Respuesta :
Answer:
Option 1 is correct.
Step-by-step explanation:
The given polynomial is
[tex]f(x)=20x^4+x^3+8x^2+x-12[/tex]
we have to find all the rational roots of the polynomial f(x)
The Rational Root Theorem states that the all possible roots of a polynomial are in the form of a rational number i.e in the form of [tex]\frac{p}{q}[/tex]
where p is a factor of constant term and q is the factor of coefficient of leading term.
In the given polynomial the constant is -12 and the leading coefficient is 20.
[tex]\text{All possible factor of -12 are }\pm1,\pm2, \pm3, \pm4,\pm6,\pm12[/tex]
[tex]\text{All possible factor of 20 are }\pm1,\pm2,\pm4,\pm5,\pm10,\pm20[/tex]
So, the all possible rational roots of the given polynomial are,
[tex]\pm1,\pm2, \pm3, \pm4,\pm6,\pm12,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{4},\pm\frac{3}{4},\pm\frac{1}{10},\pm\frac{1}{5},\pm\frac{3}{5},\pm\frac{3}{10},\pm\frac{2}{5},\pm\frac{6}{5},\pm\frac{1}{20},\pm\frac{3}{20},\pm\frac{4}{5},\pm\frac{12}{5}[/tex]
Now, the rational roots of polynomial satisfy the given polynomial
[tex]f(-\frac{4}{5})=20(-\frac{4}{5})^4+(-\frac{4}{5})^3+8(-\frac{4}{5})^2-\frac{4}{5}-12=\frac{256}{625}\times 20-\frac{64}{125}+\frac{128}{125}-\frac{4}{5}-12[/tex]
[tex]=\frac{1024}{125}-\frac{64}{125}+\frac{128}{25}-\frac{4}{5}-12[/tex]
[tex]=\frac{960}{125}+\frac{128}{25}-\frac{4}{5}-12=12-12=0[/tex]
Hence, rational root.
[tex]f(\frac{3}{4})=20(\frac{3}{4})^4+(\frac{3}{4})^3+8(\frac{3}{4})^2+\frac{3}{4}-12=\frac{405}{64}+\frac{27}{64}+\frac{9}{2}+\frac{3}{4}-12=0[/tex]
rational root
[tex]f(1)=20(1)^4+(1)^3+8(1)^2+1-12=20+1+8-11=18\neq 0[/tex]
not a rational root.
hence, option 1 is correct
