Basic Two-Factor Equation

Solve $(x - 5)(x + 3) = 0$

Apply Zero Product Property: If $(x - 5)(x + 3) = 0$, then one factor must equal zero = $x - 5 = 0$ or $x + 3 = 0$

Solve first equation: $x - 5 = 0 \Rightarrow x = 5$ = $x = 5$

Solve second equation: $x + 3 = 0 \Rightarrow x = -3$ = $x = -3$

State both solutions: The equation has two solutions = $x = 5$ or $x = -3$

Answer: $x = 5$ or $x = -3$

Equation with Common Factor

Solve $x(x - 7) = 0$

Identify the factors: The factors are $x$ and $(x - 7)$ = Two factors

Apply Zero Product Property: If $x(x - 7) = 0$, then $x = 0$ or $x - 7 = 0$ = Set each factor to zero

Solve each equation: $x = 0$ (already solved) and $x - 7 = 0 \Rightarrow x = 7$ = $x = 0$ or $x = 7$

Verify solutions: $0(0-7) = 0(-7) = 0$ \checkmark and $7(7-7) = 7(0) = 0$ \checkmark = Both solutions check

Answer: $x = 0$ or $x = 7$

Equation Needing Factoring First

Solve $x^2 - 4x - 21 = 0$

Factor the trinomial: Find factors of $-21$ that add to $-4$: $(-7) + 3 = -4$ = $(x - 7)(x + 3) = 0$

Apply Zero Product Property: $(x - 7)(x + 3) = 0$ means $x - 7 = 0$ or $x + 3 = 0$ = Two equations

Solve first equation: $x - 7 = 0 \Rightarrow x = 7$ = $x = 7$

Solve second equation: $x + 3 = 0 \Rightarrow x = -3$ = $x = -3$

State both solutions: Check: $(7)^2 - 4(7) - 21 = 49 - 28 - 21 = 0$ \checkmark = $x = 7$ or $x = -3$

Answer: $x = 7$ or $x = -3$

Three-Factor Equation

Solve $x(x + 2)(x - 4) = 0$

Apply Zero Product Property: With three factors, at least one must be zero = $x = 0$ or $x + 2 = 0$ or $x - 4 = 0$

Solve each equation: $x = 0$, $x = -2$, $x = 4$ = Three solutions

State all solutions: The cubic equation has three roots = $x = 0$, $x = -2$, or $x = 4$

Answer: $x = 0$, $x = -2$, or $x = 4$

Equation with Coefficient

Solve $2x^2 + 10x = 0$

Factor out the GCF: Both terms have $2x$ as a common factor = $2x(x + 5) = 0$

Apply Zero Product Property: If $2x(x + 5) = 0$, then $2x = 0$ or $x + 5 = 0$ = Two equations

Solve first equation: $2x = 0 \Rightarrow x = 0$ = $x = 0$

Solve second equation: $x + 5 = 0 \Rightarrow x = -5$ = $x = -5$

Verify: $2(0)^2 + 10(0) = 0$ \checkmark and $2(-5)^2 + 10(-5) = 50 - 50 = 0$ \checkmark = $x = 0$ or $x = -5$

Answer: $x = 0$ or $x = -5$

Mistake: Dividing both sides by a variable factor

Why: If you divide by $x$, you lose the solution $x = 0$. For example, in $x(x-3) = 0$, dividing by $x$ gives only $x = 3$, missing $x = 0$.

Correct: Never divide by a variable. Instead, use the zero product property to get ALL solutions.

Mistake: Applying to equations not equal to zero

Why: The property ONLY works when the product equals zero. $(x-2)(x+1) = 6$ does NOT mean $x-2 = 6$ or $x+1 = 6$.

Correct: First move all terms to one side to get $= 0$, then factor, then apply the property.

Mistake: Forgetting to factor first

Why: The equation $x^2 - 5x + 6 = 0$ is not yet in factored form. You cannot apply the property directly.

Correct: Factor to $(x-2)(x-3) = 0$ first, then apply the zero product property.

Mistake: Missing solutions when one factor is just $x$

Why: In $x(x-4) = 0$, students sometimes only find $x = 4$ and forget that $x = 0$ is also a solution.

Correct: Remember: $x$ by itself is a factor too! If $x = 0$, the product is zero.

Projectile Motion

When you throw a ball, its height follows a quadratic path. Finding when height equals zero tells you when it lands.

A ball's height is modeled by $h = -16t^2 + 48t$ feet. Setting $h = 0$ and factoring: $-16t(t - 3) = 0$, so $t = 0$ (launch) or $t = 3$ seconds (landing).

Business Break-Even Analysis

Companies use quadratic equations to model profit. The zero product property finds break-even points where profit equals zero.

If profit $P = x^2 - 100x$ euros where $x$ is units sold, then $x(x - 100) = 0$ gives break-even at $x = 0$ or $x = 100$ units.

Q: Why doesn't this work for products that equal other numbers?

A: Zero is special! If $ab = 12$, we can't conclude anything about $a$ or $b$ individually (could be $3 \times 4$, $2 \times 6$, $1 \times 12$, etc.). But for $ab = 0$, at least one factor MUST be zero.

Q: What if the same factor appears twice?

A: Like $(x-3)^2 = 0$? You still set $x - 3 = 0$, giving $x = 3$. This is called a 'repeated root' or 'double root' - the parabola touches but doesn't cross the x-axis there.

Q: Can there be more than two solutions?

A: Yes! A polynomial of degree $n$ can have up to $n$ solutions. For example, $x(x-1)(x+2)(x-3) = 0$ has four solutions: $x = 0, 1, -2, 3$.