Step-by-step explanation:
Given the equation
[tex]3x^3\:+10x^2-x-12=0[/tex]
Steps
[tex]3x^3+10x^2-x-12[/tex]
Rational Root Theorem definition
[tex]\mathrm{For\:a\:polynomial\:equation\:with\:integer\:coefficients:\quad }a_nx^n+a_{n-1}x^{n-1}+\ldots +a_0[/tex]
[tex]\mathrm{If\:}a_0\mathrm{\:and\:}a_n\mathrm{\:are\:integers,\:then\:if\:there\:is\:a\:rational\:solution}[/tex]
[tex]it\:could\:be\:found\:by\:checking\:all\:the\:numbers\:produced\:for\:\pm \frac{dividers\:of\:a_0}{dividers\:of\:a_n}[/tex]
[tex]a_0=12,\:\quad a_n=3[/tex]
[tex]\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:3,\:4,\:6,\:12,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:3[/tex]
[tex]\mathrm{The\:following\:rational\:numbers\:are\:candidate\:roots:}\quad \pm \frac{1,\:2,\:3,\:4,\:6,\:12}{1,\:3}[/tex]
Plugging x = -1 is False
[tex]3\left(-1\right)^3+10\left(-1\right)^2-\left(-1\right)-12=0[/tex]
[tex]-4=0[/tex] ∵L.H.S ≠ R.H.S
Plugging x = 1 is True
[tex]3\cdot \:1^3+10\cdot \:1^2-1-12=0[/tex]
[tex]0=0[/tex] ∵ L.H.S = R.H.S
Similarly,
x = -2 does not satisfy the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
x = 2 does not satisfy the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
x = -3 satisfies the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
As
[tex]3\left(-3\right)^3+10\left(-3\right)^2-\left(-3\right)-12=0[/tex]
[tex]0=0[/tex] ∵ L.H.S = R.H.S
x = 3 does not satisfy the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
x = 4 does not satisfy the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
x = -4 does not satisfy the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
x = 6 does not satisfy the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
x = -6 does not satisfy the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
x = -12 does not satisfy the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
x = 12 does not satisfy the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
x = -1/3 does not satisfy the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
x = 1/3 does not satisfy the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
x = -2/3 does not satisfy the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
x = 2/3 does not satisfy the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
x = -4/3 satisfies the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
As
[tex]3\left(-\frac{4}{3}\right)^3+10\left(-\frac{4}{3}\right)^2-\left(-\frac{4}{3}\right)-12=0[/tex] ∵ L.H.S = R.H.S
x = 4/3 does not satisfy the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
x = -2 does not satisfy the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
x = 2 does not satisfy the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
x = -4 does not satisfy the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
x = 4 does not satisfy the equation [tex]3x^3\:+10x^2-x-12=0[/tex].
Hence, out of all the mentioned rational roots of the equation [tex]3x^3\:+10x^2-x-12=0[/tex], the rational roots which validate the equation are [tex]x=-\frac{4}{3},\:x=1,\:x=-3[/tex].
Keywords: rational roots, rational root theorem
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