Question

find all zeros of the polynomial 2 x to the power 4 - 9 x cube + 5 x square + 3 x -1 if two of its zeros are 2 + root 3 and 2 -root 3 ​

Answer 1

f(x) =2x^4-9x^3+5x^2+3x-1

For which the factors are (x-2-√3) and (x-2+√3)

We get a common factor=x^2-2x+√3x-2x+4-2√3-3-√3x+2√3

=x^2-4x+1

Divide f(x) by x^2-4x+1

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Answer 2

Answer :-

The other zeroes of the given polynomial are 1, -1/2, (2 + √3) and (2 - √3).

Explanation :-

Let f (x) = $2 {x}^{4} - 9 {x}^{3} + 5 {x}^{2} + 3x - 1$

Let ,

$\alpha = 2 + \sqrt{3}$

$\beta = 2 - \sqrt{3}$

Now,

$\alpha + \beta = (2 + \sqrt{3} ) + (2 - \sqrt{3} ) \\ \: \: \: \: \: \: \: \: \: \: \: \: = 4$

$\alpha \beta = (2 + \sqrt{3} )(2 - \sqrt{3}) \\ \: \: \: \: \: \: \: = 1$

For quadratic equation,

${x}^{2} - ( \alpha + \beta )x + \alpha \beta \\ = > {x}^{2} - 4x + 1$

Here the quadratic equation is a factor of the given polynomial f (x).

On dividing the polynomial with the quadratic equation, we get

q (x) = $2 {x}^{2} - x - 1$

The given q (x) is also a factor of the polynomial f (x).

$2 {x}^{2} - x - 1 \\ = > 2 {x}^{2} - 2x + x - 1 \\ = > 2x(x -1) + 1(x -1) \\ = > (2x + 1)(x - 1) \\ = > x = 1 \: or \: \frac{ - 1}{2}$

Hence, the factors of the polynomial f (x) are : 1, -1/2, (2 + √3) and (2 - √3).

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