Factoring trinomials of the form x² + bx + c: 3 worked examples

To factor the quadratic expression $x^{2} + 7x + 12$, we look for two numbers that multiply to the constant term ($12$) and add to the coefficient of the linear term ($7$). We test pairs of factors of $12$:
- $1$ and $12$: $1 + 12 = 13$
- $2$ and $6$: $2 + 6 = 8$
- $3$ and $4$: $3 + 4 = 7$

Since $3$ and $4$ multiply to $12$ and add to $7$, we can write the quadratic as the product of two binomials: $(x + 3)(x + 4)$.

To factor the quadratic expression $x^{2} - 5x - 24$, we look for two numbers that multiply to the constant term $-24$ and add to the coefficient of the linear term $-5$.
We list the factor pairs of $-24$:
$1$ and $-24$ (sum: $-23$)
$-1$ and $24$ (sum: $23$)
$2$ and $-12$ (sum: $-10$)
$-2$ and $12$ (sum: $10$)
$3$ and $-8$ (sum: $-5$)
$-3$ and $8$ (sum: $5$)
$4$ and $-6$ (sum: $-2$)
$-4$ and $6$ (sum: $2$)

The pair that adds to $-5$ is $3$ and $-8$. Therefore, we can write the expression as the product of two binomials: $(x + 3)(x - 8)$.

To factor the quadratic expression $x^{2} - 10x + 21$, we need to look for two numbers that multiply to the constant term ($21$) and add up to the coefficient of the middle term ($-10$).

We list the factor pairs of $21$:
$1 \times 21 = 21$ ($1 + 21 = 22$)
$(-1) \times (-21) = 21$ ($-1 - 21 = -22$)
$3 \times 7 = 21$ ($3 + 7 = 10$)
$(-3) \times (-7) = 21$ ($-3 - 7 = -10$)

The pair that satisfies both conditions is $-3$ and $-7$. Therefore, the quadratic expression can be factored as $(x - 3)(x - 7)$.