Factoring Harder Quadratics: Practice

The product $2 \cdot (-3) = -6$ is negative, so the middle splits into $6$ and $-1$, which sum to $5$. Build a candidate with those cross products.

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

The cross products $6x$ and $-x$ add to $5x$, so this is correct.

Now $a = 6$ can split as $3 \cdot 2$ as well as $6 \cdot 1$. The middle splits into two numbers with product $6 \cdot 2 = 12$ and sum $7$, namely $3$ and $4$, which the pair $(3x + 2)(2x + 1)$ produces.

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

The cross products $3x$ and $4x$ add to $7x$. The form $(6x + 1)(x + 2)$ gives $13x$ instead.

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

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

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

Every coefficient is a multiple of $3$, so pull the $3$ out first. That drops the leading coefficient inside from $6$ to $2$.

$6x^2 + 15x + 6 = 3(2x^2 + 5x + 2) = 3(2x + 1)(x + 2)$

Inside, $2x^2 + 5x + 2 = (2x + 1)(x + 2)$. Dropping the $3$ leaves the factoring incomplete.

The pair with product $-18$ and sum $7$ is $9$ and $-2$, so the middle strips are $9x$ and $-2x$. Pairing $3x$ with $3$ and $2x$ with $-1$ gives $(3x - 1)(2x + 3)$.

$(3x - 1)(2x + 3) = 6x^2 + 9x - 2x - 3 = 6x^2 + 7x - 3$

The cross products $9x$ and $-2x$ match the split. The sign swap $(3x + 1)(2x - 3)$ gives $-7x$.

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 $x = -2$.

Split $a = 4$ as $2 \cdot 2$. The middle splits into two numbers with product $4 \cdot 3 = 12$ and sum $8$, namely $6$ and $2$, which the pair $(2x + 1)(2x + 3)$ produces.

$(2x + 1)(2x + 3) = 4x^2 + 6x + 2x + 3 = 4x^2 + 8x + 3$

The cross products $6x$ and $2x$ add to $8x$. The form $(4x + 3)(x + 1)$ gives $7x$.

The product $4$ is positive and the sum $-7$ is negative, so both constants are negative. Two numbers with product $3 \cdot 4 = 12$ and sum $-7$ are $-3$ and $-4$.

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

The cross products $-3x$ and $-4x$ add to $-7x$. The swap $(3x - 1)(x - 4)$ gives $-13x$.

Two numbers with product $2 \cdot (-6) = -12$ and sum $1$ are $4$ and $-3$, giving $(2x - 3)(x + 2)$.

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

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

Move every term to one side: $3x^2 + 2x - 5 = 0$. Two numbers with product $3 \cdot (-5) = -15$ and sum $2$ are $5$ and $-3$, giving $(3x + 5)(x - 1)$.

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

Setting each factor to $0$ gives $x = -\tfrac{5}{3}$ or $x = 1$.

The area is the product of the two sides, so factor it and read off the other factor.

$6x^2 + 11x + 3 = (3x + 1)(2x + 3)$

Since one side is $3x + 1$, the other side is $2x + 3$.

Two numbers with product $3 \cdot 2 = 6$ and sum $-7$ are $-1$ and $-6$, giving $(3x - 1)(x - 2)$.

$3x^2 - 7x + 2 = (3x - 1)(x - 2) = 0$

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