Question

Obtain all the zeroes of $3x^4+6x^3-2x^2-10x-5$, if of its zeroes are $\sqrt{\frac{5}{3}}$ & $-\sqrt{\frac{5}{3}}$.

Answer 1

Since, it is given that $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$ are the zeroes of the polynomial $3x^4+6x^3-2x^2-10x-5$, therefore, $(x-\sqrt{\frac{5}{3}})$ and $(x+\sqrt{\frac{5}{3}})$ are also the zeroes of the given polynomial. Now, consider the product of zeroes as follows:

$(x-\sqrt{\frac{5}{3}})(x+\sqrt{\frac{5}{3}})=(x{)}^2-{\left(\sqrt{\frac{5}{3}}\right)}^2(\because a^2-b^2=(a+b)(a-b))=x^2-\frac{5}{3}$

We now divide $3x^4+6x^3-2x^2-10x-5$ by $(x^2-\frac{5}{3})$ as shown in the above image:

From the division, we observe that the quotient is $3x^2+6x+3$ and the remainder is $0$.

Now, we factorize the quotient $3x^2+6x+3$ by equating it to $0$ to find the other zeroes of the given polynomial:

$3x^2+6x+3=0\Rightarrow 3x^2+3x+3x+3=0\Rightarrow 3x(x+1)+3(x+1)=0\Rightarrow (3x+3)(x+1)=0\Rightarrow (3x+3)=0,(x+1)=0\Rightarrow 3x=-3,x=-1\Rightarrow x=-\frac{3}{3},x=-1\Rightarrow x=-1,x=-1$

Hence, the other two zeroes of $3x^4+6x^3-2x^2-10x-5$ are $-1,-1$.