Question

Using the Rational Root Theorem, what are all the rational roots of the polynomial f(x) = 20x4 + x3 + 8x2 + x – 12?

Answer 1

Answer:

A.) -4/5 and 3/4

Step-by-step explanation:

Answer 2

Answer: The all possible rational roots are $x=\pm1,\pm2,\pm3,\pm4,\pm6,\pm12,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{4},\pm\frac{3}{4},\pm\frac{1}{10},\pm\frac{1}{5},\pm\frac{3}{5}\pm\frac{3}{10},\pm\frac{2}{5},\pm\frac{6}{5},\pm\frac{1}{20},\pm\frac{3}{20},\pm\frac{4}{5},\pm\frac{12}{5}$.

Explanation:

The given polynomial is,

$f(x)=20x^4+x^3+8x^2+x-12$

The Rational Root Theorem states that the all possible roots of a polynomial are in the form of a rational number,

$x=\frac{p}{q}$

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.

All possible factor of -12 are $\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$.

All possible factor of 20 are $\pm1,\pm2,\pm4,\pm5,\pm10,\pm20$.

So, the all possible rational roots of the given polynomial are,

$x=\pm1,\pm2,\pm3,\pm4,\pm6,\pm12,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{4},\pm\frac{3}{4},\pm\frac{1}{10},\pm\frac{1}{5},\pm\frac{3}{5}\pm\frac{3}{10},\pm\frac{2}{5},\pm\frac{6}{5},\pm\frac{1}{20},\pm\frac{3}{20},\pm\frac{4}{5},\pm\frac{12}{5}$