Solving Quadratics by Factoring: Practice

The product of the two real numbers $x - 3$ and $x + 5$ is zero, so the Zero Product Property says at least one factor is zero.

$x - 3 = 0 \Rightarrow x = 3, \qquad x + 5 = 0 \Rightarrow x = -5$

Each factor $x - r$ vanishes at $r$, so the solutions are $3$ and $-5$, and the solution set is $\{-5, 3\}$.

This is a difference of squares, $x^2 - 3^2$, so it factors and the Zero Product Property applies.

$x^2 - 9 = (x - 3)(x + 3) = 0$

Then $x = 3$ or $x = -3$, giving the solution set $\{-3, 3\}$.

A graph meets the $x$-axis where its height is zero, so the $x$-intercepts are the zeros of the function.

$(x - 1)(x - 6) = 0 \Rightarrow x = 1 \ \text{ or } \ x = 6$

Zeros of the function, roots of the equation, and $x$-intercepts of the graph are three names for the same numbers: $1$ and $6$.

Look for two numbers whose product is $12$ and whose sum is $7$.

$3 \times 4 = 12, \qquad 3 + 4 = 7$

So $x^2 + 7x + 12 = (x + 3)(x + 4)$. The pair $3, 4$ matches both the constant term and the middle coefficient.

Factor the left side, then apply the Zero Product Property.

$x^2 + 7x + 12 = (x + 3)(x + 4) = 0$

Then $x + 3 = 0$ or $x + 4 = 0$, so $x = -3$ or $x = -4$. The solution set is $\{-4, -3\}$.

Find two numbers with product $6$ and sum $-5$: those are $-2$ and $-3$.

$x^2 - 5x + 6 = (x - 2)(x - 3) = 0$

So $x = 2$ or $x = 3$, giving the solution set $\{2, 3\}$.

A zero at $x = r$ corresponds to the factor $x - r$, because that factor vanishes exactly at $r$.

$f(5) = 0 \ \Rightarrow \ (x - 5) \text{ divides } f(x)$

Substituting $x = 5$ into $x - 5$ gives $0$, which drives the whole product to $0$. So $x - 5$ is the matching factor.

Move everything to one side so the Zero Product Property can be used, then factor the difference of squares.

$x^2 - 16 = (x - 4)(x + 4) = 0$

So $x = 4$ or $x = -4$. A positive square has two square roots, so the solution set is $\{-4, 4\}$, not just $\{4\}$.

The Zero Product Property applies only when a product of factors equals zero.

$(x - 2)(x + 3) = 0$

The first choice equals $5$, not $0$; the second is not factored and does not equal $0$; the last is an expression, not an equation. Only $(x - 2)(x + 3) = 0$ is ready to solve.

Do not divide by $x$; factor it out instead so no solution is lost.

$x^2 - 3x = x(x - 3) = 0$

Then $x = 0$ or $x = 3$. Dividing both sides by $x$ would have thrown away the solution $x = 0$. The solution set is $\{0, 3\}$.

Find two numbers with product $-6$ and sum $+1$.

$3 \times (-2) = -6, \qquad 3 + (-2) = 1$

So $x^2 + x - 6 = (x + 3)(x - 2)$. The other choices give the wrong middle term when expanded.

The product of the two real factors is zero, so at least one factor is zero.

$x + 6 = 0 \Rightarrow x = -6, \qquad x - 6 = 0 \Rightarrow x = 6$

This is a difference of squares, $x^2 - 36$, with the symmetric solution set $\{-6, 6\}$.