Vikolers6

2021-12-15

In the given equation,, find all zeros of the given polynomial.

$3{x}^{3}-2{x}^{2}-7x-2$

otoplilp1

Beginner2021-12-16Added 41 answers

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.

$3{x}^{3}-2{x}^{2}-7x-2=0$

$3{x}^{3}+3{x}^{2}-5{x}^{2}-5x-2x-2=0$

$3{x}^{2}(x+1)-5(x+1)-2(x+1)=0$

$(x+1)(3{x}^{2}-5x-2)=0$

$(x+1)(3{x}^{2}-6x+x-2)=0$

(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$x=-\frac{1}{3}$

Step 2

Hence, zeroes of the given polynomial are$-\frac{1}{3}$ , -1 and 2.

Given polynomial is

To find all zeros of the given polynomial.

Solution:

Finding zeroes of the given polynomial.

(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

Step 2

Hence, zeroes of the given polynomial are

Mary Goodson

Beginner2021-12-17Added 37 answers

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

$f\left(x\right)=3{x}^{3}-2{x}^{2}-7x-2$

$f\left(x\right)=3{x}^{3}+4{x}^{2}-6{x}^{2}+x-8x-2$

$f\left(x\right)=x(3{x}^{2}+4x+1)-2(3{x}^{2}+4x+1)$

$f\left(x\right)=(x-2)(3{x}^{2}+4x+1)$

Step 2

On solving further by grouping the terms again, we get the result as

$f\left(x\right)=(x-2)(3{x}^{2}+4x+1)$

$f\left(x\right)=(x-2)(3{x}^{2}+3x+x+1)$

f(x)=(x-2)(3x(x+1)+1(x+1))

f(x)=(x-2)(x+1)(3x+1)

Now, zeros of the equation$f\left(x\right)=3{x}^{2}-2{x}^{2}-7x-2$ 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 $\frac{-1}{3}$ .

We have the given equation as

On solving further by grouping the terms, we get the result as

Step 2

On solving further by grouping the terms again, we get the result as

f(x)=(x-2)(3x(x+1)+1(x+1))

f(x)=(x-2)(x+1)(3x+1)

Now, zeros of the equation

f(x)=0

(x-2)(x+1)(3x+1)=0

x=2, -1, and

Hence, zeros of

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