 QUESTION

# Solve the equation ${x^5} - 5{x^4} - 5{x^3} + 25{x^2} + 4x - 20 = 0$, whose roots are of form a, – a, b, – b, c.

Hint: For a fifth-degree polynomial of the standard form $a{x^5} + b{x^4} + c{x^3} + d{x^2} + ex + f = 0$, the sum of the roots is given as $- \dfrac{b}{a}$. The sum of the products of the roots taken two at a time is $\dfrac{c}{a}$ and so on. The product of the roots of the polynomial is given by $- \dfrac{f}{a}$.

We need to solve a fifth-degree polynomial ${x^5} - 5{x^4} - 5{x^3} + 25{x^2} + 4x - 20 = 0$ where the roots are of form a, b, - a, - b, c.

The standard form of a fifth-degree polynomial equation is $a{x^5} + b{x^4} + c{x^3} + d{x^2} + ex + f = 0$.

The sum of the roots of the fifth-degree polynomial is $- \dfrac{b}{a}$. Here, we have b as – 5 and a as 1. Hence, we have:

$a + b - a - b + c = - \dfrac{{ - 5}}{1}$

Simplifying, we have:

$c = 5.............(1)$

Hence, one of the roots of the polynomial is 5.

The sum of the products of the roots of a fifth-degree polynomial taken two at a time is given as $\dfrac{c}{a}$. Then, we have:

$a( - a) + ab + a( - b) + ac + ( - a)b + ( - a)( - b) + ( - a)c + b( - b) + bc - bc = \dfrac{{ - 5}}{1}$

Simplifying, we have:

$- {a^2} + ab - ab + ac - ab + ab - ac - {b^2} + bc - bc = - 5$

Canceling common terms, we have:

$- {a^2} - {b^2} = - 5$

Multiplying both sides by – 1, we have:

${a^2} + {b^2} = 5$

${a^2} = 5 - {b^2}..............(2)$

The product of the roots of a fifth-degree polynomial is given as $- \dfrac{f}{a}$. Hence, we have:

$a( - a)b( - b)c = - \dfrac{{ - 20}}{1}$

Simplifying, we have:

${a^2}{b^2}c = 20$

Substituting the value of c in equation (1), we have:

${a^2}{b^2}(5) = 20$

Dividing both sides of the equation by 5, we have:

${a^2}{b^2} = 4$

Substituting equation (2) in the above equation, we have:

$(5 - {b^2}){b^2} = 4$

This is true for ${b^2} = 1$. Hence, we have:

$b = 1;b = - 1$

Hence, the other two roots are 1 and – 1.

Now, we substitute the value of b in equation (2), we get:

${a^2} = 5 - 1$

${a^2} = 4$

$a = 2;a = - 2$

Hence, the other two roots are 2 and – 2.

The five roots of the equation are 1, - 1, 2, - 2, and 5.

Note: After finding a root, you can also divide by this root to get a simpler polynomial of lesser degree and then solve for it to find the other roots.