1. **Stating the problem:** We want to understand and work with trinomials of the form $ax^2 + bx + c$, where $a$, $b$, and $c$ are constants and $a \neq 0$. 2. **Formula and explanation:** This is a quadratic expression. The general form is: $$ax^2 + bx + c$$ where: - $a$ is the coefficient of $x^2$ (quadratic term), - $b$ is the coefficient of $x$ (linear term), - $c$ is the constant term. 3. **Important rules:** - The degree of the polynomial is 2 because the highest power of $x$ is 2. - The trinomial can be factored, solved, or graphed. - The quadratic formula to find roots (solutions) is: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ - The discriminant $\Delta = b^2 - 4ac$ tells us about the nature of roots: - If $\Delta > 0$, two distinct real roots. - If $\Delta = 0$, one real root (repeated). - If $\Delta < 0$, two complex roots. 4. **Example 1:** Factor $2x^2 + 7x + 3$ - Find two numbers that multiply to $2 \times 3 = 6$ and add to $7$: these are $6$ and $1$. - Rewrite middle term: $2x^2 + 6x + 1x + 3$ - Group terms: $(2x^2 + 6x) + (1x + 3)$ - Factor each group: $2x(x + 3) + 1(x + 3)$ - Factor out common binomial: $(x + 3)(2x + 1)$ 5. **Example 2:** Solve $x^2 - 4x - 5 = 0$ using quadratic formula - Identify $a=1$, $b=-4$, $c=-5$ - Calculate discriminant: $\Delta = (-4)^2 - 4(1)(-5) = 16 + 20 = 36$ - Find roots: $$x = \frac{-(-4) \pm \sqrt{36}}{2(1)} = \frac{4 \pm 6}{2}$$ - So, $$x_1 = \frac{4 + 6}{2} = 5$$ $$x_2 = \frac{4 - 6}{2} = -1$$ 6. **Summary:** Trinomials of the form $ax^2 + bx + c$ are quadratic expressions that can be factored or solved using the quadratic formula. Understanding the discriminant helps determine the nature of the roots.