Question

Obtain all the zeroes of the polynomial $3x^4-12x^3+5x^2+16x-12,$ if two of its zeroes are $\frac{2}{\sqrt{3}}and\frac{-2}{\sqrt{3}}.$

• A.
  $\left(\frac{2}{\sqrt{3}}\right),(-\frac{2}{\sqrt{3}}),3,1$
• B.
  $\left(\frac{2}{\sqrt{3}}\right),(-\frac{2}{\sqrt{3}}),3,2$
• C.
  $\left(\frac{2}{\sqrt{3}}\right),(-\frac{2}{\sqrt{3}}),4,1$
• D.
  $\left(\frac{2}{\sqrt{3}}\right),(-\frac{2}{\sqrt{3}}),1,1$

Answer 1

The correct option is A

$\left(\frac{2}{\sqrt{3}}\right),(-\frac{2}{\sqrt{3}}),3,1$
Two zeroes of given polynomial are $\frac{2}{\sqrt{3}}and\frac{-2}{\sqrt{3}}(\sqrt{3}x-2)and(\sqrt{3}x+2)arethe\text{factors of the given polynomial.}\Rightarrow (\sqrt{3}x-2)(\sqrt{3}x+2)=(3x^2-4)\text{is also a factor of the given polynomial.}$


$\Rightarrow 3x^4-12x^3+5x^2+16x-12\Rightarrow (3x^2-4)(x^2-4x+3)\Rightarrow (3x^2-4)(x^2-3x-x+3)\Rightarrow (3x^2-4)\left[x\right(x-3)-1(x-3\left)\right]\Rightarrow (3x^2-4)(x-3)(x-1)\Rightarrow (\sqrt{3}x-2)(\sqrt{3}x+2)(x-3)(x-1)\text{Zeroes of given polynomial are}\frac{2}{\sqrt{3}},-\frac{2}{\sqrt{3}},3,1$

So, the correct option is a.