Simple Trinomial

Solve: $x^2 + 7x + 12 = 0$

Identify the equation form: Already in standard form: $x^2 + 7x + 12 = 0$ = $a = 1$, $b = 7$, $c = 12$

Factor the trinomial: Find two numbers that multiply to 12 and add to 7: $3 \times 4 = 12$ and $3 + 4 = 7$ = $(x + 3)(x + 4) = 0$

Apply zero product property: If $(x + 3)(x + 4) = 0$, then $x + 3 = 0$ or $x + 4 = 0$ = Two equations to solve

Solve each equation: $x + 3 = 0 \Rightarrow x = -3$ and $x + 4 = 0 \Rightarrow x = -4$ = $x = -3$ or $x = -4$

Answer: $x = -3$ or $x = -4$

Difference of Squares

Solve: $x^2 - 25 = 0$

Recognize the pattern: $x^2 - 25 = x^2 - 5^2$ is a difference of squares = Use formula $a^2 - b^2 = (a+b)(a-b)$

Factor using the formula: $x^2 - 25 = (x + 5)(x - 5)$ = $(x + 5)(x - 5) = 0$

Apply zero product property: $x + 5 = 0$ or $x - 5 = 0$ = Two linear equations

Solve each equation: $x = -5$ or $x = 5$ = $x = \pm 5$

Answer: $x = 5$ or $x = -5$

Equation Not in Standard Form

Solve: $x^2 = 3x + 10$

Rewrite in standard form: $x^2 - 3x - 10 = 0$ (subtract $3x$ and $10$ from both sides) = Standard form achieved

Factor the trinomial: Find numbers that multiply to $-10$ and add to $-3$: $(-5) \times 2 = -10$ and $(-5) + 2 = -3$ = $(x - 5)(x + 2) = 0$

Apply zero product property: $x - 5 = 0$ or $x + 2 = 0$ = Two equations

Solve and verify: $x = 5$ or $x = -2$. Check: $5^2 = 25$ and $3(5) + 10 = 25$ ✓ = Both solutions verified

Answer: $x = 5$ or $x = -2$

Leading Coefficient Not 1

Solve: $2x^2 + 5x - 3 = 0$

Identify coefficients: $a = 2$, $b = 5$, $c = -3$ = Product $ac = 2 \times (-3) = -6$

Factor by grouping: Find numbers that multiply to $-6$ and add to $5$: $6 \times (-1) = -6$ and $6 + (-1) = 5$ = Split middle term: $2x^2 + 6x - x - 3$

Group and factor: $2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3)$ = $(2x - 1)(x + 3) = 0$

Solve each factor: $2x - 1 = 0 \Rightarrow x = \frac{1}{2}$ and $x + 3 = 0 \Rightarrow x = -3$ = $x = \frac{1}{2}$ or $x = -3$

Answer: $x = \frac{1}{2}$ or $x = -3$

GCF First

Solve: $3x^2 - 12x = 0$

Factor out GCF: $3x^2 - 12x = 3x(x - 4)$ = $3x(x - 4) = 0$

Apply zero product property: $3x = 0$ or $x - 4 = 0$ = Two equations

Solve first equation: $3x = 0 \Rightarrow x = 0$ = $x = 0$

Solve second equation: $x - 4 = 0 \Rightarrow x = 4$ = $x = 4$

Answer: $x = 0$ or $x = 4$

Mistake: Dividing both sides by $x$ (losing the solution $x = 0$)

Why: When you divide by $x$, you assume $x \neq 0$. But $x = 0$ might be a valid solution!

Correct: Factor out $x$ instead: $x^2 - 5x = 0 \Rightarrow x(x - 5) = 0$, giving $x = 0$ or $x = 5$

Mistake: Forgetting to set the equation equal to zero first

Why: The zero product property only works when the product equals zero, not any other number.

Correct: Always rewrite as $\text{expression} = 0$ before factoring. For example: $x^2 = 9$ becomes $x^2 - 9 = 0$

Mistake: Writing only one solution when there are two

Why: Quadratic equations can have two distinct solutions, and both are valid.

Correct: Always check both factors. $(x - 2)(x - 3) = 0$ gives TWO solutions: $x = 2$ AND $x = 3$

Mistake: Sign errors when factoring

Why: The signs in the factored form determine the signs of the solutions.

Correct: $(x + 3) = 0$ gives $x = -3$ (opposite sign). $(x - 3) = 0$ gives $x = 3$ (same sign)

Projectile Motion

When an object is thrown upward, its height follows a quadratic equation. Finding when it hits the ground means solving for when height equals zero.

A ball's height is $h = -t^2 + 4t + 5$ meters after $t$ seconds. When does it hit the ground? Solve $-t^2 + 4t + 5 = 0$, or equivalently $t^2 - 4t - 5 = 0$. Factoring: $(t - 5)(t + 1) = 0$, so $t = 5$ seconds (we ignore $t = -1$ since time cannot be negative).

Area Problems

Finding dimensions when you know the area often leads to quadratic equations.

A rectangular garden has length 3 meters more than its width. If the area is 40 square meters, find the dimensions. Let width = $x$. Then: $x(x + 3) = 40$, so $x^2 + 3x - 40 = 0$. Factoring: $(x + 8)(x - 5) = 0$, giving $x = 5$ meters (width). Length = 8 meters.

Q: What if the equation doesn't factor nicely?

A: Not all quadratic equations can be factored with integers. In those cases, use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Factoring works best when the solutions are rational numbers.

Q: Can a quadratic equation have just one solution?

A: Yes! This happens when the trinomial is a perfect square. For example, $x^2 - 6x + 9 = 0$ factors as $(x - 3)^2 = 0$, giving only $x = 3$ (a repeated root).

Q: Why do we set factors equal to zero?

A: Zero is special: it's the only number where if a product equals it, at least one factor must equal it. If $ab = 6$, there are infinite possibilities. But if $ab = 0$, either $a = 0$ or $b = 0$.