Question

Obtain all the zeroes of the polynomial f(x)= 3x^4 + 6x^3- 2x^2- 5 if two of it's zeros root 5 upon 3 and minus root 5 upon 3​

Answer 1

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-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.

$</p><p>\sqrt{\frac{5}{3}} \text { and }-\sqrt{\frac{5}{3}}$

$should \: satisfy \: f(x)=0$

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

$x^{2}-\frac{5}{3}$

$should \: satisfy \: f(x)= 0$

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.

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