In the given equation,, find all zeros of the given polynomial. 3x3−2x2−7x−2

Step 1
Given polynomial is ${\displaystyle 3x^3-2x^2-7x-2}$
To find all zeros of the given polynomial.
Solution:
Finding zeroes of the given polynomial.
${\displaystyle 3x^3-2x^2-7x-2=0}$
${\displaystyle 3x^3+3x^2-5x^2-5x-2x-2=0}$
${\displaystyle 3x^2(x+1)-5(x+1)-2(x+1)=0}$
${\displaystyle (x+1)(3x^2-5x-2)=0}$
${\displaystyle (x+1)(3x^2-6x+x-2)=0}$
(x+1)(3x(x-2)+1(x-2))=0
(x+1)(x-2)(3x+1)=0
(x+1)=0 or (x-2)=0 or (3x+1)=0
x=-1 or x=2 or ${\displaystyle x=-\frac{1}{3}}$
Step 2
Hence, zeroes of the given polynomial are ${\displaystyle -\frac{1}{3}}$, -1 and 2.

Step 1
We have the given equation as
${\displaystyle f\left(x\right)=3x^3-2x^2-7x-2}$
On solving further by grouping the terms, we get the result as
${\displaystyle f\left(x\right)=3x^3-2x^2-7x-2}$
${\displaystyle f\left(x\right)=3x^3+4x^2-6x^2+x-8x-2}$
${\displaystyle f\left(x\right)=x(3x^2+4x+1)-2(3x^2+4x+1)}$
${\displaystyle f\left(x\right)=(x-2)(3x^2+4x+1)}$
Step 2
On solving further by grouping the terms again, we get the result as
${\displaystyle f\left(x\right)=(x-2)(3x^2+4x+1)}$
${\displaystyle f\left(x\right)=(x-2)(3x^2+3x+x+1)}$
f(x)=(x-2)(3x(x+1)+1(x+1))
f(x)=(x-2)(x+1)(3x+1)
Now, zeros of the equation ${\displaystyle f\left(x\right)=3x^2-2x^2-7x-2}$ is calculated as
f(x)=0
(x-2)(x+1)(3x+1)=0
x=2, -1, and ${\displaystyle \frac{-1}{3}}$
Hence, zeros of ${\displaystyle 3x^3-2x^2-7x-2}$ are 2,-1, and ${\displaystyle \frac{-1}{3}}$.