Question

find the zeros of 2x^4-13x^3+19x^2+7x-3, given that two zeros are 2+root3 and 2-root 3

Answer 1

Answer:

Two of it zeroes are (x-1) (2x+4)

-1 ,2 ,-√3/2,√3/2 are factors.

Answer 2

The other two zeros are 3 and $-\dfrac{1}{2}$

Step-by-step explanation:

Given: $2x^4-13x^3+19x^2+7x-3$

Two zeros are $2+\sqrt{3}$ and $2-\sqrt{3}$

The factor of the given polynomial,

$(x-(2+\sqrt{3}))(x-(2-\sqrt{3}))$

$x^2-4x+1$

Now divide the polynomial with their factor

$\Rightarrow \dfrac{2x^4-13x^3+19x^2+7x-3}{x^2-4x+1}$

$\Rightarrow 2x^2-5x-3$

Now factor the above quadratic

$\Rightarrow 2x^2-6x+x-3$

$\Rightarrow 2x(x-3)+1(x-3)$

$\Rightarrow (x-3)(2x+1)$

The zeros are,

x - 3 = 0  and  2x + 1 = 0

$x=3,x=-\dfrac{1}{2}$

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