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.
ANSWER
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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 \[ - \dfrac{f}{a}\].
Complete step-by-step answer: 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.