)$.

Visual Strategy: Use a box method or a diamond problem to visualize factor pairs.
Error-Proofing: Double-check the signs and coefficients after factoring.
Concept Reinforcement: Practice with various values of $b$ and $c$ to reinforce understanding.