# f(x) = 2x^4-3x^3-5x^2+9x-3 Given that sqrt(3) forms two of the roots, find all the roots of the polynomial?

Jul 18, 2016

The roots of the polynonial are: $\pm \sqrt{3} , 1 , \frac{1}{2}$

#### Explanation:

$f \left(x\right) = 2 {x}^{4} - 3 {x}^{3} - 5 {x}^{2} + 9 x - 3$

The roots of the polynonial are those values of $x$ for which $f \left(x\right) = 0$

Since $f \left(x\right)$ is a polynomial of degree 4 we know it will have 4 roots (real and/or imaginary).

The question tells us the $\sqrt{3}$ forms two of the roots so we deduce that $\left({x}^{2} - 3\right)$ must be a factor, and that two of the roots must be given by:

$\left({x}^{2} - 3\right) = 0 \to x = \pm \sqrt{3}$

Long dividing $2 {x}^{4} - 3 {x}^{3} - 5 {x}^{2} + 9 x - 3$ by $\left({x}^{2} - 3\right)$

$\to 2 {x}^{2} - 3 x + 1$

This factorises as: $\left(2 x - 1\right) \left(x - 1\right)$

Hence the two remaining roots are: $1 , \frac{1}{2}$

Therefore the four roots of $f \left(x\right)$ are $\pm \sqrt{3} , 1 , \frac{1}{2}$ (all real this case)

Jul 18, 2016

$x = 1 , \frac{1}{2} , \sqrt{3} , - \sqrt{3}$

#### Explanation:

There appears to be error in posting the question.
As mentioned the two zeros are not $\sqrt{3} , \mathmr{and} \sqrt{3}$.
I found that the two zeros must be $\sqrt{3} , \mathmr{and} - \sqrt{3}$
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Given equation is
$f \left(x\right) = 2 {x}^{4} - 3 {x}^{3} - 5 {x}^{2} + 9 x - 3$ ......(1)
Given that two of its zeros are $x = \sqrt{3} , - \sqrt{3}$
$\implies \left(x + \sqrt{3}\right) , \left(x - \sqrt{3}\right)$ are two factors of equation (1)
$\implies \left(x + \sqrt{3}\right) \left(x - \sqrt{3}\right) = \left({x}^{2} - 3\right)$ is a factor of the polynomial.

If we divide the equation (1) by the above quadratic by long division method we get another quadratic which is a factor of equation (1)
$\therefore \frac{2 {x}^{4} - 3 {x}^{3} - 5 {x}^{2} + 9 x - 3}{{x}^{2} - 3}$, we get dividend as
$2 {x}^{2} - 3 x + 1$

To find factors of second quadratic we use split the middle term method
$2 {x}^{2} - 2 x - x + 1$, paring and taking out the common factors we get
$2 x \left(x - 1\right) - \left(x - 1\right)$
$\implies \left(x - 1\right) \left(2 x - 1\right)$
Setting each factor $= 0$, we obtain remaining two zeros as
$x = 1 , \frac{1}{2}$