Solving Quadratics by Factoring: Practice

Both directions hold. If $f$ has a real zero $r$, then $x - r$ divides $f$ with a real leftover factor, so $f$ factors over the reals. Conversely, a real factorization $(px + q)(sx + t)$ gives the real zero $x = -\tfrac{q}{p}$.

$\text{real factorization} \ \Longleftrightarrow \ \text{real zero}$

Integer or unit leading coefficients are irrelevant, and "two distinct zeros" fails the repeated-root case $(x - r)^2$, which factors with only one zero.

Use $ac = 2 \cdot (-12) = -24$ with sum $5$: the pair is $8$ and $-3$.

$2x^2 + 8x - 3x - 12 = 2x(x + 4) - 3(x + 4) = (2x - 3)(x + 4)$

Then $2x - 3 = 0$ or $x + 4 = 0$, so $x = \tfrac{3}{2}$ or $x = -4$. The solution set is $\left\{-4, \tfrac{3}{2}\right\}$.

A zero makes $f$ vanish, so substitute $x = -3$ and solve for $k$.

$(-3)^2 + k(-3) - 6 = 9 - 3k - 6 = 3 - 3k = 0 \ \Rightarrow \ k = 1$

With $k = 1$, $f(x) = x^2 + x - 6 = (x + 3)(x - 2)$, confirming the zero at $-3$.

Factor out $x$, then treat the remaining difference of squares; do not divide by $x$.

$x^3 - 4x = x(x^2 - 4) = x(x - 2)(x + 2) = 0$

The Zero Product Property extends to three factors, so $x = 0$, $x = 2$, or $x = -2$. The solution set is $\{-2, 0, 2\}$.

A repeated root means the trinomial is a perfect square $(x - r)^2 = x^2 - 2rx + r^2$. Match the middle term: $-2r = -8$, so $r = 4$.

$c = r^2 = 4^2 = 16$

Then $x^2 - 8x + 16 = (x - 4)^2$, which has the single repeated root $x = 4$.

Move the $4$ across so one side is zero, then factor the difference of squares $(3x)^2 - 2^2$.

$9x^2 - 4 = (3x - 2)(3x + 2) = 0$

Then $3x = 2$ or $3x = -2$, so $x = \pm\tfrac{2}{3}$. The solution set is $\left\{-\tfrac{2}{3}, \tfrac{2}{3}\right\}$.

Dividing by $x$ is only valid when $x \neq 0$, but $x = 0$ satisfies the equation and is silently thrown away.

$x^2 - 9x = x(x - 9) = 0 \ \Rightarrow \ x = 0 \ \text{ or } \ x = 9$

Factoring keeps both solutions, so the full solution set is $\{0, 9\}$. Dividing by a quantity that might be zero destroys a factor.

The axis of symmetry is the midpoint of the two zeros.

$x = \frac{-4 + 6}{2} = \frac{2}{2} = 1$

The zeros are mirror images across the axis, so the axis sits halfway between them at $x = 1$.

Symmetric zeros $\pm k$ occur when the middle coefficient is zero, giving a difference of squares.

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

The others have zeros $\{4\}$ (repeated), $\{8, -2\}$, and $\{0, -4\}$, none of which is a symmetric pair about $0$.

Move the $9$ across to get a difference of squares equal to zero.

$(2x + 1)^2 - 9 = \big((2x + 1) - 3\big)\big((2x + 1) + 3\big) = (2x - 2)(2x + 4) = 0$

Then $2x - 2 = 0$ or $2x + 4 = 0$, so $x = 1$ or $x = -2$. The solution set is $\{-2, 1\}$.

Test each claim. $x^2 + 1 > 0$ always, so it has no real zeros (true). $x^2 - 2 = \big(x - \sqrt{2}\big)\big(x + \sqrt{2}\big)$ over the reals (true). A repeated root $(x - r)^2$ makes the graph tangent at $(r, 0)$ (true).

$x^2 - 2 \ \text{ has irrational zeros } \pm\sqrt{2}$

So $x^2 - 2$ does NOT factor over the rationals, which makes "every quadratic factors over the rationals" the false statement.

Use $ac = 3 \cdot 8 = 24$ with sum $10$: the pair is $6$ and $4$.

$3x^2 + 6x + 4x + 8 = 3x(x + 2) + 4(x + 2) = (3x + 4)(x + 2)$

Then $3x + 4 = 0$ or $x + 2 = 0$, so $x = -\tfrac{4}{3}$ or $x = -2$. The solution set is $\left\{-2, -\tfrac{4}{3}\right\}$.