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: In this question, we have to find the roots of the equation. Thus, we will use the algebraic identity and the long division method to get the solution. First, it is given that the two roots are $\sqrt{\dfrac{5}{3}}$ and $-\sqrt{\dfrac{5}{3}}$ , therefore, we get two factors of the given polynomial are $\left( x-\sqrt{\dfrac{5}{3}} \right)\left( x+\sqrt{\dfrac{5}{3}} \right)$ . Now, we will apply the algebraic identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ in the factors of the polynomial. After, that we will apply the long division method, where the dividend is equal to $3{{x}^{4}}+6{{x}^{3}}-2{{x}^{2}}-10x-5$ and divisor is equal to ${{x}^{2}}-\dfrac{5}{3}$ , thus we will get a quadratic equation in the quotient. In the last, we will apply splitting the middle term method, to get the remaining roots of the equation.

Complete step by step answer:
According to the question, we have to find the remaining roots of the given polynomial.
Thus, we will use the algebraic identity, the long division method, and splitting the middle term method to get the solution.
The equation given to us is $3{{x}^{4}}+6{{x}^{3}}-2{{x}^{2}}-10x-5$ --------- (1)
Now, the two roots of the equation (1) as given to us are $\sqrt{\dfrac{5}{3}}$ and $-\sqrt{\dfrac{5}{3}}$ . So, the two factors of equation (1) are $\left( x-\sqrt{\dfrac{5}{3}} \right)\left( x+\sqrt{\dfrac{5}{3}} \right)$ -------- (2)
Now, we will apply the algebraic identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ in equation (2), we get
$\Rightarrow \left( x-\sqrt{\dfrac{5}{3}} \right)\left( x+\sqrt{\dfrac{5}{3}} \right)={{x}^{2}}-\dfrac{5}{3}$ ------- (3)
So, we will apply the long division method, where dividend is equal to $3{{x}^{4}}+6{{x}^{3}}-2{{x}^{2}}-10x-5$ and divisor is equal to ${{x}^{2}}-\dfrac{5}{3}$ , thus we get
${{x}^{2}}-\dfrac{5}{3}\overset{3{{x}^{2}}+6x+3}{\overline{\left){\begin{align}
  & 3{{x}^{4}}+6{{x}^{3}}-2{{x}^{2}}-10x-5 \\
 & \underline{{}_{-}3{{x}^{4}}+0{{x}^{3}}\underset{+}{\mathop{-}}\,5{{x}^{2}}} \\
 & \text{ 6}{{x}^{3}}+3{{x}^{2}}-10x-5 \\
 & \text{ }{}_{-}\underline{6{{x}^{3}}+0{{x}^{2}}\underset{+}{\mathop{-}}\,10x} \\
 & \text{ }3{{x}^{2}}-5 \\
 & \text{ }{}_{-}\underline{3{{x}^{2}}\underset{+}{\mathop{-}}\,5} \\
 & \text{ }\underline{\underline{\text{ 0}}} \\
\end{align}}\right.}}$
Therefore, the quotient is equal to $3{{x}^{2}}+6x+3$ and the remainder is equal to 0. Thus, we see that the quotient is a quadratic equation, thus we will apply splitting the middle term method in the quadratic equation, which is
$\Rightarrow 3{{x}^{2}}+6x+3$
So, we will split the middle term of the above equation as a sum of 3x and 3x because $ac=3\times 3=9$ and $a+c=3+3=6$ , thus we get
$\Rightarrow 3{{x}^{2}}+3x+3x+3$
Now, we will take common 3x from first two terms and common 3 from the last two terms, we get
$\Rightarrow 3x\left( x+1 \right)+3\left( x+1 \right)$
Now, we will take common (x+1) from the above equation, we get
$\Rightarrow \left( x+1 \right)\left( 3x+3 \right)$
Thus, we get the two remaining factors of the given polynomial is
$\left( x+1 \right)=0$ -------- (4) and
$\left( 3x+3 \right)=0$ -------- (5)
So, now we will solve equation (4) by subtracting 1 on both sides of the equation, we get
$\Rightarrow x+1-1=0-1$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow x=-1$ ----- (6)
So, now we will solve equation (5) by subtracting 3 on both sides of the equation, we get
$\Rightarrow 3x+3-3=0-3$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow 3x=-3$
Now, we will divide 3 on both sides in the above equation, we get
$\Rightarrow \dfrac{3x}{3}=\dfrac{-3}{3}$
Therefore, we get
$\Rightarrow x=-1$ --------- (7)

Therefore, from equation (6) and (7), we get that the remaining two roots of the polynomial $3{{x}^{4}}+6{{x}^{3}}-2{{x}^{2}}-10x-5$ are -1.

Note: While solving this problem, do mention all the steps properly to avoid mathematical error. One of the alternative methods to solve the quadratic equation is, you can use the ht and trial method or the discriminant method to get the solution.