Rational zeros of f(x) = 8x³-6x²-23x +6 is equals to [tex]2, \frac{-3}{2} , \frac{1}{4}[/tex].
What are rational zeros?
"A rational zero of a polynomial P(x) is defined as when the variable x = p/q and P (p/q) = 0 where p and q are any integers."
According to the question,
Given,
f(x) = 8x³-6x²-23x +6
Factor of 6 'p' =±1, ±2, ±3, ±6
Factor of 8 'q' = ±1 , ±2, ±4 , ±8
Using trial and error method x=2 is a common factor
Substitute x=2 in f(x) we get,
f(2) = 8(2)³ -6(2)² -23(2) +6
= 64-24-46+6
=0
(x-2) is a factor of f(x).
Divide f(x) by (x-2)
[tex]\frac{8x^{3}-6x^{2} -23x+6 }{(x-2)}[/tex]
= 8x²+10x-3
Factorize it we get,
8x²+12x-2x-3
= 4x(2x+3) - 1(2x+3)
= (4x-1)(2x+3)
For rational zeros ,Substitute f(x=(p/q)) =0
x-2=0
⇒ x=2
2x+3=0
⇒x= -3/2
= [tex]-1\frac{1}{2}[/tex]
4x-1 =0
⇒x=1/4
Hence, Option (C) is the correct answer.
Learn more about rational zeros here
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