Factoring Quadratics: Additional Practice (Set 1)

Two positive numbers with sum $9$ and product $18$ are $3$ and $6$.

$x^2 + 9x + 18 = (x + 3)(x + 6)$

The pair $2$ and $9$ multiplies to $18$ but adds to $11$, so only $3$ and $6$ work.

The product $-30$ is negative and the sum $-1$ is negative, so the larger number is negative. The numbers are $-6$ and $5$.

$x^2 - x - 30 = (x - 6)(x + 5) = 0$

Setting each factor to $0$ gives $x = 6$ or $x = -5$.

The product $-27$ is negative and the sum $-6$ is negative, so the larger number is negative. The numbers are $3$ and $-9$.

$x^2 - 6x - 27 = (x + 3)(x - 9)$

Expanding gives $x^2 - 9x + 3x - 27 = x^2 - 6x - 27$.

There is no constant term, so pull out the shared $x$.

$x^2 + 4x = x(x + 4) = 0$

Setting each factor to $0$ gives $x = 0$ or $x = -4$. Dividing by $x$ would lose the root $x = 0$.

The product $-14$ is negative and the sum is $5$, so the larger number is positive. The numbers are $7$ and $-2$.

$x^2 + 5x - 14 = (x + 7)(x - 2) = 0$

Setting each factor to $0$ gives $x = -7$ or $x = 2$.

The product $-42$ is negative and the sum is $+1$, so the larger number is positive. The numbers are $7$ and $-6$.

$x^2 + x - 42 = (x + 7)(x - 6)$

Expanding gives $x^2 - 6x + 7x - 42 = x^2 + x - 42$.

Pull out the common factor $4$ first, then factor the monic trinomial.

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

The factor $4$ contributes no root, so $x = 2$ or $x = -1$.

Two numbers with sum $0$ and product $-16$ are opposites, so they are $4$ and $-4$.

$x^2 - 16 = (x + 4)(x - 4)$

Expanding gives $x^2 - 4x + 4x - 16 = x^2 - 16$, since the middle terms cancel.

Move every term to one side: $x^2 - 9x + 20 = 0$. Two numbers with product $20$ and sum $-9$ are $-4$ and $-5$.

$x^2 - 9x + 20 = (x - 4)(x - 5) = 0$

Setting each factor to $0$ gives $x = 4$ or $x = 5$.

The product $-35$ is negative and the sum $-2$ is negative, so the larger number is negative. The numbers are $-7$ and $5$.

$x^2 - 2x - 35 = (x - 7)(x + 5) = 0$

Setting each factor to $0$ gives $x = 7$ or $x = -5$.

Every term is even, so pull out the common factor $2$ first.

$2x^2 - 2x - 24 = 2(x^2 - x - 12) = 2(x - 4)(x + 3)$

Inside, two numbers with product $-12$ and sum $-1$ are $-4$ and $3$. Dropping the $2$ leaves the factoring incomplete.

The right side is not $0$, so first expand and rearrange: $x^2 + x - 12 = 8$ becomes $x^2 + x - 20 = 0$.

$x^2 + x - 20 = (x + 5)(x - 4) = 0$

Setting each factor to $0$ gives $x = -5$ or $x = 4$. Writing $x - 3 = 8$ directly would be wrong, since the product equals $8$, not $0$.