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

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

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Answer 1 · 2016-07-18T01:51:48

The roots of the polynonial are: $\pm \sqrt{3}, 1, \frac{1}{2}$ $+-sqrt(3), 1, 1/2$

Explanation:

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

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

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

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

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

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

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

This factorises as: $(2 x - 1) (x - 1)$ $(2x-1)(x-1)$

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

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

Answer 2 · 2016-07-18T02:40:38

$x = 1, \frac{1}{2}, \sqrt{3}, - \sqrt{3}$ $x=1, 1/2, sqrt3, -sqrt3$

Explanation:

There appears to be error in posting the question.
As mentioned the two zeros are not $\sqrt{3}, and \sqrt{3}$ $sqrt3, and sqrt 3$ .
I found that the two zeros must be $\sqrt{3}, and - \sqrt{3}$ $sqrt3, and -sqrt 3$
.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.--.-.-.
Given equation is
$f (x) = 2 x^{4} - 3 x^{3} - 5 x^{2} + 9 x - 3$ $f(x)=2x^4-3x^3-5x^2+9x-3$ ......(1)
Given that two of its zeros are $x = \sqrt{3}, - \sqrt{3}$ $x=sqrt3, -sqrt3$
$\Rightarrow (x + \sqrt{3}), (x - \sqrt{3})$ $=>(x+sqrt3), (x-sqrt3)$ are two factors of equation (1)
$\Rightarrow (x + \sqrt{3}) (x - \sqrt{3}) = (x^{2} - 3)$ $=>(x+sqrt3)(x-sqrt3)=(x^2-3)$ 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)
$:. (2x^4-3x^3-5x^2+9x-3)/(x^2-3)$ , we get dividend as
$2x^2-3x+1$

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

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

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Explanation:

Explanation:

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f(x) = 2x^4-3x^3-5x^2+9x-3f(x)=2x4−3x3−5x2+9x−3f(x) = 2x^4-3x^3-5x^2+9x-3 Given that sqrt(3)√3sqrt(3) forms two of the roots, find all the roots of the polynomial?

2 Answers

Explanation:

Explanation:

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