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

Question 1

Question 2

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

Question 3

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

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

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

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

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

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

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

$(x^3-12x^2+47x-60)/(x-3)=x^2-9x+20 \ \  By \ long \ division.$

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

Therefore,

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

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

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

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

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

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

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