Question

If \[\sqrt {\dfrac{5}{3}} \] and \[ - \sqrt {\dfrac{5}{3}} \] are the roots of the polynomial\[3{x^4} + 6{x^3} - 2{x^2} - 10x - 5\], then find its other roots.

Answer 1

Hint: Here first we will form the equation from the given roots and divide the given polynomial by that equation using a long division method and then finally find the roots of the quotient so obtained.

Complete step-by-step answer:
It is given that, \[\sqrt {\dfrac{5}{3}} \] and \[ - \sqrt {\dfrac{5}{3}} \]are the roots of the polynomial\[3{x^4} + 6{x^3} - 2{x^2} - 10x - 5\]
This implies, \[x = \sqrt {\dfrac{5}{3}} \] and \[x = - \sqrt {\dfrac{5}{3}} \]
\[ \Rightarrow x - \sqrt {\dfrac{5}{3}} = 0\]…………….. (1)
\[ \Rightarrow x + \sqrt {\dfrac{5}{3}} = 0\]…………….. (2)
Multiplying equation1 and 2 we get:-
\[\left( {x - \sqrt {\dfrac{5}{3}} } \right)\left( {x + \sqrt {\dfrac{5}{3}} } \right) = 0\]
Now applying the following identity:
\[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\]
We get:-
\[{x^2} - {\left( {\sqrt {\dfrac{5}{3}} } \right)^2} = 0\]
Simplifying it we get:-
\[{x^2} - \dfrac{5}{3} = 0\]
Taking LCM we get:-
\[3{x^2} - 5 = 0\]……………………. (3)
Now we will divide the given polynomial by \[3{x^2} - 5\] using long division method.
On dividing we get:-
                \[\underline {{x^2} - 2x + 1{\text{ }}} \]
\[3{x^2} - 5)3{x^4} + 6{x^3} - 2{x^2} - 10x - 5\]
               \[\underline { - 3{x^4} + 5{x^2}} \]
                \[6{x^3} + 3{x^2} - 10x - 5\]
                \[\underline { - 6{x^3} + 10x{\text{ }}} {\text{ }}\]
                 \[3{x^2} - 5\]
                \[\underline { - 3{x^2} + 5} \]
                       \[0\]
Hence, we get the quotient as:-
\[{x^2} - 2x + 1 = 0\]
Now we know that,
\[{\left( {x - 1} \right)^2} = {x^2} - 2x + 1\]
Hence substituting the value we get:-
\[{\left( {x - 1} \right)^2} = 0\]
\[ \Rightarrow x = 1;x = 1\]

Hence the other two roots of the given polynomial are 1 and 1.

Note: Students can also factorize the quotient obtained and then obtain the required roots as:-
\[{x^2} - x - x + 1 = 0\]
Taking the terms common we get:-
\[x\left( {x - 1} \right) - 1\left( {x - 1} \right) = 0\]
Simplifying it we get:-
\[\left( {x - 1} \right)\left( {x - 1} \right) = 0\]
Solving for x we get:-
\[x - 1 = 0;x - 1 = 0\]
\[ \Rightarrow x = 1;x = 1\]