Question

15/35
factor
$x^2 - 16x + 48$
$(x - 16)(x - 3)$
$(x + 6)(x - 8)$
$(x - 12)(x - 4)$
$(x + 12)(x - 4)$

Explanation:

Step1: Recall factoring form

For a quadratic \(x^2 + bx + c\) (or \(x^2 - bx + c\) here), we need two numbers that multiply to \(c\) (48) and add to \(-b\) (16, since the middle term is \(-16x\)).

Step2: Find two numbers

We need two numbers \(m\) and \(n\) such that \(m \times n = 48\) and \(m + n = 16\). The numbers 12 and 4 work: \(12 \times 4 = 48\) and \(12 + 4 = 16\). Since the middle term is \(-16x\) and the constant term is positive, both numbers are negative. So the factors are \((x - 12)(x - 4)\).

Step3: Verify (optional, but to check)

Multiply \((x - 12)(x - 4)\): \(x^2 - 4x - 12x + 48 = x^2 - 16x + 48\), which matches the original quadratic.

Answer:

C. \((x - 12)(x - 4)\) (assuming the orange option is labeled C; if the labels are A, B, C, D with A being green, B purple, C orange, D teal, then the answer is C. \((x - 12)(x - 4)\))