Factoring Harder Quadratics: Practice

With $a = 2$ and $c = 3$ both prime, place $3$ and $1$ in the skeleton $(2x + \underline{\ })(x + \underline{\ })$ and expand each arrangement.

$(2x + 1)(x + 3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3$

The cross products $6x$ and $x$ add to $7x$, so this is the factorization. The swap $(2x + 3)(x + 1)$ gives $5x$ instead.

Since $3$ is prime the skeleton is $(3x + \underline{\ })(x + \underline{\ })$, and $c = 2$ offers $1$ and $2$. Test the arrangements.

$(3x + 1)(x + 2) = 3x^2 + 6x + x + 2 = 3x^2 + 7x + 2$

The cross products $6x$ and $x$ add to $7x$. The swap $(3x + 2)(x + 1)$ gives $5x$ instead.

By the zero-product property, set each factor to $0$.

$2x + 1 = 0 \quad \text{or} \quad x - 3 = 0$

The first gives $x = -\tfrac{1}{2}$, a fraction from dividing out the leading $2$; the second gives $x = 3$.

Set each factor equal to $0$.

$3x - 2 = 0 \quad \text{or} \quad x + 4 = 0$

The first gives $x = \tfrac{2}{3}$; the second gives $x = -4$. Each root is the opposite sign of the constant in its factor, divided by the leading number.

Place $1$ and $4$ in $(2x + \underline{\ })(x + \underline{\ })$ and expand.

$(2x + 1)(x + 4) = 2x^2 + 8x + x + 4 = 2x^2 + 9x + 4$

The cross products $8x$ and $x$ add to $9x$. The swap $(2x + 4)(x + 1)$ gives only $6x$.

Place $1$ and $5$ in $(2x + \underline{\ })(x + \underline{\ })$ and expand.

$(2x + 1)(x + 5) = 2x^2 + 10x + x + 5 = 2x^2 + 11x + 5$

The cross products $10x$ and $x$ add to $11x$. The swap $(2x + 5)(x + 1)$ gives only $7x$.

Every term is even, so pull out the common factor $2$ first, which shrinks the leading coefficient inside from $4$ to $2$.

$4x^2 + 14x + 6 = 2(2x^2 + 7x + 3) = 2(2x + 1)(x + 3)$

Inside, $2x^2 + 7x + 3 = (2x + 1)(x + 3)$. Dropping the $2$ leaves the factoring incomplete.

You need two numbers with product $6$ and sum $5$.

$2 + 3 = 5 \quad \text{and} \quad 2 \cdot 3 = 6$

So the pair is $2$ and $3$. The pair $1$ and $6$ multiplies to $6$ but adds to $7$, and $-2$ and $-3$ add to $-5$.

Two numbers with product $3 \cdot (-2) = -6$ and sum $-5$ are $-6$ and $1$, giving $(3x + 1)(x - 2)$.

$3x^2 - 5x - 2 = (3x + 1)(x - 2) = 0$

Setting each factor to $0$ gives $x = -\tfrac{1}{3}$ or $x = 2$.

Two numbers with product $3 \cdot 1 = 3$ and sum $4$ are $3$ and $1$, giving $(3x + 1)(x + 1)$.

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

The cross products $3x$ and $x$ add to $4x$, so this is correct.

Match the two expressions term by term. The squared term gives $pq = a$, the constant gives $rs = c$, and the middle term gives the cross-product sum.

$pq = a, \qquad rs = c, \qquad ps + qr = b$

The leading coefficient $a$ is the product of the two leading numbers, which is what makes a non-monic trinomial harder than a monic one.

Set each factor equal to $0$.

$2x + 3 = 0 \quad \text{or} \quad x + 2 = 0$

The first gives $x = -\tfrac{3}{2}$, a fraction from the leading $2$; the second gives $x = -2$.