Answer :
We want to match each quadratic expression in standard form with its factored form. Here is a step‐by‐step process:
1. For
[tex]$$
x^2 + 2x - 35,
$$[/tex]
we look for two numbers whose product is [tex]$-35$[/tex] and whose sum is [tex]$2$[/tex]. Notice that [tex]$-5$[/tex] and [tex]$7$[/tex] work because
[tex]$$
(-5)(7) = -35 \quad \text{and} \quad -5 + 7 = 2.
$$[/tex]
Thus, we can write:
[tex]$$
x^2 + 2x - 35 = (x-5)(x+7).
$$[/tex]
2. For
[tex]$$
x^2 + 12x + 35,
$$[/tex]
we need two numbers with a product of [tex]$35$[/tex] and a sum of [tex]$12$[/tex]. The numbers [tex]$5$[/tex] and [tex]$7$[/tex] meet the requirements since:
[tex]$$
5 \times 7 = 35 \quad \text{and} \quad 5 + 7 = 12.
$$[/tex]
So, the factorization is:
[tex]$$
x^2 + 12x + 35 = (x+5)(x+7).
$$[/tex]
3. For
[tex]$$
x^2 - 12x + 35,
$$[/tex]
we want two numbers with a product of [tex]$35$[/tex] and a sum of [tex]$-12$[/tex]. The numbers [tex]$-5$[/tex] and [tex]$-7$[/tex] satisfy this because:
[tex]$$
(-5)(-7) = 35 \quad \text{and} \quad -5 + (-7) = -12.
$$[/tex]
Thus,
[tex]$$
x^2 - 12x + 35 = (x-5)(x-7).
$$[/tex]
4. For
[tex]$$
x^2 - 2x - 35,
$$[/tex]
we look for two numbers whose product is [tex]$-35$[/tex] and whose sum is [tex]$-2$[/tex]. Here, [tex]$-7$[/tex] and [tex]$5$[/tex] work since:
[tex]$$
-7 \times 5 = -35 \quad \text{and} \quad -7 + 5 = -2.
$$[/tex]
Therefore, we have:
[tex]$$
x^2 - 2x - 35 = (x-7)(x+5).
$$[/tex]
In summary, the matchings are:
[tex]\[
\begin{array}{rcl}
x^2+2x-35 &\longrightarrow& (x-5)(x+7),\\[6mm]
x^2+12x+35 &\longrightarrow& (x+5)(x+7),\\[6mm]
x^2-12x+35 &\longrightarrow& (x-5)(x-7),\\[6mm]
x^2-2x-35 &\longrightarrow& (x-7)(x+5).
\end{array}
\][/tex]
1. For
[tex]$$
x^2 + 2x - 35,
$$[/tex]
we look for two numbers whose product is [tex]$-35$[/tex] and whose sum is [tex]$2$[/tex]. Notice that [tex]$-5$[/tex] and [tex]$7$[/tex] work because
[tex]$$
(-5)(7) = -35 \quad \text{and} \quad -5 + 7 = 2.
$$[/tex]
Thus, we can write:
[tex]$$
x^2 + 2x - 35 = (x-5)(x+7).
$$[/tex]
2. For
[tex]$$
x^2 + 12x + 35,
$$[/tex]
we need two numbers with a product of [tex]$35$[/tex] and a sum of [tex]$12$[/tex]. The numbers [tex]$5$[/tex] and [tex]$7$[/tex] meet the requirements since:
[tex]$$
5 \times 7 = 35 \quad \text{and} \quad 5 + 7 = 12.
$$[/tex]
So, the factorization is:
[tex]$$
x^2 + 12x + 35 = (x+5)(x+7).
$$[/tex]
3. For
[tex]$$
x^2 - 12x + 35,
$$[/tex]
we want two numbers with a product of [tex]$35$[/tex] and a sum of [tex]$-12$[/tex]. The numbers [tex]$-5$[/tex] and [tex]$-7$[/tex] satisfy this because:
[tex]$$
(-5)(-7) = 35 \quad \text{and} \quad -5 + (-7) = -12.
$$[/tex]
Thus,
[tex]$$
x^2 - 12x + 35 = (x-5)(x-7).
$$[/tex]
4. For
[tex]$$
x^2 - 2x - 35,
$$[/tex]
we look for two numbers whose product is [tex]$-35$[/tex] and whose sum is [tex]$-2$[/tex]. Here, [tex]$-7$[/tex] and [tex]$5$[/tex] work since:
[tex]$$
-7 \times 5 = -35 \quad \text{and} \quad -7 + 5 = -2.
$$[/tex]
Therefore, we have:
[tex]$$
x^2 - 2x - 35 = (x-7)(x+5).
$$[/tex]
In summary, the matchings are:
[tex]\[
\begin{array}{rcl}
x^2+2x-35 &\longrightarrow& (x-5)(x+7),\\[6mm]
x^2+12x+35 &\longrightarrow& (x+5)(x+7),\\[6mm]
x^2-12x+35 &\longrightarrow& (x-5)(x-7),\\[6mm]
x^2-2x-35 &\longrightarrow& (x-7)(x+5).
\end{array}
\][/tex]
