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Question

Obtain all other zeroes of $$3x^4+6x^3-2x^2-10x-5$$, if two of its zeros are $$\sqrt{\dfrac{5}{3}}$$ and $$-\sqrt{\dfrac{5}{3}}$$.


Solution

Since the two zeroes of the given polynomial $$p(x)=3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5$$ are $$\sqrt { \dfrac { 5 }{ 3 }  } ,-\sqrt { \dfrac { 5 }{ 3 }  }$$, therefore, 

$$\left( x-\sqrt { \dfrac { 5 }{ 3 }  }  \right) \left( x+\sqrt { \dfrac { 5 }{ 3 }  }  \right) =x^{ 2 }-\dfrac { 5 }{ 3 }$$ is a factor of the given polynomial.

now, we divide the polynomial $$p(x)=3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5$$ by $$x^{ 2 }-\dfrac { 5 }{ 3 }$$ as shown below:

8.png

Therefore, $$3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5=\left( x^{ 2 }-\dfrac { 5 }{ 3 }  \right) (3x^{ 2 }+6x+3)$$

Now, we factorize $$3x^{ 2 }+6x+3$$ as follows:

$$3x^{ 2 }+6x+3=0\\ \Rightarrow 3(x^{ 2 }+2x+1)=0\\ \Rightarrow x^{ 2 }+2x+1=0\\ \Rightarrow (x+1)^{ 2 }=0\quad \quad \quad \quad (\because \quad (a+b)^{ 2 }=a^{ 2 }+b^{ 2 }+2ab)\\ \Rightarrow (x+1)(x+1)=0\\ \Rightarrow x+1=0,\quad x+1=0\\ \Rightarrow x=-1,\quad x=-1$$

Hence, the zeroes of the polynomial $$p(x)=3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5$$ are $$\sqrt { \dfrac { 5 }{ 3 }  } ,-\sqrt { \dfrac { 5 }{ 3 }  } ,-1,-1$$.

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