Solving Quadratics by Factoring: Practice

With $a \neq 1$, split the middle term using a pair with product $ac = 2 \cdot 3 = 6$ and sum $7$, namely $6$ and $1$.

$2x^2 + 6x + x + 3 = 2x(x + 3) + (x + 3) = (2x + 1)(x + 3)$

Then $2x + 1 = 0$ or $x + 3 = 0$, so $x = -\tfrac{1}{2}$ or $x = -3$. The solution set is $\left\{-3, -\tfrac{1}{2}\right\}$.

The axis of symmetry runs midway between the two zeros, at their average.

$x = \frac{-1 + 7}{2} = \frac{6}{2} = 3$

The zeros $-1$ and $7$ are mirror images across the axis, so the axis is their midpoint, $x = 3$.

Use $ac = 3 \cdot (-2) = -6$ with sum $-5$: the pair is $-6$ and $1$.

$3x^2 - 6x + x - 2 = 3x(x - 2) + (x - 2) = (3x + 1)(x - 2)$

Then $3x + 1 = 0$ or $x - 2 = 0$, so $x = -\tfrac{1}{3}$ or $x = 2$. The solution set is $\left\{-\tfrac{1}{3}, 2\right\}$.

Factor the trinomial, using product $-15$ and sum $-2$, which the pair $-5, 3$ gives.

$x^2 - 2x - 15 = (x - 5)(x + 3)$

Since $x - 5$ is one factor, the other is $x + 3$. It gives the second zero, $x = -3$.

This is a difference of squares, $(2x)^2 - 3^2$.

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

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

Each factor changes sign at its zero. Between the zeros, the factors have opposite signs, so their product is negative.

$x - 2 > 0 \ \text{ and } \ x - 6 < 0 \ \text{ for } \ 2 < x < 6$

So $f(x) < 0$ exactly on $2 < x < 6$. Outside the zeros both factors share a sign, making $f$ positive.

Pull out the greatest common factor $2$ first, then factor the simpler trinomial.

$2x^2 - 8x - 24 = 2(x^2 - 4x - 12) = 2(x - 6)(x + 2)$

The inner pair $-6, 2$ has product $-12$ and sum $-4$. Expanding $2(x - 6)(x + 2)$ returns the original, confirming the factorization.

A single zero buys a factor: if $f\left(\tfrac{2}{3}\right) = 0$, then $\left(x - \tfrac{2}{3}\right)$ divides $f$, and the leftover factor has rational coefficients.

$f(x) = \left(x - \tfrac{2}{3}\right)(\text{rational linear factor})$

So $f$ factors over the rationals. A rational zero forces a rational factorization; nothing here forces the other zero to be irrational.

The left side is a perfect square, since $25 = 5^2$ and $10x = 2 \cdot 5 \cdot x$.

$x^2 + 10x + 25 = (x + 5)^2 = 0$

Then $x + 5 = 0$, so $x = -5$ is a repeated root and the solution set is the single element $\{-5\}$.

Factored form displays the zeros directly, one per factor.

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

The zeros $3$ and $-5$ are visible at once. The vertex is not (it takes the midpoint and a substitution), and the $y$-intercept is $2(-3)(5) = -30$, not $0$.

Outside the two zeros, both factors share a sign, so their product is positive.

$x < 2: \ (-)(-) > 0; \qquad x > 6: \ (+)(+) > 0$

Between the zeros the factors disagree in sign, making $f$ negative. So $f(x) > 0$ for $x < 2$ or $x > 6$.

Move all terms to one side and factor; do not divide by $x$.

$5x^2 - 20x = 5x(x - 4) = 0$

The constant $5$ is never zero, so $x = 0$ or $x = 4$. The solution set is $\{0, 4\}$.