Solving Quadratics by Factoring: Additional Practice (Set 1)

Use $ac = 4 \cdot (-3) = -12$ with sum $-4$: the pair is $-6$ and $2$.

$4x^2 - 6x + 2x - 3 = 2x(2x - 3) + (2x - 3) = (2x + 1)(2x - 3)$

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

Set each factor to zero.

$3x - 2 = 0 \Rightarrow x = \tfrac{2}{3}, \qquad x + 7 = 0 \Rightarrow x = -7$

The solutions are $\tfrac{2}{3}$ and $-7$. Of the choices, only $x = \tfrac{2}{3}$ is one of them.

The zeros are $-4$ and $1$, and the parabola opens upward, so it is above the axis outside the zeros.

$x < -4: \ (-)(-) > 0; \qquad x > 1: \ (+)(+) > 0$

Between $-4$ and $1$ the factors disagree in sign, so $f < 0$ there. Thus $f > 0$ for $x < -4$ or $x > 1$.

A zero at $4$ gives the factor $x - 4$; find the partner from product $20$ and sum $-9$.

$x^2 - 9x + 20 = (x - 4)(x - 5)$

The pair $-4, -5$ has product $20$ and sum $-9$, so the second zero is $x = 5$.

A real-but-not-rational factorization needs real irrational zeros. Check $x^2 - 5$.

$x^2 - 5 = 0 \ \Rightarrow \ x = \pm\sqrt{5}$

Those are real and irrational, so $x^2 - 5 = \big(x - \sqrt{5}\big)\big(x + \sqrt{5}\big)$ over the reals only. The others have no real zeros ($x^2 + 1$) or rational zeros ($(x + 1)^2$, $(x - 2)(x - 3)$).

Factor out the common $2$, then use the difference of squares.

$2x^2 - 50 = 2(x^2 - 25) = 2(x - 5)(x + 5) = 0$

The factor $2$ is never zero, so $x = 5$ or $x = -5$. The solution set is $\{-5, 5\}$.

Build the factored form and clear the fraction: $2\left(x - \tfrac{1}{2}\right)(x + 3) = (2x - 1)(x + 3)$.

$(2x - 1)(x + 3) = 2x^2 + 6x - x - 3 = 2x^2 + 5x - 3$

The leading coefficient $2$ absorbs the $\tfrac{1}{2}$ into an integer factor, giving standard form $2x^2 + 5x - 3$.

The expression stays positive for every real $x$, so it never equals zero. For instance, for $x \ge 0$ each term is nonnegative, and for $-1 < x < 0$ we can write it as $x^2 + (x + 1)$ with both parts positive.

$x^2 + x + 1 > 0 \ \text{ for all real } x$

With no real zero, it has no real linear factorization, so it is not a perfect square and does not factor over the rationals.

Use $ac = 6 \cdot (-10) = -60$ with sum $11$: the pair is $15$ and $-4$.

$6x^2 + 15x - 4x - 10 = 3x(2x + 5) - 2(2x + 5) = (3x - 2)(2x + 5)$

Then $3x - 2 = 0$ or $2x + 5 = 0$, so $x = \tfrac{2}{3}$ or $x = -\tfrac{5}{2}$. The solution set is $\left\{\tfrac{2}{3}, -\tfrac{5}{2}\right\}$.

Apply the Zero Product Property to the two quadratic factors separately.

$x^2 - 4 = 0 \Rightarrow x = \pm 2; \qquad x^2 + 4 = 0 \Rightarrow \text{no real } x$

Only $x^2 - 4$ contributes real solutions, since $x^2 + 4 > 0$ for every real $x$. So there are two real solutions, $-2$ and $2$.

Between the zeros $1$ and $4$, the two factors have opposite signs, so their product is negative.

$1 < x < 4: \ (x - 1) > 0 \ \text{ and } \ (x - 4) < 0 \ \Rightarrow \ f(x) < 0$

Outside the zeros both factors share a sign, so $f > 0$ there. Thus $f(x) < 0$ exactly on $1 < x < 4$.

Recognize a perfect square: $25x^2 = (5x)^2$, $4 = 2^2$, and $20x = 2 \cdot 5x \cdot 2$.

$25x^2 - 20x + 4 = (5x - 2)^2 = 0 \ \Rightarrow \ 5x - 2 = 0$

Then $x = \tfrac{2}{5}$ is a repeated root, so the solution set is $\left\{\tfrac{2}{5}\right\}$.