Question

Solve 2x^3-x^2-5x+2 ,x=1/2,1,-2

Answer 1

2x³ - x² - 5x + 2

x = $\frac{1}{2}$, 1, - 2

Whenever you receive such questions which says - "to solve" it means that, we've to substitute thw given values of x in the place of x in the given equation. After substituting, if the equation is equal to zero, then that number is the zero of that polynomial.

1. Substituting x = $\frac{1}{2}$ in 2x³ - x² - 5x + 2

$\implies 2(\frac{1}{2})^3 - (\frac{1}{2})^2 - 5(\frac{1}{2}) + 2 = 0$

$\implies 2(\frac{1}{8}) - \frac{1}{4} - 5(\frac{1}{2}) + 2 = 0$

$\implies \cancel{2}(\frac{1}{\cancel{8}}) - \frac{1}{4} - 5(\frac{1}{2}) + 2 = 0$

$\implies \frac{1}{4} - \frac{1}{4} - \frac{5}{2} + 2 = 0$

$\implies (\frac{1}{4} - \frac{1}{4}) - (\frac{5}{2} + 2) = 0$

$\implies \cancel{\frac{1}{4}} - \cancel{\frac{1}{4}} - (\frac{5}{2} + 2) = 0$

$\implies - \frac{5}{2} + 2 = 0$

$\implies \frac{-5 + 4}{2} = 0$

$\implies \frac{-1}{2} = 0$

But we know that, $\frac{-1}{2}$ can never be equal to 0.

Therefore, $\bold{\frac{1}{2}}$ is not the zero of the given polynomial.

2. Substituting x = 1 in the given polynomial 2x³ - x² - 5x + 2.

2(1)³ - (1)² - 5(1) + 2 = 0

2(1) - (1) - 5 + 2 = 0

2 - 1 - 5 + 2 = 0

1 - 3 = 0

- 2 = 0

But we know that, - 2 never can be equal to zero.

Therefore, 1 is not the zero of the given polynomial.

3. Substituting x = - 2 in given equation 2x³ - x² - 5x + 2.

2(-2)³ - (-2)² - 5(-2) + 2 = 0

2(-8) - (4) + 10 + 2 = 0

-16 - 4 + 10 + 2 = 0

-20 + 12 = 0

- 8 = 0

But we know that, - 8 can never be equal to zero. Therefore, - 2 is not a zero of the given equation.

Therefore, $\bold{\frac{1}{2}}$, 1 and - 2 are not the zeros of the given polynomial 2x³-x²-5x+2.

Answer 2

Step-by-step explanation:

1. Substituting x = \frac{1}{2}

2

1

in 2x³ - x² - 5x + 2

\implies 2(\frac{1}{2})^3 - (\frac{1}{2})^2 - 5(\frac{1}{2}) + 2 = 0⟹2(

2

1

)

3

−(

2

1

)

2

−5(

2

1

)+2=0

\implies 2(\frac{1}{8}) - \frac{1}{4} - 5(\frac{1}{2}) + 2 = 0⟹2(

8

1

)−

4

1

−5(

2

1

)+2=0

\implies \cancel{2}(\frac{1}{\cancel{8}}) - \frac{1}{4} - 5(\frac{1}{2}) + 2 = 0⟹

2

(

8

1

)−

4

1

−5(

2

1

)+2=0

\implies \frac{1}{4} - \frac{1}{4} - \frac{5}{2} + 2 = 0⟹

4

1

−

4

1

−

2

5

+2=0

\implies (\frac{1}{4} - \frac{1}{4}) - (\frac{5}{2} + 2) = 0⟹(

4

1

−

4

1

)−(

2

5

+2)=0

\implies \cancel{\frac{1}{4}} - \cancel{\frac{1}{4}} - (\frac{5}{2} + 2) = 0⟹

4

1

−

4

1

−(

2

5

+2)=0

\implies - \frac{5}{2} + 2 = 0⟹−

2

5

+2=0

\implies \frac{-5 + 4}{2} = 0⟹

2

−5+4

=0

\implies \frac{-1}{2} = 0⟹

2

−1

=0

But we know that, \frac{-1}{2}

2

−1

can never be equal to 0.

Therefore, \bold{\frac{1}{2}}

2

1

is not the zero of the given polynomial.

2. Substituting x = 1 in the given polynomial 2x³ - x² - 5x + 2.

2(1)³ - (1)² - 5(1) + 2 = 0

2(1) - (1) - 5 + 2 = 0

2 - 1 - 5 + 2 = 0

1 - 3 = 0

- 2 = 0

But we know that, - 2 never can be equal to zero.

Therefore, 1 is not the zero of the given polynomial.

3. Substituting x = - 2 in given equation 2x³ - x² - 5x + 2.

2(-2)³ - (-2)² - 5(-2) + 2 = 0

2(-8) - (4) + 10 + 2 = 0

-16 - 4 + 10 + 2 = 0

-20 + 12 = 0

- 8 = 0

But we know that, - 8 can never be equal to zero. Therefore, - 2 is not a zero of the given equation.

Therefore, \bold{\frac{1}{2}}

2

1

, 1 and - 2 are not the zeros of the given polynomial 2x³-x²-5x+2.