Question

obtain all other zeros 3x4+6x3-2x2-10x-5 if two of its zeros are 5/3and-5/3

Answer 1

Hey .....

I am sure that zeroes are √5/3 and -√5/3.

See the attachment

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Answer 2

-1 , -1 are the another two zeros of the polynomial $\bold{f(x)=3 x^{4}+6 x^{3}-2 x^{2}-10 x-5=0}$

Given:

$f(x)=3 x^{4}+6 x^{3}-2 x^{2}-10 x-5=0$

Two roots \frac{\sqrt{5}}{3} \text { and }-\frac{\sqrt{5}}{3}

To find:

Another two roots = ?

Solution:

Let the given polynomial be $f(x)=3 x^{4}+6 x^{3}-2 x^{2}-10 x-5=0$

To find the two of its roots or zeroes are given, we need to find two other roots.  

The two given roots are $\sqrt{\frac{5}{3}} \text { and }-\sqrt{\frac{5}{3}}$

If any value is said to be the root, it has to satisfy the polynomial.

$\sqrt{\frac{5}{3}} \text { and }-\sqrt{\frac{5}{3}}$ should satisfy f(x)=0  

$\left(x-\sqrt{\frac{5}{3}}\right)\left(x+\sqrt{\frac{5}{3}}\right)$

$x^{2}-\frac{5}{3}$ should satisfy f(x)= 0    

When $3 x^{2}-5$ is divided by f(x), the remainder should be 0

Let us check find the quotient using the division method are attached below:

$\begin{array}{l}{f(x)=3 x^{4}+6 x^{3}-2 x^{2}-10 x-5=\left(3 x^{2}-5\right)\left(x^{2}+2 x+1\right)} \\ {=\left(3 x^{2}-5\right)(x+1)^{2}}\end{array}$

Therefore, the other roots are -1, -1.