Answer

Verified

459.3k+ views

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

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

Recently Updated Pages

Onehalf of a convex lens is covered with a black paper class 12 physics CBSE

Derive an expression for electric potential at point class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Which are the Top 10 Largest Countries of the World?

What is 67 as a simplified fraction class 8 maths ICSE

What is the value of 1875 in fractional form class 8 maths ICSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Derive an expression for electric potential at point class 12 physics CBSE

Differentiate between polar and nonpolar dielectri class 12 physics CBSE

ABCD is a square of side 2m Charges of 5nC + 10nC and class 12 physics CBSE

Distinguish between asexual and sexual reproduction class 12 biology CBSE

Drive an expression for the electric field due to an class 12 physics CBSE

How was the Civil Disobedience Movement different from class 12 social science CBSE