Question

According to the Rational Root Theorem, what are all the potential rational roots of f(x) = 15x11 – 6x8 + x3 – 4x + 3?

Answer 1

Answer-

All the potential rational roots are,

$\pm \frac{1}{15}, \pm \frac{1}{5},\pm \frac{1}{3},\pm \frac{3}{5},\pm 1, \pm 3,$

Solution-

Rational Root Theorem

$f(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+.......+a_1x+a_0\ \ \ and\ a_n\neq 0$

All the potential rational roots are,

$\pm (\dfrac{\text{factors of}\ a_0}{\text{factors of}\ a_n})$

The given polynomial is,

$f(x) = 15x^{11}-6x^8+x^3-4x+3$

Here,

$a_n=15,\ a_0=3$

$\text{factors of}\ 15=1,3,5,15\\\\\text{factors of}\ 3=1,3$

The potential rational roots are,

$=\pm \frac{1}{1},\pm \frac{1}{3}, \pm \frac{1}{5}, \pm \frac{1}{15}, \pm \frac{3}{1}, \pm \frac{3}{3}, \pm \frac{3}{5}, \pm \frac{3}{15}$

$=\pm 1,\pm \frac{1}{3}, \pm \frac{1}{5}, \pm \frac{1}{15}, \pm 3, \pm 1, \pm \frac{3}{5}, \pm \frac{1}{5}$

$= \pm \frac{1}{15}, \pm \frac{1}{5},\pm \frac{1}{3},\pm \frac{3}{5},\pm 1, \pm 3,$

Answer 2

The rational root theorem is used to determine the possible roots of a polynomial

The potential rational roots are: $\mathbf{1,3,5,15,\frac{1}{3},\frac{5}{3},5}$

The polynomial is given as:

$\mathbf{f(x) = 15x^{11} - 6x^8 + x^3 + 4x + 3}$

The leading coefficient is:

$\mathbf{k_1= 15}$

The factors of 15 are:

$\mathbf{k_1= 1, 3, 5, 15}$

Also:

The trailing coefficient is:

$\mathbf{k_2= 3}$

The factors of 3 are:

$\mathbf{k_2= 1, 3,}$

So, the possible roots of the polynomial is:

$\mathbf{k = \frac{k_1}{k_2}}$

This gives:

$\mathbf{k = \frac{1,3,5,15}{1,3}}$

Spit

$\mathbf{k = \frac{1}{1},\frac{3}{1},\frac{5}{1},\frac{15}{1},\frac{1}{3},\frac{3}{3},\frac{5}{3},\frac{15}{3}}$

$\mathbf{k = 1,3,5,15,\frac{1}{3},1,\frac{5}{3},5}$

Remove repetition

$\mathbf{k = 1,3,5,15,\frac{1}{3},\frac{5}{3},5}$

Hence, the potential rational roots are: $\mathbf{1,3,5,15,\frac{1}{3},\frac{5}{3},5}$

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