# Solve the equation: $2{x^5} + {x^4} - 12{x^3} - 12{x^2} + x + 2 = 0$

Last updated date: 29th Mar 2023

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Answer

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Hint: For these types of questions, we start by finding one of the factors and then dividing the given equation with the factor.

So let us start by finding the first factor. If we observe the coefficients of the equation they are$2,1, - 12, - 12,1,2$. Now, if you observe you can see that we have same coefficients for ${x^5}$ and ${x^0}$, ${x^1}$ and ${x^4}$, ${x^2}$ and ${x^3}$.Therefore we can say that $\left( {x + 1} \right)$ , is a factor of the given polynomial.

Therefore we can write,

$\left( {x + 1} \right)\left( {2{x^4} - {x^3} - 11{x^2} - x + 2} \right) = 0$

Now, we will concentrate on $\left( {2{x^4} - {x^3} - 11{x^2} - x + 2} \right) = 0$

We’ll start by dividing the above equation throughout with${x^2}$, we get,

$2{x^2} + \frac{2}{{{x^2}}} - x - \frac{1}{x} - 11 = 0$

To simplify the above equation, let us substitute$x + \frac{1}{x} = a$, we have,

$2\left( {{a^2} - 2} \right) - a - 11 = 0$

If we solve it, we get,

$2{a^2} - a - 15 = 0$

$2{a^2} - 6a + 5a - 15 = 0$

$\left( {2a + 5} \right)\left( {a - 3} \right) = 0$

Next step is to re-equate $a = x + \frac{1}{x}$ and to simplify we are going to multiply the entire equation with${x^2}$, we get,

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

Note: Do not forget to re-equate the variables.

On solving the brackets, we get,

$2{x^5} + {x^4} - 12{x^3} - 12{x^2} + x + 2 = \left( {x + 1} \right)\left( {2{x^2} + 5x + 2} \right)\left( {{x^2} - 3x + 1} \right) = 0$

On simplifying we get,

\[\left( {x + 1} \right)\left( {2x + 1} \right)\left( {x + 2} \right)\left( {{x^2} - 3x + 1} \right) = 0\]

So from the above equation,

The values of x we get are

\[x = - 1, - \frac{1}{2}, - 2,\frac{{3 + \sqrt 5 }}{2},\frac{{3 - \sqrt 5 }}{2}\]

So let us start by finding the first factor. If we observe the coefficients of the equation they are$2,1, - 12, - 12,1,2$. Now, if you observe you can see that we have same coefficients for ${x^5}$ and ${x^0}$, ${x^1}$ and ${x^4}$, ${x^2}$ and ${x^3}$.Therefore we can say that $\left( {x + 1} \right)$ , is a factor of the given polynomial.

Therefore we can write,

$\left( {x + 1} \right)\left( {2{x^4} - {x^3} - 11{x^2} - x + 2} \right) = 0$

Now, we will concentrate on $\left( {2{x^4} - {x^3} - 11{x^2} - x + 2} \right) = 0$

We’ll start by dividing the above equation throughout with${x^2}$, we get,

$2{x^2} + \frac{2}{{{x^2}}} - x - \frac{1}{x} - 11 = 0$

To simplify the above equation, let us substitute$x + \frac{1}{x} = a$, we have,

$2\left( {{a^2} - 2} \right) - a - 11 = 0$

If we solve it, we get,

$2{a^2} - a - 15 = 0$

$2{a^2} - 6a + 5a - 15 = 0$

$\left( {2a + 5} \right)\left( {a - 3} \right) = 0$

Next step is to re-equate $a = x + \frac{1}{x}$ and to simplify we are going to multiply the entire equation with${x^2}$, we get,

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

Note: Do not forget to re-equate the variables.

On solving the brackets, we get,

$2{x^5} + {x^4} - 12{x^3} - 12{x^2} + x + 2 = \left( {x + 1} \right)\left( {2{x^2} + 5x + 2} \right)\left( {{x^2} - 3x + 1} \right) = 0$

On simplifying we get,

\[\left( {x + 1} \right)\left( {2x + 1} \right)\left( {x + 2} \right)\left( {{x^2} - 3x + 1} \right) = 0\]

So from the above equation,

The values of x we get are

\[x = - 1, - \frac{1}{2}, - 2,\frac{{3 + \sqrt 5 }}{2},\frac{{3 - \sqrt 5 }}{2}\]

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