Question

Which of the following represents the zeros of f(x) = x3 − 12x2 + 47x − 60

Answer 1

Answer: $x=3,\:x=4,\:x=5$.

Step-by-step explanation: Given polynomial function $f(x) = x^3-12x^2+47x-60$.

$\mathrm{Use\:the\:rational\:root\:theorem}$

$p=60,\:\quad q=1$

$\mathrm{The\:dividers\:of\:}p:\quad 1,\:2,\:3,\:4,\:5,\:6,\:10,\:12,\:15,\:20,\:30,\:60,\:\quad \mathrm{The\:dividers\:of\:}q:\quad 1$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:3,\:4,\:5,\:6,\:10,\:12,\:15,\:20,\:30,\:60}{1}$

$\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x-3$

$\frac{x^3-12x^2+47x-60}{x-3}=x^2-9x+20 \ \ By \ long \ division.$

$\mathrm{Factor}\:x^2-9x+20:\quad \left(x-4\right)\left(x-5\right)$

Therefore,

$x^3-12x^2+47x-60=\left(x-3\right)\left(x-4\right)\left(x-5\right)$

$\mathrm{Solve\:}\:x-3=0:\quad x=3$

$\mathrm{Solve\:}\:x-4=0:\quad x=4$

$\mathrm{Solve\:}\:x-5=0:\quad x=5$

$\mathrm{The\:final\:solutions\:to\:the\:equation\:are:}$

$x=3,\:x=4,\:x=5$

Answer 2

F(x) = x • 3 • 2 + 47x - 60
fx =3x - 24x + 47x - 60
fx = 26x - 60
x = - 60/f - 26