Question

factor:
$x^3 - 12x^2 + 47x - 60 = 0$ when 5 is a root.

$(x - 3)(x - 4)(x - 5) = 0$

$(x - 3)(x - 4)(2x - 5) = 0$

$(x - 5)(x - 4)(x - 5) = 0$

$(x + 1)(x - 4)(x - 5) = 0$

Explanation:

Step1: Use root to form factor

Given root $x=5$, so factor is $(x-5)$.

Step2: Polynomial division

Divide $x^3-12x^2+47x-60$ by $(x-5)$:
$$\frac{x^3-12x^2+47x-60}{x-5}=x^2-7x+12$$

Step3: Factor quadratic

Factor $x^2-7x+12$:
$x^2-7x+12=(x-3)(x-4)$

Step4: Combine all factors

Full factored form: $(x-3)(x-4)(x-5)=0$

Answer:

A. $(x-3)(x-4)(x-5)=0$