Question

25 factor the following expression. $x^2 + 14x + 48$ a $(x + 6)(x - 8)$ b $(x + 8)(x + 6)$ c $(x - 6)(x - 8)$ d $(x + 8)(x - 6)$

Explanation:

Step1: Recall factoring trinomial

For \(x^2 + bx + c\), find two numbers that multiply to \(c\) and add to \(b\). Here, \(b = 14\), \(c = 48\).

Step2: Find numbers for 48 and 14

Numbers 6 and 8: \(6\times8 = 48\), \(6 + 8 = 14\).

Step3: Factor the trinomial

So \(x^2 + 14x + 48=(x + 6)(x + 8)\) (or \((x + 8)(x + 6)\)).

Answer:

B. \((x + 8)(x + 6)\)