Factoring Harder Quadratics: Additional Practice (Set 2)

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

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

Setting each factor to $0$ gives $x = \tfrac{3}{2}$ or $x = -\tfrac{1}{2}$. Both roots are fractions because each factor has a leading $2$.

For $4x^2 + 5x + 3$ the middle would split into two integers with product $4 \cdot 3 = 12$ and sum $5$, but the pairs multiplying to $12$ are $1, 12$; $2, 6$; and $3, 4$, with sums $13$, $8$, and $7$.

$\text{no integer pair has product } 12 \text{ and sum } 5$

So it is irreducible. The others factor as $(2x + 1)(2x + 3)$, $(2x + 3)(2x - 1)$, and $(4x + 3)(x + 2)$.

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

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

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

The product $3$ is positive and the sum $-11$ is negative, so both constants are negative. Two numbers with product $6 \cdot 3 = 18$ and sum $-11$ are $-9$ and $-2$.

$(3x - 1)(2x - 3) = 6x^2 - 9x - 2x + 3 = 6x^2 - 11x + 3$

The cross products $-9x$ and $-2x$ add to $-11x$. The form $(3x - 3)(2x - 1)$ hides a factor of $3$ and gives $-9x$.

The middle splits into two numbers with product $6 \cdot 6 = 36$ and sum $13$, namely $9$ and $4$. Pairing $3x$ with $2$ and $2x$ with $3$ gives cross products $9x$ and $4x$.

$(3x + 2)(2x + 3) = 6x^2 + 9x + 4x + 6 = 6x^2 + 13x + 6$

The cross products add to $13x$. The form $(6x + 3)(x + 2)$ hides a factor of $3$ and gives $15x$.

Two numbers with product $5 \cdot (-2) = -10$ and sum $9$ are $10$ and $-1$, giving $(5x - 1)(x + 2)$.

$5x^2 + 9x - 2 = (5x - 1)(x + 2) = 0$

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

Move every term to one side: $2x^2 - 5x - 12 = 0$. Two numbers with product $2 \cdot (-12) = -24$ and sum $-5$ are $-8$ and $3$, giving $(2x + 3)(x - 4)$.

$2x^2 - 5x - 12 = (2x + 3)(x - 4) = 0$

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

Move every term to one side: $2x^2 + 3x - 2 = 0$. Two numbers with product $2 \cdot (-2) = -4$ and sum $3$ are $4$ and $-1$, giving $(2x - 1)(x + 2)$.

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

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

The middle splits into two numbers with product $6 \cdot (-5) = -30$ and sum $-7$, namely $-10$ and $3$. Pairing $3x$ with $-5$ and $2x$ with $1$ gives cross products $3x$ and $-10x$.

$(3x - 5)(2x + 1) = 6x^2 + 3x - 10x - 5 = 6x^2 - 7x - 5$

The cross products add to $-7x$. The sign swap $(3x + 5)(2x - 1)$ gives $+7x$.

Two numbers with product $6 \cdot 6 = 36$ and sum $-13$ are $-9$ and $-4$, giving $(3x - 2)(2x - 3)$.

$6x^2 - 13x + 6 = (3x - 2)(2x - 3) = 0$

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

Move every term to one side: $4x^2 + 4x - 3 = 0$. Two numbers with product $4 \cdot (-3) = -12$ and sum $4$ are $6$ and $-2$, giving $(2x + 3)(2x - 1)$.

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

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

Two numbers with product $10 \cdot (-6) = -60$ and sum $11$ are $15$ and $-4$, giving $(2x + 3)(5x - 2)$.

$10x^2 + 11x - 6 = (2x + 3)(5x - 2) = 0$

Setting each factor to $0$ gives $x = -\tfrac{3}{2}$ or $x = \tfrac{2}{5}$.