Question

Which of the following represents the zeros of f(x) = 6x3 + 25x2 − 24x + 5?

Answer 1

The zeros of f(x) are {-5, 1/3, 1/2}.

Answer 2

The given polynomial is

$f(x) = 6x^3 + 25x^2 - 24x + 5$

By rational root theorem

For a  cubic polynomial of the type :

a x³+ b x² + c x + d=0

→ x³ + $\frac{bx^2}{a} +  \frac{c x}{a}+\frac{d}{a}=0$

The possible roots of this polynomial is , $\pm 1, \pm d, \pm \frac{1}{a} , \pm\frac{d}{a}$.

The polynomial f(x) can be written as: $x^3 +\frac{25x^2}{6}-4 x+\frac{5}{6}$

The possible roots or Zeroes of the polynomial are:  

$\pm 1,\pm 3, \pm 5,\pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{5}{2}\pm \frac{5}{3},\pm \frac{5}{6}$

The values of x for which f(x) =0 , are roots or zeroes of the polynomial.

f(1)= 6 +25 -24+5= 12

f(-1)= -6 +25 +24 +5=48

f(3)=162 +225-72+5≠ 0

f(-3)=-162+225+72+5≠0

$f(\frac{-1}{2}) =\frac{-6}{8}+\frac{25}{4}+12+5=\frac{93}{4}- \frac{6}{8}\neq 0$

$f(\frac{1}{2}) =\frac{6}{8}+\frac{25}{4}-12+5= 7-7=0$

$f(\frac{1}{3})=\frac{6}{27}+\frac{25}{9}-8+5=0 \\\\f(\frac{-1}{3})=\frac{-6}{27}+\frac{25}{9}+8+5 \neq 0$

$f(5)=750 +625-120+5\neq 0\\ f(-5)= -750+625 +120+5=0\\f(\frac{5}{2})=\frac{750}{8}+\frac{625}{4}-60+5\neq 0\\f(\frac{-5}{2})=\frac{-750}{8}+\frac{625}{4}+60+5\neq 0$

$f(\frac{5}{3})= \frac{750}{27}+\frac{625}{9}-40+5\neq 0\\\\\\f(\frac{-5}{3})= \frac{-750}{27}+\frac{625}{9}+40+5\neq 0\\\\\\f(\frac{5}{6})= \frac{125}{36}+\frac{625}{36}-20+5\neq 0 \\\\\\\\f(\frac{-5}{6})= \frac{-125}{36}+\frac{625}{36}+20+5\neq 0$

So , zeroes of f(x) are -5,$\frac{1}{2}$,$\frac{1}{3}$.