Factoring Quadratics: Practice

The product $-63$ is negative and the sum $-2$ is negative, so the larger number is negative. The numbers are $7$ and $-9$.

$x^2 - 2x - 63 = (x + 7)(x - 9)$

Expanding gives $x^2 - 9x + 7x - 63 = x^2 - 2x - 63$.

Two numbers with product $-48$ and sum $2$ are $8$ and $-6$.

$x^2 + 2x - 48 = (x + 8)(x - 6) = 0$

Setting each factor to $0$ gives $x = -8$ or $x = 6$.

Every term is a multiple of $4$, so pull out the $4$ first.

$4x^2 - 24x + 32 = 4(x^2 - 6x + 8) = 4(x - 2)(x - 4)$

Inside, two numbers with sum $-6$ and product $8$ are $-2$ and $-4$. The form $2(x-2)(x-4)$ pulls out too little.

For $x^2 + 4x + 1$ you need two integers with product $1$ and sum $4$, but the only integer pairs multiplying to $1$ are $1, 1$ and $-1, -1$, with sums $2$ and $-2$.

$\text{no integer pair has product } 1 \text{ and sum } 4$

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

The product $45$ is positive and the sum $-14$ is negative, so both numbers are negative, namely $-5$ and $-9$.

$x^2 - 14x + 45 = (x - 5)(x - 9)$

The pair $-3$ and $-15$ multiplies to $45$ but adds to $-18$, so it fails.

Pull out the common factor $3$ first, then factor the monic trinomial.

$3x^2 - 3x - 36 = 3(x^2 - x - 12) = 3(x - 4)(x + 3) = 0$

The factor $3$ contributes no root, so $x = 4$ or $x = -3$.

Pull out the common factor $2$ first, then factor the monic trinomial.

$2x^2 - 16x + 30 = 2(x^2 - 8x + 15) = 2(x - 3)(x - 5) = 0$

The factor $2$ contributes no root, so $x = 3$ or $x = 5$.

The product $-21$ is negative and the sum $-4$ is negative, so the larger number is negative. The numbers are $3$ and $-7$.

$x^2 - 4x - 21 = (x + 3)(x - 7)$

Expanding gives $x^2 - 7x + 3x - 21 = x^2 - 4x - 21$, so this is the only match.

Two numbers with sum $15$ and product $56$ are $7$ and $8$.

$x^2 + 15x + 56 = (x + 7)(x + 8) = 0$

Setting each factor to $0$ gives $x = -7$ or $x = -8$.

The right side is not $0$, so first expand and rearrange to standard form: $x^2 - 3x - 10 = 8$ becomes $x^2 - 3x - 18 = 0$.

$x^2 - 3x - 18 = (x - 6)(x + 3) = 0$

Setting each factor to $0$ gives $x = 6$ or $x = -3$. Writing $x + 2 = 8$ directly would be wrong, since the product equals $8$, not $0$.

Read this as $x^2 + 0x - 100$, so the two numbers have sum $0$ and product $-100$, namely $10$ and $-10$.

$x^2 - 100 = (x + 10)(x - 10) = 0$

Setting each factor to $0$ gives $x = 10$ or $x = -10$.

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

$x^2 + 7x + 10 = (x + 2)(x + 5)$

Since one side is $x + 2$, the other side is $x + 5$.