Vikolers6

2021-12-15

In the given equation,, find all zeros of the given polynomial.
$3{x}^{3}-2{x}^{2}-7x-2$

otoplilp1

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

Mary Goodson

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

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