Factoring Harder Quadratics: Practice

The middle splits into two numbers with product $6 \cdot 4 = 24$ and sum $11$, namely $8$ and $3$. Pairing $3x$ with $1$ and $2x$ with $4$ gives cross products $3x$ and $8x$.

$(3x + 4)(2x + 1) = 6x^2 + 3x + 8x + 4 = 6x^2 + 11x + 4$

The cross products add to $11x$. The form $(6x + 1)(x + 4)$ gives $25x$.

The middle splits into two numbers with product $6 \cdot (-4) = -24$ and sum $-5$, namely $-8$ and $3$. Pairing $3x$ with $-4$ and $2x$ with $1$ gives cross products $3x$ and $-8x$.

$(3x - 4)(2x + 1) = 6x^2 + 3x - 8x - 4 = 6x^2 - 5x - 4$

The cross products add to $-5x$. The sign swap $(3x + 4)(2x - 1)$ gives $+5x$.

Every term is even, so pull out the $2$ first, leaving $3x^2 + 11x + 10$ inside.

$6x^2 + 22x + 20 = 2(3x^2 + 11x + 10) = 2(3x + 5)(x + 2)$

Inside, two numbers with product $3 \cdot 10 = 30$ and sum $11$ are $6$ and $5$, giving $(3x + 5)(x + 2)$. Dropping the $2$ leaves it incomplete.

For $6x^2 + 4x + 1$ the middle would split into two integers with product $6 \cdot 1 = 6$ and sum $4$, but the pairs multiplying to $6$ are $1, 6$ (sum $7$) and $2, 3$ (sum $5$).

$\text{no integer pair has product } 6 \text{ and sum } 4$

So it is irreducible. The others factor as $(3x + 2)(2x + 1)$, $(3x + 1)(2x + 3)$, and $(3x + 1)(2x + 1)$.

A root $x = \tfrac{2}{3}$ comes from the factor $3x - 2$, since $3x - 2 = 0$ gives $x = \tfrac{2}{3}$. A root $x = -4$ comes from $x + 4$.

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

Check: $x = \tfrac{2}{3}$ makes the first factor $0$ and $x = -4$ makes the second $0$.

The product $5$ is positive and the sum $-12$ is negative, so both constants are negative. Two numbers with product $4 \cdot 5 = 20$ and sum $-12$ are $-2$ and $-10$.

$(2x - 1)(2x - 5) = 4x^2 - 10x - 2x + 5 = 4x^2 - 12x + 5$

The cross products $-10x$ and $-2x$ add to $-12x$. The sign swap $(2x - 1)(2x + 5)$ gives $+8x$.

Two numbers with product $5 \cdot (-2) = -10$ and sum $-3$ are $-5$ and $2$, giving $(5x + 2)(x - 1)$.

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

Setting each factor to $0$ gives $x = -\tfrac{2}{5}$ or $x = 1$.

Every term is even, so pull out the $2$ first, leaving $6x^2 - x - 2$ inside.

$12x^2 - 2x - 4 = 2(6x^2 - x - 2) = 2(3x - 2)(2x + 1)$

Inside, two numbers with product $6 \cdot (-2) = -12$ and sum $-1$ are $-4$ and $3$, giving $(3x - 2)(2x + 1)$. Dropping the $2$ leaves it incomplete.

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

$6x^2 - 5x + 1 = (3x - 1)(2x - 1) = 0$

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

The middle splits into two numbers with product $6 \cdot 5 = 30$ and sum $17$, namely $15$ and $2$. Pairing $3x$ with $1$ and $2x$ with $5$ gives cross products $2x$ and $15x$.

$(3x + 1)(2x + 5) = 6x^2 + 15x + 2x + 5 = 6x^2 + 17x + 5$

The cross products add to $17x$. The form $(3x + 5)(2x + 1)$ gives $13x$.

Since $5$ is prime the skeleton is $(5x + \underline{\ })(x + \underline{\ })$. Two numbers with product $5 \cdot (-6) = -30$ and sum $13$ are $15$ and $-2$.

$(5x - 2)(x + 3) = 5x^2 + 15x - 2x - 6 = 5x^2 + 13x - 6$

The cross products $15x$ and $-2x$ add to $13x$, so this is correct.

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

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

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