Factoring Harder Quadratics: Additional Practice (Set 1)

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 $-5x$.

Two numbers with product $3 \cdot (-8) = -24$ and sum $2$ are $6$ and $-4$, giving $(3x - 4)(x + 2)$.

$3x^2 + 2x - 8 = (3x - 4)(x + 2) = 0$

Setting each factor to $0$ gives $x = \tfrac{4}{3}$ or $x = -2$.

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

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

The cross products add to $11x$. The form $(4x + 2)(x + 3)$ hides a factor of $2$ and gives $14x$.

Since $5$ is prime the skeleton is $(5x + \underline{\ })(x + \underline{\ })$. Two numbers with product $5 \cdot 3 = 15$ and sum $8$ are $5$ and $3$.

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

The cross products $5x$ and $3x$ add to $8x$. The swap $(5x + 1)(x + 3)$ gives $16x$.

Two numbers with product $6 \cdot 2 = 12$ and sum $-7$ are $-3$ and $-4$, giving $(3x - 2)(2x - 1)$.

$6x^2 - 7x + 2 = (3x - 2)(2x - 1) = 0$

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

The middle splits into two numbers with product $6 \cdot 2 = 12$ and sum $7$, namely $3$ and $4$. Pairing $3x$ with $2$ and $2x$ with $1$ gives cross products $3x$ and $4x$.

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

The cross products add to $7x$. The form $(6x + 2)(x + 1)$ hides a factor of $2$ and gives $8x$.

Two numbers with product $4 \cdot 3 = 12$ and sum $-8$ are $-2$ and $-6$, giving $(2x - 1)(2x - 3)$.

$4x^2 - 8x + 3 = (2x - 1)(2x - 3) = 0$

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

The product $8$ is positive and the sum $-10$ is negative, so both constants are negative. Two numbers with product $3 \cdot 8 = 24$ and sum $-10$ are $-6$ and $-4$.

$(3x - 4)(x - 2) = 3x^2 - 6x - 4x + 8 = 3x^2 - 10x + 8$

The cross products $-6x$ and $-4x$ add to $-10x$. The swap $(3x - 2)(x - 4)$ gives $-14x$.

A root $x = -\tfrac{1}{2}$ comes from the factor $2x + 1$, since $2x + 1 = 0$ gives $x = -\tfrac{1}{2}$. A root $x = 5$ comes from $x - 5$.

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

Check: $x = -\tfrac{1}{2}$ makes the first factor $0$ and $x = 5$ makes the second $0$.

Every term contains an $x$, so pull out the common factor $x$ first, leaving $2x^2 + 7x + 3$ inside.

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

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

The middle splits into two numbers with product $4 \cdot (-15) = -60$ and sum $-4$, namely $-10$ and $6$.

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

The cross products $6x$ and $-10x$ add to $-4x$. The sign swap $(2x + 5)(2x - 3)$ gives $+4x$.

Expand each candidate and keep the one that matches. The pair with product $8 \cdot (-3) = -24$ and sum $2$ is $6$ and $-4$, giving cross products $6x$ and $-4x$.

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

The swap $(2x + 1)(4x - 3)$ gives $-2x$, so only $(2x - 1)(4x + 3)$ matches.